CHAPTER TWO FIRST LAW OF THERMODYNAMICS Table of Contents INTRODUCTION ..................................................................................................................... 1 Learning Objectives ........................................................................................................................1
THE FIRST LAW ...................................................................................................................... 2 Mechanical Energy .........................................................................................................................2 Conservation of Energy: The First Law ............................................................................................3 Definitions .....................................................................................................................................5 Types of Thermodynamic Systems.......................................................................................................................5 Types of Thermodynamics Processes ..................................................................................................................6
MORE ON HEAT TRANSFER .................................................................................................... 6 General..........................................................................................................................................6 Convection ....................................................................................................................................7 Conduction ....................................................................................................................................7 Radiation .......................................................................................................................................9 Example Problem 2.1: Heat Transfer from a Human Body...............................................................9
MORE ON WORK ................................................................................................................. 11 General........................................................................................................................................ 11 Thermodynamic Work ....................................................................................................................................... 11 Power .................................................................................................................................................................12 Notation .............................................................................................................................................................13
Expansion/Compression Work ..................................................................................................... 14 Quasi-equilibrium Processes ........................................................................................................ 15 Use of Ideal Gas Equation............................................................................................................. 17 General Formulation ..........................................................................................................................................17 Isothermal Expansion .........................................................................................................................................17
Example Problem 2.2: Application of 1st Law for Isothermal Conditions ........................................ 17
INTERNAL ENERGY AND TEMPERATURE ............................................................................... 19 Background ................................................................................................................................. 19 Heat Capacities ............................................................................................................................ 20 General Approach ..............................................................................................................................................20 Heat Capacity at Constant Volume ....................................................................................................................21 Heat Capacity at Constant Pressure ...................................................................................................................22 Relationship between Cvm and Cpm .....................................................................................................................25 Incompressible Substance.................................................................................................................................. 25
Variation with Temperature...............................................................................................................................26
Enthalpy ...................................................................................................................................... 27 Mathematical Features ...................................................................................................................................... 27
Application of Heat Capacities to Engineering Problems ................................................................ 27 Example Problem 2.3: Application of 1st Law for Isobaric Conditions.............................................. 29 Example Problem 2.4: Application of 1st Law for Varying P, T and V. .............................................. 32 Problem Definition and Solution Results ........................................................................................................... 32 Exact and Inexact Differentials ...........................................................................................................................37
Additional Discussion ................................................................................................................... 38
ENTHALPY OF PHASE TRANSITIONS...................................................................................... 39 Introduction ................................................................................................................................ 40 Standard Enthalpies of Transitions ............................................................................................... 41 Introduction .......................................................................................................................................................41 Vaporization and Notation .................................................................................................................................43 Fusion and Sublimation ...................................................................................................................................... 44 Tabular Values ....................................................................................................................................................44
Changes with Temperature .......................................................................................................... 45 Example Problem 2.5: Cooling by Perspiration ............................................................................. 48
ENTHALPY OF CHEMICAL REACTIONS ................................................................................... 50 Introduction ................................................................................................................................ 50 General Concepts ...............................................................................................................................................50 Molar Extent of Reaction ...................................................................................................................................52
Enthalpy of Reaction .................................................................................................................... 54 Standard Enthalpy of Formation ................................................................................................... 55 Reference State ..................................................................................................................................................55 Experimentally Derived Values ..........................................................................................................................55 Hess’ Law............................................................................................................................................................56
Changes with Temperature .......................................................................................................... 56 Example Problem 2.6: Standard Enthalpy of Reaction of Urea ....................................................... 58 Enthalpy of Combustion ............................................................................................................... 61 Metabolic Rate and Thermoregulation ......................................................................................... 62 Introduction .......................................................................................................................................................62 Animal Calorimetry ............................................................................................................................................ 63 Human Metabolic Rate ......................................................................................................................................67 Example Problem 2.7: Change in Room Air Temperature .................................................................................68
HEAT TRANSFER BY RADIATON ............................................................................................ 71 Basic Concepts ............................................................................................................................. 71 Blackbody ...........................................................................................................................................................71 Early Theoretical Framework ............................................................................................................................. 72 Spectral Flux Density .......................................................................................................................................... 76
Solar and Terrestrial Radiation........................................................................................................................... 78
Radiant Flux Density .................................................................................................................... 79 Blackbody radiation ........................................................................................................................................... 79 Gray bodies ........................................................................................................................................................80
Radiation Properties .................................................................................................................... 81 Measured Solar and Terrestrial Radiation .........................................................................................................81 Interactions with Incident Radiation ..................................................................................................................82 More Discussion .................................................................................................................................................83 Kirchhoff’s law ....................................................................................................................................................84 Absorptivity values .............................................................................................................................................84 Reflectivity values .............................................................................................................................................. 85
RADIATATIVE BALANCE FOR THERMAL EQUILIBRIUM .......................................................... 85 Earth’s Orbit around the Sun ........................................................................................................ 85 Radiation and Thermal Equilibrium Relationships ......................................................................... 86 Extraterrestrial Radiation ...................................................................................................................................86 Thermal Equilibrium for Non-absorbing Atmospheric Gases ............................................................................87
Example Problem 2.9: Thermal Equilibrium and Earth’s Temperature ............................................ 88 Discussion.................................................................................................................................... 90
THERMODYNAMIC PROCESSES OF PLANT CANOPIES ............................................................ 91 Photosynthesis and Transpiration ................................................................................................ 91 Transpiration and the First Law .................................................................................................... 93 System Definition ...............................................................................................................................................94 Use of Bowen Ratio ............................................................................................................................................ 94
Example Problem 2.10: Transpiration Depth from Plant Canopies ................................................ 95
PROBLEM ASSIGMENTS ..................................................................................................... 100
Working Notes
CHAPTER TWO1 FIRST LAW OF THERMODYNAMICS Indeed the phenomena of nature, whether mechanical, chemical or vital, consist almost entirely in a continual conversion of attraction through space (potential energy), living force (kinetic energy) and heat into one another. Thus it is that order is maintained in the universe – nothing is deranged, nothing ever lost, but the entire machinery, complicated as it is, works smoothly and harmoniously. And though, as in the awful vision of Ezekiel, “wheel may be in the middle of wheel”, and everything may appear complicated and involved in the apparent confusion and intricacy of an almost endless variety of causes, effects, conversion, and arrangements, yet is the most perfect regularity preserved – the whole being governed by the sovereign will of God. James P. Joule, Co-discover of the 1st Law, 1818-1889
INTRODUCTION Learning Objectives After reading the lecture materials and completing the homework assignments, you are expected to understand and to be able to apply:
The First Law of thermodynamics,
Radiation and the heating of Earth, and Evapotranspiration from plant canopies.
1
© 2018. Bruce N. Wilson and the Regents of the University of Minnesota. All Rights Reserved. These notes have not been professionally reviewed.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
THE FIRST LAW Mechanical Energy Let’s revisit the definition of mechanical work using Newton’s second law where force is equal to the rate of change in momentum (𝐹⃗ = 𝑚 𝑑𝑣⃗ ⁄𝑑𝑡 for constant mass). Since our focus in to illustrate energy-work concepts, we will simplify by considering movement in a single vertical direction z. Work is then defined by multiplying both sides by dz and integrating over a distance of z2-z1, that is,
Let’s now consider work only done by gravity in accelerating our system, where F z = -mg. By also using the chain rule of calculus and v=dz/dt, we obtain −mg
𝑑𝑧 = 𝑚
𝑑𝑣 𝑑𝑧
𝑑𝑧 𝑑𝑡
𝑑𝑧 = m
𝑣 𝑑𝑣
(2.1.2a)
By evaluating the integrals, we conclude that
where KE is the kinetic energy defined as KE = mv2/2. It is the energy associated with the state of motion of the system. If the final velocity is greater than the initial velocity, the work is positive. PE is the gravitational potential energy defined as PE = mgz. It is the energy of the system because of its position within a gravitational field. Eq. 2.1.2a is a special form of the work-energy theorem evaluated for work done by gravity. Mechanical energy is defined as the sum of potential and kinetic energies. The conservation of mechanical energy can be obtained from Eq. 2.1.2b as (2.1.3)
∆KE + ∆PE = 0
where ΔPE = PEf – PEi and ΔKE = KEf- KEi. Mechanical energy is defined for conservative forces, and it doesn’t include frictional and drag forces.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Conservation of Energy: The First Law Some engineering problems with negligible friction forces and heat transfer can be solved by using the conservation of mechanical energy. However, these problems could have alternatively been solved by directly applying Newton’s second law of motion. The engineer has simply replaced the solution approach using Newton’s second law of motion with an alternative form of this law written in terms of mechanical energies. The First Law is a more general and powerful statement than the conservation of mechanical energy. It ties together the seemingly different processes of heat transfer and work as forms of transfer of a fundamental property of the Universe called energy. In fact, combining work, heat transfer and other important processes within a single energy-based framework is one of the greatest contributions in all of science. Let’s revisit the energies for the universe, system and surroundings shown below that were first introduced in Chapter 1.
Figure 2.1. Important Components of the First Law. The First Law of thermodynamics is succinctly stated by Rudolf Clausius as "Die Energie der Welt ist constant" (the energy of the Universe is a constant). To conserve total energy, a change in the energy of the system must be balanced by an equal and opposite energy change in the surroundings. This change is consequence of the transfer of energy between the system and the surroundings. The First Law given in Chapter 1 can be written more generally by including the transfer of energy by the flow of matter into and out-of the system as shown Fig. 2.1. Once again, energy transfer determined by temperatures of the system and the surroundings is called heat (q) transfer. All other non-mass energy transfer is done by work (w). The conservation of energy for the system can then be written as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where ∆𝐸 = 𝑚̇ 𝐸 , ∆𝑡 − 𝑚̇ 𝐸 ∆𝑡 is the transport of energy by the flow of matter, 𝑚̇ 𝑎𝑛𝑑 𝑚̇ is the mass flow rate (mass per time) into and out of the system, respectively, Es and Es,sur are the energies per unit mass of matter leaving the system and entering from the surroundings, respectively, and Δt is the time interval over which the change in energy is of interest. The mass flow rate is defined by the product of the density of matter, its velocity and the surface area of flow. Our system boundaries often don’t allow the flow of matter through them, or they have conditions of 𝑚̇ 𝐸 , ∆𝑡 ≈ 𝑚̇ 𝐸 ∆𝑡. ΔEtransport is then zero or negligible, and the transfer of energy is limited to heat transfer and work. We will define heating
In Fig. 2.1, the system energy is evaluated with kinetic energy (KE), gravitational potential energy (PE), internal energy (U) and other possible energy (O) such as electrostatic potential energy and possible surface energies. Changes in kinetic energy and potential energy are defined by changes in the position and motion of the system. These forms of energies are external to the changes within the system itself. As discussed in Chapter 1, internal energy is the energy associated with the motion, interactions and bonding of atoms, ions and molecules in the system. Internal energy is an extensive property of the system, and therefore an infinitesimal change is represented by an exact differential. By focusing on energy changes within the system, the First Law can then be simplified for a system of fixed mass as
Infinitesimal changes in heat transfer and work are represented by inexact differentials. We have divided the change in internal energy into possible changes with temperature (ΔU tpr), phase transitions (ΔUphase), and chemical reactions (ΔUr). Changes between two equilibrium states are shown graphically in Fig. 2.3. Each type of change is introduced by holding constant the internal energy associated with the other two. Sometimes, especially in older textbooks, ΔUtpr, ΔUphase, and ΔUr are considered by using terminology of heat energy (or even sensible heat), latent heat and heat of reaction, respectively. Our terminolgy provides an
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
easier theoretical framework for discussing thermodynamics and limits the definition of “heat” to the transfer of energy by a temperature difference.
Figure 2.3. Heat Transfer and Work Done Between Two Equilibrium States. Definitions Types of Thermodynamic Systems Let’s consider the following systems:
Figure 2.4. Types of System Bound in Energy and Mass Exchange at Their Boundaries. Open system: Exchanges mass and energy with its surroundings. Hence the mass flow rates of Eq. 2.1.2 are not zero. Closed system: Has no mass flow into and out of the system and therefore it only exchanges energy with its surroundings. The mass flow rates are zero. Isolated system: Does not exchange mass or energy with the surroundings. The Universe is an isolated system.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Types of Thermodynamics Processes Let’s consider the following system and surroundings:
Figure 2.5. Adiabatic and Isothermal Processes. Adiabatic process: Boundaries do not allow heating as mode of energy transfer, even if there is a temperature difference between system and the surroundings. Adiabatic conditions can often be used if the system changes are rapid so that heat transfer is insignificant in comparison to other energy terms. Isothermal process: Process where the temperature of the system is constant. Isothermal conditions are frequently an adequate approximation when a gas is given sufficient time (slow change) to expand (or contract) when the system is exposed to a different temperature than the surrounding. Many of the processes in animals are approximately isothermal. For example, humans maintain a nearly constant body temperature of approximately 37 C. Isobaric process: Process where the pressure of the system is constant. Many environmental processes occur under nearly constant ambient pressure of the atmosphere, which is often taken as P = 1 atm = 101.325 kPa. Isochoric process: Process where the volume of the system is constant.
MORE ON HEAT TRANSFER General Relationships for heat transfer are often written in terms of heat flux. Heat flux is the rate of heat transfer per unit area. Heat transfer is obtained by integrating over the surface area and over the time interval of interest. This integration will usually be done by assuming that the heat flux is constant with space and constant for the time between equilibrium states. The heat transfer between initial and final equilibrium states is obtained from its rate as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where 𝑞̇ is the rate of heat transfer with units of J s-1 or W, tf and ti are the initial and final times of the transfer, and Δt=tf-ti. Convection Convection is a process of heat transfer that is driven transport processes of gas or liquid velocity (or bulk movement) moving over the surface area of the system. The concept is shown below for velocity over a flat plate.
Figure 2.6. Convection Processes. Convection is often predicted using Newton’s law of cooling defined as 𝑞̇
= 𝑘
𝐴
2.2.2
(𝑇 − 𝑇 )
where 𝑞̇ is the rate of heat transfer by convection, kconv is the heat transfer coefficient of convection, Aconv is the surface area, Tp is the temperature (K) of the plate and Tf is the temperature (K) of the fluid. Note that the rate of heat transfer is positive for a temperature of the plate is greater than that of the fluid. For a system corresponding to the plate temperature, heat transfer is from the system to surroundings, and you will need to use a negative heat transfer in the First Law. Conduction Conduction is a process of heat transfer by the interactions of more energetic molecules at higher temperatures with those at lower temperatures. These interactions occur in solids, liquids and gases. To distinguish conduction from convection for fluids, conduction corresponds to the condition of no net transport by a fluid velocity. This concept is shown below for the heat transfer through a wall.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.7. Conduction Process. Conduction is often predicted by the well-known Fourier’s Law, where the rate of heat transfer is directly proportional to the temperature gradient. Fourier’s Law can be written as 𝑞̇
=−𝑘
𝐴
𝑑𝑇 ≈ −𝑘 𝑑𝑥
𝐴
𝑇 −𝑇 𝑥 −𝑥
2.2.3a
where 𝑞̇ is the rate of heat transfer for conduction, kcond is thermal conductivity, Acond is the surface area, dT/dx is the temperature gradient, and T and x are temperature (K) and distance as defined in the above figure. Heat transfer in Fig. 2.7 is positive from higher temperature to a lower temperature. If the system is outside the house then the sign of heat transfer is correct; however, if the system is inside the house, then you need to use a negative heat transfer in the First Law. The R-value is often used as a measure of the thermal conductivity of insulation properties of clothing and materials used in the construction. R-values are determining by directly measuring the heat transfer rate though materials of thickness Δx for a known temperature difference (ΔT). The R-value (rq) is defined as 𝑟 =
∆T −∆𝑥 = 𝑞̇ ∕𝐴 𝑘
2.2.3b
The rate of heat conduction is then defined using R-value as 𝑞̇
=
1 𝐴 ∆𝑇 𝑟
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2.2.3c
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Radiation Radiation is a process where energy is transported by photons (or as electromagnetic waves) as a consequence of changes in the configurations of atoms, including the detachment of electrons and changes in the transition states of rotational and vibrational energies. Radiation is of fundamental importance in understanding energy transfer in many biological processes. It will be discussed in detail later in the chapter and also in Chapter 3. To introduce the concept of heat transfer by radiation, consider two large plates of the same area and material as shown below.
Figure 2.8 Radiation Process. Under certain conditions, the rate of energy transfer from the lower plate is defined by using the Stefan-Boltzmann law as 𝑞̇
=𝜎𝐴
2.2.4
(𝜀 𝑇 −𝜀 𝑇 )
where 𝑞̇ is the rate of radiation heat transfer, ε2 and ε1 are the emissivities of the radiating matter for the lower and upper plates, respectively,, σ is the Stefan-Boltzmann constant (5.67 x 10-8 W m-2 K-4) and T is the temperature (K) of the two bodies. For the same emissivities, Eq. 2.2.4 is positive if T2>T1. If the lower plate corresponds to the system temperature, then you will need to use a negative heat transfer in the First Law. Processes of heat transfer by radiation are considerably different than convection and conduction. For example, the net heat transfer can be from a cooler to a warmer body if the emissivity of the warmer plate is smaller than that of the cooler plate. We will use the subscript “rad” for these processes to avoid possible confusion in the solution of problems. Example Problem 2.1: Heat Transfer from a Human Body
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Problem Statement: For inactive nude human body resting on a bench, determine the heat transfer rate from the body by:
An illustrative sketch of the problem and key parameters are shown below. The human body is our system.
The core temperature of humans is approximately 37 C, but the surface skin temperature is closer to Tskin = 34 C for the conditions of this problem. Temperatures of the air and the walls are Tair = Twall = 23 C. The surface area of humans can be estimated using mass and height 2. We will use a surface area of Askin = 1.8 m2. For this problem, the effective percentage of this surface area for radiation is estimated as 65%, 85% for convection, and 5% for conduction (direct contact area with bench). Rigorous analysis of the effective surface area of the walls, floor and ceiling corresponding to incoming radiation to our human is beyond the scope of this problem. To simplify, we will use the same surface area for incoming and outgoing radiation. Emissivity of the skin and walls is taken as ε = 0.95, the heat transfer coefficient for convection is taken as 3 J s-1 m-2 K-1, and the R-value for conduction is taken as rq = 0.166 m2 K s J-1. Preliminary calculations are given below.
2
The following relationship for skin surface was used: 𝐴 = 0.2 𝑚 units of kg, and h has units of m.
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.
ℎ
.
, where As has units of m2, m has
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝐴
=𝑓
,
𝐴
= 0.85 (1.8) = 1.53 𝑚
𝐴
=𝑓
,
𝐴
= 0.05 (1.8) = 0.09 𝑚
Solution: For radiation, we have:
𝑞̇
= (0.95) 5.67𝑥10
𝐽 𝑠𝑚 𝐾
(1.17 𝑚 )( (304.15 𝐾) − (296.15 𝐾) ) = 54.5 𝐽𝑠
For convection:
𝑞̇
= 3
𝐽 𝑠𝑚 𝐾
(1.53 𝑚 )( 304.15 𝐾 − 296.15 𝐾) = 36.7 𝐽𝑠
For conduction:
𝑞̇
=
(1.53 𝑚 )( 304.15 𝐾 − 296.15 𝐾) = 4.32 𝐽𝑠 0.16 𝑚 𝐾 𝑠 𝐽
Since our thermodynamic system is the human body, all of these heat transfer rates are negative. The largest mode of heat transfer is by radiation.
MORE ON WORK General Thermodynamic Work In Chapter 1, work was defined as the energy transfer by coherent or non-random motion of particles and heat transfer as that done by a motion that includes an incoherent or random component. We have also previously defined mechanical work as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑤=
𝐹⃗ ⋅ 𝑑𝑥⃗
2.3.1
Thermodynamic work requires the type of motion of mechanical work be placed within our framework of a system and a surroundings. Within this framework, we will define thermodynamic work as interactions between the system and surroundings such that the sole external effect could have been done by the coherent motion associated with the raising of a weight. To illustrate the definition of thermodynamic work, we will use the conditions shown in Fig. 2.9. Here we are first interested in the transfer of the energy from a system with a battery and motor to a fan operating in the surroundings. The transfer of energy from the change in internal energy of the battery is not done by heat transfer. The transfer of energy by thermodynamic work can be obtained by replacing the fan with a pulley supporting a weight. This weight is raised by the rotation of the motor’s shaft from which we can compute the thermodynamic work. Let’s further explore the definition of thermodynamic work by moving the motor out of the system as shown in the right-sided schematic of Fig. 2.9. We are now interested in thermodynamic work obtained by the flow of electrical energy across the system boundaries. This transfer of energy results in the sole effect could be the raising of a weight for a perfect motor. The issue of a perfect motor requires an analysis of the surroundings, and it is not really a concern for us. We can rationalize that someday, perhaps in the very distant future, a nearly perfect motor could be designed.
Figure 2.9. Insight into Thermodynamic Work. Power
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Relationships for work are sometimes given as rate of work done. Rate of work done is called power. Similar to heat transfer rate, the work is obtained from power between initial and final equilibrium states as
where 𝑤̇ is the power with units of J s-1 or W and tf and ti are the initial and final times of the transfer. The last expression on the right-hand side is only valid if the power is approximately constant. Let’s discuss the work and power for the rotating shifts for the systems of Fig. 2.9. The analysis of the key components is shown below.
Shaft
+ Electrical Source
Ft
Shaft ω
Motor
r
Figure 2-10. Schematic of Work Done by a Shaft.
Angular velocity (ω) and torque (τω) are features of interest in the design of machine elements with shafts. For a radius of r, we have the following definitions 2.3.3a 2.3.3b
𝜏 =𝐹 𝑟 𝑣 =𝜔𝑟
where Ft is the tangential force and v is the velocity at the boundary of the shaft. By using the definition of power, we obtain 𝑤̇ =
𝑤 𝐹 ∆𝑥 𝜏 (𝜔𝑟)=𝜏 𝜔 = =𝐹𝑉= ∆𝑡 ∆𝑡 𝑟
2.3.3c
Notation As previously discussed, work is defined as positive for work done by the surroundings on the system. This notation is consistent with that used for heat transfer. However, prior to 1990, many engineering textbooks defined the work done by the system on the surroundings as positive. Engineers are often interested in the amount of useful work available from the system, which is negative using our notation but is positive using the old notation. The First Law can be analyzed using either notation as long as it is consistently applied (e.g. ∆𝑈 = 𝑞 − 𝑤 ).
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Expansion/Compression Work The most important type of work in this course is the work done by pressure forces resulting in the expansion or compression of the system. The system of interest is shown in the following figure.
Figure 2.15. Schematic Illustrating Expansion Work. By using the definition of mechanical work, and for approximately constant pressure, we obtain that the work done between the initial and final position as
where P is the pressure, A is the surface area and Δz = z f – zi. The negative sign is used in the above equation because it is work done by the system on the surroundings. Expansion work doesn’t include any friction forces between the piston and the cylinder. For a non-constant pressure, we then obtain the following definition of work
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where volume is defined by exact differentials. Quasi-equilibrium Processes Internal energy is defined by exact differentials, and therefore the change in internal energy is independent of the path. In contrast, expansion work is defined by inexact differentials and is a function of the path between initial and final states. To compute this work (and heat transfer), the path is often assumed to be obtained by a quasi-equilibrium (or quasi-static) process. Quasi-equilibrium processes are defined such that the path can be treated at thermodynamic equilibrium at every step of the change between two endpoint states. Similar to ideal gases, quasi-equilibrium processes are an idealized representation of actual thermodynamic processes. They provide insight into thermodynamic processes and are widely used in engineering designs. Insight into quasi-equilibrium processes for expansion work can be obtained by considering the gas under compression as series of equilibrium steps shown in Figs. 2.16a and 2.16b. For the non quasi-equilibrium process, the piston is moved rapidly in the downward direction for step. For this type of movement, the internal pressure in the cylinder does not have sufficient time to adjust its pressure over the entire chamber. The pressure at the top of the piston is then different than at its bottom, and there is not a uniform pressure to use in the ideal gas equation. A noticeable time interval is required for the system to establish equilibrium conditions. After the new equilibrium conditions is established the piston is then again moved rapidly in the downward direction. Let’s consider a piston that is pressed downward slowly as shown in Fig. 2.16b. The system now has time to adjust the pressure throughout the chamber at each step of the process. The internal pressure is effectively the same throughout the entire system during the compression. The time interval required to establish new equilibrium conditions is negligible. Because of problems in defining “pressure” (and other characteristics) for rapid moving piston, our definition of expansion work is only valid for quasi-equilibrium processes. The solution for quasi-equilibrium processes can be obtained by integrated Eq. 2.4.2 (which assumes equilibrium at each infinitesimal volume).
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.16a. Non Quasi-equilibrium Process.
Figure 2.16b. Quasi-equilibrium Process. Gases generally respond quickly to changes in the boundaries of their system. As shown in Chapter 1, the root-mean-square speed of common gas molecules is of the magnitude of 400 m s-1 (~ 1000 mph) at room temperatures. The impact of changes in boundary location can then be quickly converted into changes in thermodynamic properties such as pressure, temperature and internal energy. Another feature of equilibrium condition is that there is no work acting on the system at equilibrium. Therefore for quasi-equilibrium condition, the pressure force of the gas inside the piston is essentially equal to the external pressure force of the surroundings on the system.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Use of Ideal Gas Equation General Formulation Assumption of quasi-equilibrium process allows us to use the ideal gas equation to represent pressure within the system. For an ideal gas, P = nRT/V, we then have the following relationship for expansion work (for constant number of moles):
Isothermal Expansion Isothermal expansion refers to expansion work that occurs while maintaining a constant gas temperature in the system. This can be done by controlling the heat transfer to closely balance the work done. In practice, it corresponds to gases that are expanding slowly. By using expansion work with the ideal gas equation, isothermal expansion is defined for an ideal gas as
An isotherm is often used to indicate a line on a graph of constant temperature. Isothermal expansion work corresponds to the area under this curve. Example Problem 2.2: Application of 1st Law for Isothermal Conditions Problem Statement: Consider 450 g of air that is compressed slowly in a piston-cylinder assembly from 100 kPa at 26 C to final pressure of 620 kPa. Molar mass of air is approximately 29 g mol-1 (as shown in Chapter 3). During the compression, the heat transfer to surroundings is at rate so that the process is isothermal. We will further assume a quasiequilibrium process for an ideal gas. There are no phase transitions or chemical reactions within our system. The process is shown in the following schematic.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Determine:
Solution: Change in internal energy. Let’s consider the isothermal process for an ideal gas. In Chapter 1, we have previously consider the internal energy as
where a = 3/2, a=5/2, and a=3 for monatomic, diatomic, and polyatomic gases, respectively. As discussed in Chapter 1, the internal energy of ideal gas is only a function of temperature. Since the initial and final temperatures of the gas are the same, and there is no phase transitions or chemical reactions, we will conclude that internal energy is constant, that is,
Solution: Compression work. For isothermal processes, we know that
For this problem, we have R = 8.314 J K-1 mol-1
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
T = 26 + 273.15 = 299.15 K m 450 g n= = = 15.51 mol m 29 g mol V P P V = nRT = P V → = V P and therefore we can compute the work done as w = −nRT ln
P = −(15.51 mol)(8.314 JK P
mol
)(299.15 K) ln
100 𝑘𝑃𝑎 620 𝑘𝑃𝑎
The work is positive because it is the surroundings that does work on the system to compress the gas. Solution: Heat transfer. Heat transfer can be determined from the First Law of thermodynamics. For no change in internal energy, we have
and therefore q = - w = -70.426 kJ that is, 70.426 kJ is need to be transferred from the system to the surroundings to have isothermal compression.
INTERNAL ENERGY AND TEMPERATURE Background In Chapter 1, we used the kinetic theory of gas to derive relationships between the internal energy to the gas’s temperature. These relationships were developed to provide insight into the ideal gas equation and internal energy. They are too imprecise for design purposes. From this analysis, we conclude that internal energy is dependent on:
Last Modified: March, 2018
2-19
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
To better understand the observed internal energies of solids, Einstein proposed that energies of the oscillators were limited to discrete energy levels as required by quantum mechanics. Theoretical internal energies are then in good agreement with those observed when “Einstein” temperature is properly selected. Details of Einstein’s approach are given in Chapter 7. Instead of relying solely on theoretical models, engineers have long used experiments to study and to define parameters for different substances to quantify changes in the internal energies related to temperature change, phase change and chemical reactions. These parameters are usually summarized in tables that engineers use in their analysis and design. The important parameters for this course are given in the downloadable Thermodynamic Data Handout. Heat transfer is a relatively easy to measure experimentally. Work can also be defined by experimentally limiting it to expansion work under conditions of constant pressure (w=PΔV) or of constant volume (no expansion work). The measurable heat transfer is then related changes in internal energy from the First Law as
The focus in this section is the development and use of relationships to relate changes in internal energy for a system without phase transitions and chemical reactions (i.e. ΔU phase = ΔUr = 0). The measured heat transfer is then used to define parameters tied to changes with temperature differences, and the subscript “tpr” is used. In subsequent sections, parameters for phase change are obtained from Eq. 2.5.1 where experimentally ΔUtpr = ΔUr = 0 and for chemical reaction where ΔUtpr=ΔUphase = 0. Some of the relationships developed in this section are general, that is, they are applicable to all three types of changes in internal energy. Heat Capacities General Approach Heat capacity is broadly defined as the amount of heat transfer necessary to raise the temperature of a system by one degree for a system with no phase transitions or chemical reactions. We can therefore use the following general definition of heat capacity
Last Modified: March, 2018
2-20
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
As previously discussed, the heat capacity is related to the internal energy using the First Law with only expansion work, that is, w = -∫PdV. For no phase transitions or chemical reactions, the heat capacity can then be evaluated as
where 𝑃 is the average pressure for a change in volume. In this section, we are interested in measuring C under conditions of either constant volume or constant pressure. We are interested in (1) how the heat capacity can be determined experimentally and (2) how they can be used to solve engineering problems. Specific heat capacity (Cs = C/m) is defined as the heat capacity per unit mass. Molar heat capacity (Cm = C/n) is defined as the heat capacity per mole of substance. Heat Capacity at Constant Volume To illustrate the computations of heat capacity at constant volume, we will consider the hypothetical experiment shown below for a system of fixed (constant) volume (no phase transitions or chemical reactions). Here the heat transfer of 400 J is measured to change the temperature by 4 K for 4 moles of a gas. The heat capacity at constant volume is then determined by dividing the heat transfer by the change in temperature, that is, C v = 100 J K-1. We will use the notation of Cvs for specific heat at constant volume (Cvs = Cv/m) and Cvm for molar heat capacity at constant volume (Cvm = Cv/n). Therefore, the molar heat capacity at constant volume for our hypothetical experiment is obtained by simply dividing the experimental determined Cv by the number of moles, or Cvm = 25 J K-1mol-1. This example illustrate how heat capacity can be determined experimentally using a constant volume system.
Last Modified: March, 2018
2-21
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.17. Illustration of Heat Capacity at Constant Volume To use the experimentally-derived Cv to solving engineering problems, let’s evaluate the heat transfer using the First Law under conditions of constant volume (i.e., V=0). The heat capacity can then be related to the change in the internal energy as C =
∆𝑈
+ 𝑃 ∆𝑉 ∆𝑈 = ∆T ∆T
2.5.3
where Cv is now defined using a fundamental property of the system (internal energy) that is useful in solving thermodynamic problems. If we consider the limit as ΔT → 0, we obtain
where internal energy for our system is tied only to temperature for ideal gases and incompressible liquids and solids. For these conditions, the change in internal energy for our system can be determined directly as ∆𝑈
=
𝐶 𝑑𝑇 =
𝑛𝐶
𝑑𝑇 ≈ 𝑛 𝐶
∆𝑇
2.5.6
for a system of fixed volume with no phase transitions or chemical reactions. For a constant Cv, internal energy is the defined as Utpr = Cv T, where the integration constant is taken as zero at T=0. The qualifier “at constant volume” describes the experimental method used to measure heat capacity. The crude theoretical relationship (from kinetic theory of gases) between internal energy and temperature represented by the “3nR/2”, “5nR/2” or “3nR” terms can now be replaced by the experimentally derived Cv. As such, Cv represents a characteristic of substances that is not limited to the type of systems used to measure it. It is generally used to compute the change in internal energy for systems of both constant and varying volumes. As discussed later, this use is valid for ideal gases and incompressible liquids and solids. Heat Capacity at Constant Pressure Thermodynamic systems often have one or more expanding (or contracting) boundaries surrounded by ambient air (open vessels or parcels of air). Ambient air pressure is nearly constant and equal to the atmospheric air pressure of approximately 100 kPa. For these systems, expansion work is done under conditions of constant pressure. Illustrations of three important processes are shown in Fig. 2.18. For each of these processes, it is more convenient to evaluate their thermodynamic properties using conditions of constant pressure. We will start our discussion by considering heat capacity at constant pressure.
Last Modified: March, 2018
2-22
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.18. Systems under Ambient Air Conditions. To illustrate the experimental approach to determine the heat capacity at constant pressure, we will re-do our hypothetical experiment of the previous section, but now we will allow the system to expand with the change in temperature. An illustration of the revised experimental approach is shown in Fig. 2.19 where a heat transfer of 533 J is needed to change the temperature of the same gas by 4 K. An additional heat transfer of 133 J is needed for our expanding system because of the work done by the system on the surroundings. Heat capacity at constant pressure is determined by dividing the heat transfer by change in temperature, that is, Cp = 133.25 J K-1. The specific heat at constant pressure is Cps = Cp/m and the molar heat capacity at constant pressure is Cpm = Cp/n, which for our example is Cpm=33.31 J K-1 mol-1. The experimentally derived Cpm can now be placed in a table or other formats and used to solve engineering problems.
Figure 2.19. Illustration of Heat Capacity at Constant Pressure.
Last Modified: March, 2018
2-23
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
To apply the experimentally-derived C p to thermodynamic, the heat transfer will be evaluated using the First Law under conditions of constant pressure. We then obtain C =
∆𝑈
+ 𝑃 ∆𝑉 ∆𝐻 = ∆T ∆𝑇
2.5.7
where Cp is the heat capacity at constant pressure. Heat capacity here is now defined using Htpr, which is another property (not as fundamental as internal energy) of the system. This property is called enthalpy. The subscript of “tpr” (for temperature) has been used to indicate this definition doesn’t include possible changes in enthalpy with the number of moles from phase transitions or chemical reactions. Enthalpy is defined in general as
More on enthalpy is given in the next subsection. For the limit as ΔT → 0, the heat capacity at constant pressure is defined as C =
∂H ∂T
2.5.9
where the partial derivative is used because enthalpy is, in general, also a function of pressure. As discussed in greater detail later, enthalpy for our system is tied only to temperature for ideal gases and incompressible liquids and solids. By using dHtpr = Cp dT for ideal gases, liquids and solids, we obtain the change in enthalpy as ∆H
=
2.5.10a
𝐶 𝑑𝑇
You can obtain Cpm for different substances from the Thermodynamic Data Handout available at the BBE 3043 website. These values are for a temperature of 25 C and a standard pressure of 100 kPa. We will often assume that C pm is constant over the range of temperatures in our problems. Changes in temperature enthalpy are then obtained.as
Last Modified: March, 2018
2-24
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where Cpm values is obtained from the Thermodynamic Data Handout. Additional discussion on the use of Eqs. 2.6.10 is discussed later in this section. Relationship between Cvm and Cpm Let’s consider the relationship between Cvm and Cpm for ideal gases. We will use the following useful form of Cp evaluated for conditions of constant pressure with a fixed mass (n=constant). We then obtain:
where we have also used Cv = dUtpr/dT (valid for ideal gases for constant and varying volumes as discussed later). For an ideal gas (PV = nRT) undergoing changes for constant pressure with a fixed mass (n=constant), we have d(PV) = PdV = nRdT. We can then simplify as 2.5.13
C 𝑑𝑇 = 𝐶 𝑑𝑇 + 𝑛𝑅 𝑑𝑇
By cancelling dT and dividing by n, we obtain the following relationship between the molar heat capacities at constant pressure and constant volume as
or, alternatively, Cvm can be computed from Cpm values in the Thermodynamic Data Handout as C
=𝐶
2.5.14b
−𝑅
which is valid for ideal gases. Heat capacity is generally tabulated using Cpm instead of Cvm. Not only is enthalpy often measured directly using ambient air conditions, it is also a convenient property for computing heat transfer (or change in temperature for a given q) for the common condition of constant pressure. Heat transfer for this condition is simply equal to the change in enthalpy. An alternative approach to obtain heat transfer from the First Law by computing separately (1) the change in internal energy using Cvm obtained from Eq. 2.5.14b and (2) the expansion work under constant pressure. The heat transfer by this approach is the same as that obtained using enthalpy, but the one-step enthalpy-based approach is simpler. Incompressible Substance
Last Modified: March, 2018
2-25
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Many liquids and solids can be evaluated as incompressible substances. For incompressible substances, the internal energy is only a function of temperature. We conclude that the heat capacity at constant volume is defined as C =
dU dT
2.5.15a
Furthermore, the heat capacities at constant volume and pressure are equal for an incompressible substance, that is,
Variation with Temperature Heat capacities for gases vary with temperature (for example, see the internal energy relationship of hydrogen gas given in Chapter 1 by Fig. 1.25). To represent this variation, the heat capacities are often defined using power functions with temperature. For example, the molar heat capacity at constant pressure is sometimes represented by the following function
where the coefficients a, b, c, and d derived from data analysis. Coefficients for a few gases are shown in Table 2.1 (from Sandler, 2006; NIST, 2017). Table 2.1 Coefficients for Temperature Dependent Molar Heat Capacities.
Type
Gas
Monatomic
He H2 N2 O2 H2O CH4 CO2
Diatomic Polyatomic
a
-1
J K mol 20.786 29.088 28.883 25.460 32.218 19.875 22.243
-1
b
-2
-1
J K mol 4.851 x 10-13 -0.192 x 10-2 -0.157 x 10-2 1.519 x 10-2 0.192 x 10-2 5.021 x 10-2 5.977 x 10-2
c
-3
-1
J K mol -1.583 x 10-16 0.400 x 10-5 0.808 x 10-5 -0.715 x 10-5 1.055 x 10-5 1.268 x 10-5 -3.499 x 10-5
d
-4
J K mol-1 1.525 x 10-20 -0.870 x 10-9 -2.871 x 10-9 1.311 x 10-9 -3.593 x 10-9 -11.004 x 10-9 7.464 x 10-9
The error obtained by using a constant Cpm, instead of Eq. 2.5.16, is explored in Homework Problems 2.11 and 2.12. This error is acceptable for many engineering designs. The vast majority of the problems in this course is solved by using a constant C pm. Eq. 2.5.9 is used only for a large difference in final and initial temperatures or if a very precise solution is needed.
Last Modified: March, 2018
2-26
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Enthalpy Mathematical Features We now wish to explore more fully the characteristics and definition of enthalpy. Let’s first investigate its differential characteristics. Enthalpy is defined by exact differentials. Since enthalpy is defined as H = U + PV, we obtain
The change in enthalpy between initial and final states can be written as 2.5.18
∆H = ∫ dU + ∫ PdV + ∫ VdP For a system under constant pressure, dP = 0. We then obtain dH = dU + PdV
2.5.19a
∆H = ∆U + ∫ PdV
2.5.19b
and
Since dH is an exact differential, the change in enthalpy is only dependent on the difference in H between the final and initial states. This has important implications on possible solution approaches and is discussed in greater detail later. Enthalpy is also an extensive property. The enthalpy of the system can then be determined as the sum of the enthalpy of its components. Molar enthalpy is defined as enthalpy per mole that is Hm = H/n. Specific enthalpy is defined as enthalpy per mass that is Hs = H/m. Application of Heat Capacities to Engineering Problems A summary of the typical formulation of engineering design problems based on heat capacities is given in Figure 2.20. Here we have an ideal gas with no phase transitions or phase changes so the number of moles is constant between the initial and final equilibrium states. We will only consider expansion work. In general, the initial pressure, temperature, and volume are known (one of them is often computed using the ideal gas equation). Possible unknowns are ΔUtpr, q and w between the two equilibrium states. In addition, the engineer may be interested in the final pressure, temperature and volume of the system. If these are unknown, then the engineers has the potential of 6 unknown characteristics.
Last Modified: March, 2018
2-27
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.20. Typical Formulation of Engineering Design Problems. Let’s now consider the number of relationships that we have available to us. We need six relationships to solve for six unknowns. One of our relationships is the First Law. For an ideal gas, we have the ideal-gas equation and the change in internal energy with temperature using Cvm. A fourth relationship is obtained using definition of expansion work. However, determining expansion work requires a relationship between pressure and volume or between temperature and volume that effectively increases our unknowns to 7! We can obtain an additional equation if the process is adiabatic (q= 0), isothermal (P=nRT i/V), isobaric (P=constant), or isochoric (V=constant). If one of these processes is applicable, we have five equations for the seven unknowns. The engineer needs to reduce the unknowns to five (1) by using relationships obtained from other courses (such as a heat transfer course), (2) by measuring one or more of the final state variables (Tf, Pf, Vf) and/or q, w or ΔUtpr or (3) by setting the final state variables based on design specifications. Let’s now simplify the discussion for solving problems in this course. We are primarily interested in expansion work. Thermodynamic processes for many biological and environmental systems occur under constant pressure (isobaric process). We are then able to solve for q for a ΔT or for ΔT for a given q as
Last Modified: March, 2018
2-28
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Once again, these relationships are only valid for constant pressure systems. For example, q from Eq. 2.5.30a is zero for isothermal systems (ΔT=0), but we have previously shown in Example Problem 2.2 that it is actually q = -w. If we are interested in the change in internal energy for an ideal gas (where PdV = nRdT), we can use the definition of enthalpy to obtain ∆U
= ∆H
− P∆V = n C
∆𝑇 − 𝑛𝑅∆𝑇 = 𝑛 𝐶
− 𝑅 ∆𝑇
2.5.30c
We will occasionally be interested in thermodynamic processes in rigid containers. Rigid containers correspond to systems of constant volume (isochoric process). The heat transfer and change in temperature for rigid containers (for ΔV= 0) are obtained as q = ∆U
q ∆T = nC
≈nC
2.5.31a
∆𝑇
2.5.31b
Clearly, we need an estimate of Cpm to determine the heat transfer for some (many) systems and Cvm for others (e.g. rigid containers). Values for C pm for different substances are given in the Thermodynamic Data Handout. But what about values for Cvm? These values are frequently not tabulated. Solids and liquids can often be represented as incompressible substances for which the heat capacities are equal. As previously shown for ideal gases, we can estimate Cvm directly from values in the Thermodynamic Data Handout as 𝐶 = 𝐶 − 𝑅. Example Problem 2.3: Application of 1st Law for Isobaric Conditions Problem Statement. A vertical piston-cylinder assembly with an initial volume of 32 L is filled with 50 g of nitrogen gas. A piston is weighted so that it maintains a constant pressure on the system of 140 kPa. We will compress the gas in the cylinder until the final volume is 90% of the initial value. Estimate the (1) final temperature of nitrogen, (2) amount of energy transferred by heating, (3) the amount of energy transferred by work and (4) the change in internal energy. Assume a constant molar heat capacity and a quasi-equilibrium process of an ideal gas. At least two different methods are available to compute the some of the unknowns, and they are given to show that you obtain the same value with the different methods.
Tabular values. From the Data Handout, we obtain R = 8.314 J K-1 mol-1 = 8.314 Pa m3 K-1 mol-1 mm = 28 g mol-1
Last Modified: March, 2018
2-29
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Cpm = 29.125 J K-1 mol-1 and therefore the number of moles is
Solution: Temperature at Final State. For a quasi-equilibrium process, the system is near equilibrium conditions at each step of the path. Since at equilibrium there is no work (or heat transfer) on the system, we conclude that the piston pressure equals the external pressure, that is, 140 kPa. We can then estimate the initial temperature of the gas from the ideal gas equation, that is,
Pa m (32 L) 1000 L kPa = 301.76 K (28.6 C) 1.785 mol(8.314 Pa m mol K )
140 kPa 1000 T =
The final temperature can be obtained from the ideal gas equation, that is, Pa m (0.9)(32 L) 140 kPa 1000 PV 1000 L kPa T = = = 271.6 K nR 1.785 mol(8.314 Pa m mol K ) ∆𝑇 = T − T = 271.6 K − 301.76 K = −30.16 K An alternative (and more fun) method for obtaining Tf and ΔT is to use relationships that can be derived from the condition of constant pressure. For a constant pressure system we have P =
nRT nRT = → T =T V V
V V
and the change in temperature and final temperature are defined as 𝑉 T P (V − V ) = ( ) (V − V ) −1 = 𝑉 V nR 0.9V ∆T = 301.76 K − 1 = 301.76 K (−0.1) = −30.16 K V T = 𝑇 + ∆𝑇 = 301.76 − 30.17 = 271.6 𝐾 ∆𝑇 = 𝑇 − 𝑇 = 𝑇
Note that the change in temperature is independent of the characteristics of N 2. The same change in temperature would be obtained for any ideal gas (ΔT = PΔV/nR for constant pressure). We didn’t need to compute the final and initial temperatures to evaluate the
Last Modified: March, 2018
2-30
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
internal energy change, work and heat transfer, but they were included in our analysis to provide insight on the compression process. Solution: Heat transfer. From the First Law of thermodynamics with expansion work under constant pressure, we have
Using the definition of heat capacity at constant pressure we obtain
𝑞 ≈ (1.785 𝑚𝑜𝑙) 29.125
𝐽 (−30.17 𝐾) = −1569 𝐽 = −1.569 𝑘𝐽 𝐾 𝑚𝑜𝑙
Therefore 1.57 kJ of energy is transferred to the surroundings by heating. Solution: Work done. Since we have a constant pressure system, we can simply use the change in volume to compute work (1 kJ = 1 kPa m3), that is, w = −P∆V = −( 140 kPa
kJ m kPa
( 0.9 (32 L) − 32 L)
m = 0.448 kJ 1000 L
As an alternative method for computing work, let’s use derivative of PV for the ideal gas equation for a constant pressure, that is,
w = −P∆V = −nR∆T = −1.785 mol 8.314 w = 448 J = 0.448 kJ
J (271.58 − 301.75) K K mol
which is identical to our previous value. Expansion work is only a function of the number of moles, R and ΔT. If we replace N2 in our cylinder with the same number of moles of O 2, He, H2, CO2 or any other gas that behaves as an ideal gas, the expansion work is identical (heat transfer and internal energy would change). Solution: Changes in Internal Energy. From the First Law, we have
Last Modified: March, 2018
2-31
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
As an alternative method, let’s compute the change in internal energy using the heat capacity at constant volume. For an ideal gas, we have C
=C
− R = 29.125
J J − 8.314 = 20.811 J K K mol K mol
mol
Using the definition of heat capacity at constant volume we obtain ∆𝑈
=
𝑑𝑈
=
𝑞 ≈ (1.785 𝑚𝑜𝑙) 20.811
𝑛𝐶
𝑑𝑇 ≈ 𝑛 𝐶
𝑇 −𝑇 =𝑛𝐶
∆𝑇
𝐽 (−30.17 𝐾) = −1121 𝐽 = −1.121 𝑘𝐽 𝐾 𝑚𝑜𝑙
Even though the volume of the system is not constant, the change in the internal energy using the molar heat capacity at constant volume is equal to that obtained from the First Law. More on this result is given later in the chapter. Summary. The components of the First Law of thermodynamics are shown in the following schematic. The consequence of compression work is to increase the internal energy by 0.45 kJ. However to maintain a constant pressure of 140 kPa, 1.57 kJ of heat transfer is necessary from the system to the surroundings. Therefore the net change in the internal energy is a decrease of 1.12 kJ.
Example Problem 2.4: Application of 1st Law for Varying P, T and V. Problem Definition and Solution Results
Last Modified: March, 2018
2-32
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Problem Statement. Let’s consider a vertical piston-cylinder assembly designed so that the 500 g of oxygen can be compressed using a non-constant piston pressure. For this assembly, we are interested in thermodynamic characteristics for the two different paths shown below with the same initial states of Pi = 100 kPa and Vi = 0.3 m3 and final states of Pf = 400 kPa and Vf = 0.1 m3. Isotherms corresponding to temperatures of 231 K, 308 K and 400 K are also shown in the figure. We will assume a quasi-equilibrium process and ideal gas characteristics. For steps corresponding to an incremental change in volume of 0.02 m3, we will compute for each path: (1) temperature, (2) work done, (3) change in internal energy, (3) heat transfer, (4) temperature enthalpy (not a function of pressure for ideal gases), and (5) quantity of heat transfer divided by temperature. We will compare the results between Paths #1 and #2 to gain insight on which characteristics are represented by exact differentials. 700
Path #1:
400 K
xo = 0.1 m3, a = 400 kPa b= -7500 kPa m-3
600
Pressure (kPa)
308 K 500
Vf Pf
400
P a b ( V xo ) 2
231 K 300 200
Path #2:
100
Pi
xo = 0.3 m3, a = 100 kPa b= 7500 kPa m-3
Vi
0 0.08
0.1
0.12
0.14
0.16
0.18 0.2 0.22 Volume (m3)
0.24
0.26
0.28
0.3
0.32
Thermodynamic parameters. From the Data Handout, we obtain R = 8.314 J K-1 mol-1 = 8.314 Pa m3 K-1 mol-1 mm = 32 g mol-1 Cpm = 29.355 J K-1 mol-1 The number of moles are defined as
Last Modified: March, 2018
2-33
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For a more rigorous analysis, we will use a temperature-dependent molar heat capacities defined as
where, for oxygen, we have previously given in Table 2.1 values of a, b, c and d of 25.46, 0.01519, -7.153 x 10-6, and 1.31 x 10-9, respectively. Units for these coefficients are defined corresponding to those necessary to obtain molar heat capacity in units of J K -1 mol-1. To convert to units of kJ K-1, we will multiply the above relationships by the number of moles and use 1000 J = 1 kJ (conversion factor of 0.015625 mol kJ J -1) to obtain 𝐶 = 0.3978 + (2.373𝑥10 ) 𝑇 − (1.1176𝑥10 )𝑇 + (2.048𝑥10 𝐶 = 0.2679 + (2.373𝑥10 ) 𝑇 − (1.1176𝑥10 )𝑇 + (2.048𝑥10
)𝑇 )𝑇
Integration Relationships. The change is our variables of interest is obtained for compression steps in our cylinder of ΔV=-0.02 m3. These changes are determined by direct integration of appropriate relationships. Subscripts “j” and “j+1” refer to values at the beginning and end of the step, respectively. The subscript “tpr” for ΔU and ΔH has not been included to simplify the notation. Changes in internal energy and enthalpy are defined as:
∆H ,
=
𝐶 𝑑𝑇 = 0.3978 𝑇 +
2.373𝑥10 2
𝑇 −
1.1176𝑥10 3
𝑇 +
2.048𝑥10 4
T
Work done is computed by the changing pressure over the volumetric change. Heat transfer is obtained using the First Law. Expansion work and heat transfer are then defined as w,
=−
=− 𝑞
,
= ∆U ,
(𝑎 + 𝑏(𝑉 − 𝑥 ) )𝑑𝑉
𝑃𝑑𝑉 = − a∆V +
b 3
𝑉
−𝑥
− 𝑉 −𝑥
−w,
Last Modified: March, 2018
2-34
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
We are also interested in the heat transfer normalized by the temperature. In Chapter 4, under certain conditions, we use this ratio as a definition of entropy. The method given in this section is a rough approximation. More rigorous equations are given in the next section and in Chapter 4. The change in the ratio of heat transfer and temperature is
where 𝑇 is the average temperature corresponding to the heat transfer. Path #1 Results. Let’s evaluate the characteristics of thermodynamic system for a change of volume from 0.3 m3 to 0.28 m3. For Path #1, we have a= 400 kPa, b = -7500 kPa m-3, and xo = 0.1 m3.
Pa 100 kPa 1000 (0.3 𝑚 ) PV kPa T = = = 230.9 K nR 15.625 mol(8.314 Pa m mol K ) Pa 157 kPa 1000 (0.28 𝑚 ) PV kPa T = = = 338.4 K nR 15.625 mol(8.314 Pa m mol K ) b w , = − a∆V + 𝑉 −𝑥 − 𝑉 −𝑥 3 = − 400 (0.02) −
w,
7500 ((0.28 − 0.1) − (0.3 − 0.1) ) 3
2.373𝑥10 = 0.2679 𝑇 + 2
∆U , ∆H ,
= 0.3978 𝑇 + q, ∆S ,
2.373𝑥10 2
1.1176𝑥10 𝑇 − 3 𝑇 −
1.1176𝑥10 3
.
2.048𝑥10 𝑇 + 4
T
2.048𝑥10 4
T
𝑇 +
= 35.1 𝑘𝐽 . .
= 49.1 𝑘𝐽 .
= ∆U , =
q, 𝑇
+ 𝑇 ⁄2
=
32.5 𝑘𝐽 (338.4 𝐾 + 230.9 𝐾)⁄2
A summary of the computation steps for Path #1 is shown below. Volume
P
T
w
ΔHtpr
ΔUtpr
q
ΔS
m3
kPa
K
kJ
kJ
kJ
kJ
kJ K-1
0.30
100
230.9
0.28
157
338.4
Last Modified: March, 2018
2.58
49.1
2-35
35.1
32.5
0.114
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
0.26
208
416.3
3.66
36.8
26.7
23.0
0.061
0.24
253
467.4
4.62
24.7
18.0
13.4
0.030
0.22
292
494.5
5.46
13.2
9.7
4.3
0.009
0.20
325
500.4
6.18
2.9
2.1
-4.1
-0.008
0.18
352
487.7
6.78
-6.2
-4.5
-11.3
-0.023
0.16
373
459.4
7.26
-13.8
-10.1
-17.4
-0.037
0.14
388
418.1
7.62
-19.9
-14.5
-22.2
-0.050
0.12
397
366.7
7.86
-24.4
-17.7
-25.6
-0.065
0.10
400
307.9
7.98 60
-27.4 34.9
-19.8 24.9
-27.7 -35.1
-0.082
Total
-0.051
For the initial steps, the relative pressure increases are greater than the decreases in volume (T increases because PV increases), but later in the compression, the relative volume decreases are greater than the pressure increases (T decreases because PV decreases). Summation totals are of greatest interest to us. Of course the total change in internal energy is equal to the sum of work and heat transfer. Most of the problems in BBE 3043 are for expansion work (only) under constant pressure. The heat transfer for these problems is equal to the temperature enthalpy. Heat transfer is not equal to the temperature enthalpy for systems under varying pressure conditions. Path #2 Results. For Path #2, we have a= 100 kPa, b = 7500 kPa m -3, and xo = 0.3 m3. We can now repeat our calculations for this path. Examples of these calculations are shown below.
Pa 103 kPa 1000 (0.28 𝑚 ) PV kPa = = = 220.0 K nR 15.625 mol(8.314 Pa m mol K ) 7500 ((0.28 − 0.3) − (0.3 − 0.3) ) = − 100 (0.02) + 3
T w, ∆U ,
= 0.2679 𝑇 +
∆H ,
= 0.3978 𝑇 + q, ∆S ,
2.373𝑥10 2
𝑇 −
2.373𝑥10 2
𝑇 −
1.1176𝑥10 3
𝑇 +
1.1176𝑥10 3
𝑇 +
2.048𝑥10 4
T
2.048𝑥10 4
T
= −2.83 𝑘𝐽 .
= −4.0 𝑘𝐽 .
= −2.83 q, −4.85 𝑘𝐽 = = 𝑇 + 𝑇 ⁄2 (222 𝐾 + 230.9 𝐾)⁄2
The remaining steps are shown in the table below.
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2-36
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Volume
P
T
w
ΔHtpr
ΔUtpr
q
ΔS
m3
kPa
K
kJ
kJ
kJ
kJ
kJ K-1
0.30
100
230.9
0.28
103
222.0
2.0
-4.0
-2.8
-4.8
-0.021
0.26
112
224.2
2.1
1.0
0.7
-1.5
-0.007
0.24
127
234.6
2.4
4.7
3.3
0.9
0.004
0.22
148
250.6
2.7
7.2
5.1
2.4
0.010
0.20
175
269.4
3.2
8.5
6.1
2.8
0.011
0.18
208
288.2
3.8
8.6
6.1
2.3
0.008
0.16
247
304.2
4.5
7.3
5.3
0.7
0.002
0.14
292
314.7
5.4
4.8
3.5
-1.9
-0.006
0.12
343
316.8
6.3
1.0
0.7
-5.6
-0.018
0.10
400
307.9
7.4
-4.1 34.9
-3.0 24.9
-10.4 -15.1
-0.033
Total
40.0
-0.050
The initial and final temperatures are identical to those obtained for Path #1 (which is of course necessary for the same pressures and volumes). For Path #2, the changes in PV are initially negative (T decreases) and are positive (T increases) later in the compression process. Although the work and heat transfer for Path #2 are smaller than that of Path #1, the total change in internal energy is again equal to their sum and is the same as Path #1. Exact and Inexact Differentials An important component of this problem is to examine which of the thermodynamic characteristics are represented by exact differentials and which by inexact differentials. Let’s first compute values using the endpoint states only. The initial state corresponds to Pi = 100 kPa, Vi = 0.3 m3 and Ti = 230.9 K and final states of Pf = 400 kPa, Vf = 0.1 m3, and Tf = 307.9 K. We can then compute ΔU and ΔH between initial and final states as 2.373𝑥10 = 0.2679 𝑇 + 2
∆U ∆H
= 0.3978 𝑇 +
2.373𝑥10 2
1.1176𝑥10 𝑇 − 3 𝑇 −
1.1176𝑥10 3
.
2.048𝑥10 𝑇 + 4
T
2.048𝑥10 4
T
𝑇 +
= 24.9 𝑘𝐽 . .
= 34.9 𝑘𝐽 .
We can also compute the heat transfer between initial and final states using q
= ∆U
−w
where work is defined as w = − a(V − V ) +
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b 3
𝑉 −𝑥
− (𝑉 − 𝑥 )
2-37
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Work, however, is a function of initial and final volumes and on the coefficients of a, b, and xo. Since these coefficients are path dependent, work is also path dependent. Since ΔU is the same for both paths, heat transfer is different for the two paths as well. A summary of our calculations is shown below. Characteristic ΔU (kJ)
Path #1
Path #2
Endpoint Only
ΔHtpr (kJ) ΔS (kJ K-1) w (kJ) q (kJ)
Based on these results, we conclude the work and heat transfer are defined by inexact differentials. Under certain conditions discussed in Chapter 4, ΔS is a change in entropy of the system. A more rigorous analysis of this change is given in the next subsection. The results of the example problem support the assertion that the internal energy, enthalpy and entropy are defined by exact differentials. As discussed briefly in Chapter 1, the Second Law requires that the change in entropy is greater than or equal to zero. You may then be concerned that the change in entropy for this problem is negative. Clearly we will need to be careful about entropy concepts discussed in detail in Chapter 4. Suffice it to say for now that the Second Law applies to the change in entropy of the Universe, and for this problem, we have only computed the change in entropy of our system. Additional Discussion For the design of a constant volume system, we can straightforwardly compute the change in internal energy using dU = Cv dT. But what if our design is for an expanding or contracting (varying volume) system? For Example Problem 2.2, the change in internal energy using C v was shown to equal that obtained from the First law, even though this system was clearly contracting between initial and final states. Insight into why we can use dU = C v dT even for systems with varying volume has already been explored in Chapter 1. In Chapter 1, we defined the change in internal energy (with no phase transitions or chemical reactions) as
where partial derivatives are the changes in internal energy with respect to temperature (for a given volume) and volume (for a given temperature). We have also used our definition of heat capacity at constant volume of Cv= (∂U/∂T)V. Internal energy is, in its more general representation, a function of temperature and volume.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For a system of constant volume, dV=0. The second term on the right-hand side of Eq. 2.5.36 is then zero. We can then compute the change in internal energy as previously done in the example problems as
Let’s consider the system where dV is not zero. For the second term on the right-hand side of Eq. 2.5.36 to be zero, we now need to remember that internal energy is only a function of temperature for ideal gases. For example in Chapter 1, we used Ut,m = a+bT in describing the ideal gas equation. When internal energy is only a function of temperature, we know that
For ideal gases, the term (∂U/∂V)T dV is then always zero regardless of whether or not dV is zero. The change in internal energy can be computed directly from Eq. 2.5.37 as done in Example Problem 2.2. We conclude that Eq. 2.5.37 can be used to compute the change in internal energy for non-ideal gases for systems of constant volume and for ideal gases for systems of constant and varying volumes. Students sometime wonder why we didn’t develop relationships to adjust C p for pressure. For Example Problem 2.2, the Cpm was obtained from the Thermodynamic Handout that corresponds to Po= 100 kPa; whereas the system was compressed using P = 140 kPa. Wouldn’t pressure work, and consequently C pm, be greater for a larger pressure? Let’s consider the response of ideal gases. For a fixed mass system undergoing a constantpressure change, we have previously shown that PΔV = nRΔT (Eq. 2.5.27). Let’s now consider the pressure work for two different pressures (say P 1 = 100 kPa and P2 = 140 kPa) for the same temperature difference (same change in internal energies). Since n, R, and ΔT are constant, we conclude that
that is, the expansion work is equal for the two different pressures. An increase in pressure corresponds to an equivalent decrease in expansion volume. Consequently, C p values are not a function of pressure for ideal gases.
ENTHALPY OF PHASE TRANSITIONS
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Introduction We now wish to expand our application of the First Law to include phase transitions. Phase transition is the conversion of a substance to a different phase. There can be more than one phase for solids. Only the most basic features of phase transitions will be given here. A more rigorous analysis is given in Chapter 5. Let’s consider the heating of water and the corresponding change in temperature of water from an initial state with ice cubes to a final state of boiling water (see Fig. 2.23). Although heat transfer into the system is positive, the initial water-ice mixture maintains a temperature of approximately 0 C until the ice is melted. Since there is no change in internal energy with temperature, the heat transfer must be tied to a change in the internal energy for the phase transition from solid to liquid, that is, the internal energy of liquid water is greater than that of ice. Since the impact of heat transfer cannot be “sensed” by a changing temperature, this type of transfer of energy has historically been considered to be “hidden”. The change in internal energy was referred to as latent heat. After the ice is melted, the impact of additional heat transfer results in an increase in the temperature of the liquid. Since this impact can be easily sensed by us, this type of heat transfer has historically been called sensible heat transfer. Once the liquid starts to boil, the temperature of the liquid again is constant and the internal energy change with temperature is again zero. The heat transfer is used to drive liquid molecules apart to form water vapor. Once again, this process is sometimes referred to as latent heat. Although engineering problems can be properly solved using the terminology of latent and sensible heat, they are not needed to understand thermodynamic processes and provide a less straightforward framework for analyzing systems. Latent heat is usually defined by using the heat transfer corresponding to phase transitions. Similar to heat capacity, this heat transfer changes the characteristics of matter. In the distant past, the processes of this change were “hidden” from us, but the connection is now well established. “Latent heat” is also ambiguous as to whether it refers to energy transfer for the change in the internal energy or the change in internal energy and expansion work. We can avoid this unnecessary confusion by consistently applying general enthalpy concepts. Enthalpy is a well-establish property of thermodynamics. It is defined by exact differentials and is a thermodynamic potential (as discussed in Chapter 5). Both of these attributes are valuable in understanding the response of biological and environmental systems.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.23. Temperature Changes of Water with Phase Transitions. Possible phase transitions are summarized in Fig. 2.24. Vaporization (or evaporation) is the phase transition from a liquid to a gas and condensation is the reverse of vaporization (transition from a gas to liquids). Fusion (or melting) is the phase transition from solid to liquid and freezing is the reverse of fusion (transition from a liquid to solid). Sublimation is the phase transition from solid to gas (dry ice, snow to vapor) and vapor deposition is the reverse of sublimation (transition from a gas to solid). Vapors are gases and therefore behave as gases.
Figure 2.24. Terminology Used for Phase Transitions. Standard Enthalpies of Transitions Introduction Similar to experiments done for heat capacities, let’s consider a possible experiment that measures the heat transfer into the system to determine the change in internal energy and expansion work with phase transitions as shown in Fig. 2.25. Conceptually, data collected with this type of experiment could then be used to estimate the change in enthalpy (actual
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
measurement is based on the theory given in Chapter 5). However for phase transition, we are not interested in the change in the temperature of the system but the change in the number of moles of each phase, that is, q/Δn instead of q/ΔT as used for heat capacities (which is obviously of no value if ΔT=0). Enthalpies of transitions are therefore reported in units of energy per mole (or per mass) of phase change. For our illustration in Fig. 2.25, heat transfer into the system is used to vaporize 10 moles of liquid water and to expand the system until the water vapor pressure equals the external (atmospheric) pressure. A change of internal of energy and expansion work of 377 kJ and – 30 kJ, respectively, requires a heat transfer of 407 kJ. The measured molar enthalpy of vaporization would then be 40.7 kJ mol -1. Of course the increase in the number of moles of vapor in Fig. 2.25 is equal to the decrease in the moles of the liquid phase. However, changes in liquid volume are negligible in comparison to changes in water vapor (see Problem 1.12).
For vaporization, the initial and final states correspond to the liquid and vapor phases, respectively. Since enthalpy is defined by an exact differential, the change in enthalpy for vaporization can be evaluated from basic calculus as
where for condensation the initial state corresponds to the vapor phase and the final state to the liquid phase. We therefore conclude that the enthalpy of transition of particular direction is equal to the negative value of the transition in the reverse direction, that is,
Freezing Enthalpy = - Fusion Enthalpy Vapor deposition Enthalpy = - Sublimation Enthalpy A positive ΔH indicates that heat transfer from the surroundings to the system is necessary to drive the system. For example, vaporization requires heat transfer from the surroundings to
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
provide the energy necessary for molecules to break away from their liquid phase. These concepts are shown in the schematic below for melting and freezing processes of water.
Figure 2.26. Sign Notation for Enthalpy of Phase Transitions. Standard Enthalpy Vaporization and Notation The standard enthalpy of vaporization is defined as
Although the definition of standard enthalpy does not require that data be collected at a standard temperature, enthalpies of transitions are dependent on temperature. A corresponding temperature is therefore needed for the reported standard enthalpy of substances, usually given at their transition temperature. Molar enthalpy of vaporization is used to refer to values at non-standard pressures. We will use the notation of ∆ 𝐻 for standard enthalpy of vaporization. For expansion work by ideal gases under constant pressure and temperature, we have d(PV) = P dV = d(nRT) = RT dn. Standard enthalpy of vaporization is defined as
which has a component related to the change in the internal energy and an expansion work component. Eq. 2.6.3 uses the First Law for conditions of constant pressure and ΔUtpr=ΔUphase = 0. The “°” is used to indicate that it corresponds to the standard pressure. The symbol T is a holder for the temperature at which the data has been recorded or the temperature corresponding to the system of interest. For example, the standard enthalpy of
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2-43
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
vaporization for water at a temperature of 100 C is ∆ 𝐻 = 40.7 kJ mol-1 and is ∆ 𝐻 = 45.07 kJ mol-1 at 0 C. We will use the specific enthalpy of vaporization as the enthalpy of vaporization per mass (instead of moles) and adopt the notation of ∆ 𝐻 . Since 1 mole of water = 18 g, the specific enthalpy of vaporization for water at temperature of 0 C is defined as ∆ 𝐻 =45.07/18 = 2.5 kJ g-1 or 2500 J g-1. The change in internal energy for vaporization can be determined from a measured enthalpy value by rearranging Eq. 2.6.3 as ∆ 𝑈 = ∆ 𝐻 − 𝑅𝑇. Similar to Cpm of the previous section, we need to be careful to distinguish between the tabular values obtained from experiments and the application of these values to thermodynamic problems. The notation of ∆ 𝐻 is used for measured change in enthalpy that are summarized in table or equivalent relationships. It is a property of the substance. To solve thermodynamic problems that have vaporization, the notation for this enthalpy change is represented by ∆𝐻 = 𝑛 ∆ 𝐻 = 𝑚 ∆ 𝐻 , where n and m are the number of moles or mass, respectively, that are vaporized. Fusion and Sublimation Similar to vaporization, the standard enthalpy of fusion is defined as Heat transfer required to melt one mole of molecules of a pure substance at a constant pressure of 1 bar (100 kPa). Similar to vaporization, we will use the notation of ∆
𝐻 for standard enthalpy of fusion.
For a temperature of 0 C, it is defined for water as ∆ 𝐻 =6.01 kJ mol-1. Once again, a positive value indicates that heat transfer is into the system for melting. The standard enthalpy of sublimation is defined as: Heat transfer required to convert one mole of molecules of a pure substance at a constant pressure of 1 bar (100 kPa) from a solid to a gas. We will use the notation of ∆ 𝐻 for standard enthalpy of sublimation. Standard enthalpy of sublimation equals the sum of that for fusion and vaporization. For a temperature of 0 C of water, we then obtain ∆
H =∆
H +∆
H = 45.07 𝑘𝐽 𝑚𝑜𝑙
+ 6.01 𝑘𝐽 𝑚𝑜𝑙
= 51.08 𝑘𝐽 𝑚𝑜𝑙
Tabular Values Values for standard enthalpy of transitions are shown in the following tables. Most of the values are reported at their transition temperatures (freezing point and boiling point).
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Table 2.2. Standard Enthalpies of Transitions for Different Substances. Substance
Chemical Symbol
T: Freezing (K)
Ammonia Benzene Ethanol Helium Methane Water
NH3 C6H6 C2H5OH He CH4 H2 O
195.3 278.7 158.7 3.5 90.7 273.15
Tfus H om kJ mol-1 5.65 9.87 4.60 0.02 0.94 6.01
T: Boiling (K) 239.7 353.3 351.5 4.22 111.7 373.2
Tvap H om
kJ mol-1 23.4 30.8 38.6 0.08 8.2 40.7
Changes with Temperature We are often interested in converting the above tabular enthalpy of vaporization to different conditions. A relatively rigorous analysis of the dependency of ∆ 𝐻 with temperature and pressure is given in Chapter 5. However, greater insight into changes with temperature (only) and in the usefulness of exact differential is obtained by vaporization of water for two different paths. The change in enthalpy is defined by an exact differential, and it is independent of the path. We can therefore use the equality of the solution for each path to determine the relationship between temperature and the standard enthalpy of vaporization. Let’s consider the change in enthalpy corresponding to n moles of water at the liquid state at 25 C to n moles of vapor at a temperature of 35 C (Δn vapor = n moles). We will consider two possible paths between the initial and final states as shown conceptually in Fig. 2.29. For Path #1, all of the water is evaporated at 25 C and the vapor temperature increases to 35 C. For Path #2, the temperature of liquid water is first increased to 35 C and then all of the water is evaporated. We have to consider both the change in enthalpy for phase transition and the enthalpy changes with temperature. These two paths were selected to obtain simple solutions. It is likely that the actual path won’t correspond to either of them.
Figure 2.29. Framework Temperature Dependent Enthalpy Relationship. Let’s now consider the change in enthalpy for the following path where the n moles of water are first vaporized at 25 C and then the temperature of the water vapor is increased to 35 C.
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2-45
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.30a. ΔH Change for Liquid → Gas and then 25 C → 35 C. The change in enthalpy for this pathway can be computed as
where ΔH1 is the change in enthalpy for the first path. We have assumed a negligible change in Cpm (water vapor) with temperature. Let’s now consider the alternative path between initial and final states as shown below. For this path, the temperature of the liquid water is first increased to 35 C. The liquid water is then vaporized at 35 C to obtain the conditions corresponding to the final state.
Figure 2.30b. ΔH Change for 25 C → 35 C and then Liquid → Gas.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
The change in enthalpy here is ∆H =
𝑛 𝐶
𝑑𝑇 + n ∆
H ≈ 𝑛 𝐶
∆𝑇 + n ∆
H
2.6.5
where ΔH2 is the change in enthalpy for the second path and where again we have assumed a constant Cpm (liquid water) with temperature. Since enthalpy corresponds to an exact differential, the change in enthalpy is independent of the path. If we set the change by the two paths equal to each other (ΔH1=ΔH2), we obtain n
𝐶
𝑑𝑇 + n ∆
H = n∆
H +n
𝐶
𝑑𝑇
2.6.6
If we divide both sides by n and rearrange terms, we can solve for the standard enthalpy of vaporization at 35 C as ∆
H =∆
H + ≈ ∆
[ 𝐶 H +[ 𝐶
− 𝐶 − 𝐶
] 𝑑𝑇
2.6.7
] ∆𝑇
where for water (Cpm)gas = 33.58 J K-1 mol-1 and (Cpm)liq = 75.29 J K-1 mol-1. Let’s generalize by representing the temperature Ti = To for the tabular standard enthalpy of vaporization and the temperature of interest as Tf = T. We then obtain
Let’s consider the usefulness of enthalpies defined by exact differentials by placing a cup of water at a temperature of 25 C into a room at a temperature of 35 C.
The water in an open cup would eventually evaporate and the resulting water vapors would eventually reach a temperature of 35 C. The actual path between these initial and final states can be quite complex. For example, there would be some evaporation at 25 C, at 26 C, at 27 C and at other temperature to our final state temperature of 35 C (assuming that there is still
Last Modified: March, 2018
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
liquid water in the cup). For each of these temperatures, we need to compute the heat transfer to increase the liquid temperature from 25 C, the heat transfer to evaporate the mass, and the heat transfer to increase the gas temperature to 35 C. Additional equations for heat and mass transport are needed to determine the actual evaporation rates at each temperature. Fortunately, this level of complexity is not needed for our solution. Since enthalpy is defined by exact differentials, the change in enthalpy for the actual path is the same as our idealized Path 1 (all of the water evaporates at 25 C) or as our idealized Path 2 (all of the water evaporates at 35 C). For a system under constant pressure, we can therefore determine the heat transfer between our endpoints by selecting a path that corresponds to one of these simpler solutions. We don’t need to understand the complex processes corresponding to the actual path if the endpoint conditions are known. Example Problem 2.5: Cooling by Perspiration Problem Statement. Perspiration is an important mechanism for cooling the body. During vigorous exercise, as much as 1 kg to 2 kg of water (1 to 2 L) of water can be produced per hour. We will assume that 0.5 kg of perspired water is evaporated at a constant temperature of 20 C and that all of the heat transfer from the skin is used to the evaporate water for a closed system (no mass transfer). We wish to calculate the
Initial
Vaporization
Expansion work
Final 0.5 kg vapor T = 20 C
0.5 kg liquid T = 20 C
q
Skin
Tabular Values. From the Data Handout, we obtain mm = 18 g mol-1 (Cpm )gas = 33.58 J K-1 mol-1 (Cpm )liq = 75.29 J K-1 mol-1 As previously given
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
∆
H = 40.7 𝑘𝐽 𝑚𝑜𝑙
Preliminary Calculations. Temperature and number of moles:
Standard enthalpy of vaporization at 20 C ∆
H
≈ ∆
H +[ 𝐶
∆
H
≈ 40.7 𝑘𝐽 𝑚𝑜𝑙
∆
H = 44.04 𝑘𝐽 𝑚𝑜𝑙
− 𝐶
] ∆𝑇
+ (33.58 − 75.29)
𝐽 𝐾 𝑚𝑜𝑙
𝑘𝐽 ( −80 𝐾) 1000 𝐽
Solution: Heat Transfer. From the First Law for constant-pressure expansion work, we have for this problem
By using the standard enthalpy of vaporization to determine ΔH vap, we obtain
We conclude that 1223 kJ of heat transfer from the skin into the system is needed to evaporate 0.5 kg of perspiration. This removal of heat from the body is an important cooling mechanism for humans. Solution: Expansion work and Internal Energy. The change in enthalpy includes a change in internal energy as well as expansion work. We will assume that the water vapor behaves as an ideal gas where PV = nRT. For a constant temperature and pressure, we then obtain d(PV) = PdV = RT dn. The change in enthalpy can then be written as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where the initial number of vapor moles is zero. We can then evaluate the expansion work as P∆V = RT ∆n
= 8.314
J (293.15 𝐾)(27.8 − 0)𝑚𝑜𝑙 = 67701 𝐽 = 67.7 𝑘𝐽 K mol
Expansion work is defined as w = - PΔV = - 67.7 kJ, that is, work is done on the surroundings to create a volume for 27.8 moles of water vapor. Of the heat transfer of 1223 kJ, 67.7 kJ of the energy transfer was needed for this work or roughly 5.5% of the heat transfer (67.7/1223). The remaining heat transfer is used to change the internal energy of the system. The change in the internal energy is simply ∆U
= q + w = 1223 k𝐽 − 67.7 𝑘𝐽 = 1155.3 𝑘𝐽
ENTHALPY OF CHEMICAL REACTIONS Introduction General Concepts Biological systems are often driven by the heat transfer as a result of chemical reactions. Chemical reactions require a change in the composition of constituent atoms among various molecules. Reconfiguration possibilities are restricted by the mass balance requirements corresponding to the chemical reaction equation. The redistribution of atoms among molecules also requires a redistribution of internal energy. Changes in the binding energy can result in an increase or decrease of the overall internal energy of systems. These changes are represented by ΔUr = Uprod – Ureact, where Uprod and Ureact are the internal energies of the products and reactants, respectively. Similar to heat capacities, chemical reactions can be analyzed for a given temperature and volume or a given temperature and pressure, where the former is used to analyze the change in internal energy and the latter the change in enthalpy. Let’s start by considering the chemical reaction of urea (a widely used fertilizer) with oxygen. The chemical reaction is defined as
where the stoichiometric coefficients (necessary for mass conservation) are defined as υ
= 1, υ
= 3⁄2 , 𝜐
= 1, 𝜐
= 2, 𝑎𝑛𝑑 𝜐
=1
This chemical reaction results in the heat transfer from the system to the surroundings as shown in Figure 2.31. Specialized equipment can be used to measure this transfer. An example of this type of equipment is an adiabatic bomb calorimeter. More on its use is given in Homework Problem 2.32. As shown later in Example Problem 2.5, the urea reaction
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
increases the volume of gases resulting in expansion work for a system under constant pressure. Enthalpy is therefore a useful measure of the heat transfer because it includes both changes in internal energies and expansion work.
Figure 2.31. Conceptual Representation of Reaction Enthalpy. Enthalpy of chemical reactions is determined from reactions carried out under a constant pressure and where the products are returned to the same temperature as the reactants. For the closed system under quasi-equilibrium and constant pressure, the heat transfer is equal to the change in this enthalpy, that is,
where ΔUr and ΔHr are the changes in the internal energy and in the enthalpy, respectively, for the chemical reaction. Since the initial and final temperatures are the same and there are no phase transitions, ΔUtpr = ΔUphase = 0. The change in enthalpy is then determined by the change internal energy resulting from a new configuration of atoms of the products and the expansion work resulting from the net increase or decrease of gases. Heat transfer for each reaction of interest could conceptually be obtained from an experiment similar to that shown above for urea. Reaction enthalpy could then be computed and summarized in tables. However, there are a very large number of interesting chemical reactions and the reporting of the reaction enthalpy for each of them would be cumbersome (and unnecessary). An alternative approach is to determine the enthalpy for each of the compounds and appropriately combined them to determine the overall reaction enthalpy. For example, since the internal energy of O2 is the same as a reactant with urea and as a reactant with methane (for the same initial pressure and temperature), we can avoid duplication of effort if we determine this internal energy once and then use it for all reactions of O 2. Our approach utilizes enthalpy’s extensive properties (individual components can be added to
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
determine the total) and that the change in enthalpy is independent of the path. After the enthalpies of compounds are defined, they can be appropriately combined to determine the reaction enthalpy of chemical reactions using them. Molar Extent of Reaction Similar to the molar heat capacity and standard enthalpy of transitions, we are interested in analyzing the reaction enthalpy in units of energy per mole. The number of moles used in molar heat capacity and standard enthalpy of transitions was easily defined from the mass of the system and the mass change with phase transitions, respectively. Selection of the “number of moles” to obtain a molar energy unit needs to be done carefully for chemical reactions. This point will be illustrated using the chemical reaction equation for urea (Eq. 2.7.1). Let’s consider our urea reaction with 2 moles of consumed urea. To conserve mass, we need to maintain the same number of hydrogen, carbon, and nitrogen atoms before and after the reaction. We then conclude that 2 moles of CO2 and N2 and 4 moles of H2O must be produced by the reaction. To maintain the balance of oxygen, 3 moles of O 2 must have also been used in the reaction. A summary of the changes in moles (Δn) is shown below.
(2 moles) = (−3/2)(2 moles) = −3 moles
∆n
=υ
∆n ∆n ∆n
= υ (2 moles) = (1)(2 moles) = 2 moles (2 moles) = (2)(2 moles) = 4 moles =υ = υ (2 moles) = (1)(2 moles) = 2 moles
where υurea through υN2 are the stoichiometric coefficients. There is a decrease in the number of moles for the reactants and an increase in the number for the products. At an initial glance, it might seem impossible to select a “number of moles” for defining a molar energy unit because the number of moles varies among chemical compounds, ranging from 2 moles for CO2 to 4 moles for H2O. However, notice that a common quantity of 2 moles is used to compute the change in the number of moles for each compound. This common quantity is called the molar extent of reaction (Δξ). The change in molar enthalpy is defined as the enthalpy change per molar extent of reaction. Let’s extend the concepts established for the reaction of urea to the more general representation of chemical reactions (as given in Chapter 1) of 2.7.3a
𝑣 𝐴 + 𝜈 𝐵 + ⋯ → 𝜈 𝐿 + 𝜈 𝑀 + ⋯.
where υA through υM are the stoichiometric coefficients. The molar extent of reaction is used to compute the change in the number of moles in reaction from the stoichiometric
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
coefficients. For a given molar extent of reaction of Δξ, the change in the number of moles for each substance is defined as ∆n = υ ∆ξ , ∆n = υ ∆ξ , … ∆n = υ ∆ξ, ∆n = υ ∆ξ , …
2.7.3b
where Δn is the change in the number of moles as a consequence of a given molar extent of reaction. By notation convention, we can define the change for the reactants as negative and the change in the products as positive (alternatively, the reactant stoichiometric coefficients are sometimes defined as negative). The molar extent of reaction is constant for all substances. Clearly from the above relationships, the stoichiometric coefficient for substance j is defined as
A rigorous discussion on the definition of the molar extent of reaction will have to wait until we have a better understanding of the Second Law. However, it is possible for us to provide some insight on Δξ. Let’s assume that you want a gas mixture of 10 moles of CO 2 and 10 moles of N2. Based on the chemical reaction for urea, you intend to obtain this mixture by the reaction of 10 moles of urea with 15 moles of O 2. This plan assumes that all of the moles of urea is used. For our hypothetical example, let’s assume that after your reaction you measured a reduction in the number of O2 of only 12 moles. From your observed data and the stoichiometric coefficients, we can compute the molar extent of reaction, and the corresponding moles of the other compounds (because of mass balance), as:
∆n
= −υ
∆ξ = −(1)(8 𝑚𝑜𝑙) = −8 𝑚𝑜𝑙, ∆n = (1)(8 𝑚𝑜𝑙) = 8 𝑚𝑜𝑙, ∆n = 8 mol and ∆n = 16 mol
2.7.5b
Even though adequate O2 was available, not all of the urea was used in the reaction. The actual molar extent of the reaction is only 8 moles. Instead of obtaining your target of 10 moles of CO2 and N2, you were only able to obtain 8 moles of each of them. Incomplete consumption of reactant occurs because chemical equilibrium between the reactants and the products. For metabolic processes given later in the chapter, the molar extent of reaction is obtained (effectively) by measuring changes in the number of moles of reactants and products as computed by Eq. 2.7.5a. However, a more insightful approach is to use theoretical relationships obtained from the Second Law of thermodynamics. This approach is discussed in detail in Chapter 6. Let’s now consider reactions for powerful combustion processes. Here the organic reactant (i.e., “fuel”) is completely consumed (essentially) by the reaction. The molar extent of
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
reaction is then equal to the number of moles of the organic reactant. This concept can be illustrated for the combustion of methane (natural gas is mostly methane), represented by the following reaction CH 4 (g ) 2 O 2 ( g ) CO 2 (g ) 2 H 2 O (l)
2.7.6
If we have a system with 10 moles of methane then the molar extent of reaction would be Δξ = 10 moles. Twenty moles of oxygen gas would then also be consumed by the combustion process, and we would have products of 10 moles of carbon dioxide and 20 moles of water. Enthalpy of Reaction Since change for exact differentials is equal to the difference between the endpoints, we can readily evaluate the change in enthalpy for chemical reactions at constant pressure and temperature, that is,
where Hprod and Hreact is the enthalpies of the products and reactants, respectively. Enthalpies of the products and the reactants can be obtained by the sum of their components. The change in enthalpy for a chemical reaction can then be written as ∆𝐻 =
∆𝑛 𝐻
−
∆𝑛 𝐻 2.7.8
=
𝜈 ∆ξ 𝐻
−
𝜈 ∆ξ 𝐻
where Δnj = υj Δξ is the change in the number of moles for each product/reactant, υ j are the stoichiometric coefficients, Δξ is the molar extent of reaction and H mj is the molar enthalpies of the components in the reaction, that is, Hmj = Hj/Δnj. If we divide both sides by the molar extent of reaction, we obtain the reaction enthalpy defined as
where ΔrTHm is the molar enthalpy of reaction at temperature T and pressure of the reaction. It is defined per molar extent of reaction. The final step is to determine the molar reaction enthalpies of its components. However, the determination of these enthalpies needs to be done thoughtfully and carefully. The concept of
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
standard enthalpy of formation (ΔfTHom) is used to systematically evaluate the molar reaction enthalpies. Details on this concept are given in the next section. Standard enthalpy of reaction is defined from the standard enthalpy of formulation as ∆ 𝐻 =
𝜈∆ 𝐻
−
𝜈∆ 𝐻
2.7.10
Stoichiometric coefficients are sometimes defined as a positive for the products (increasing) and negative for the reactants (decreasing). Negative changes in the number of moles are then defined directly from Eq. 2.7.2, and the standard enthalpy of reaction can then be computed using a single summation term. We will, however, use the notation of positive stoichiometric coefficients for both products and reactants and subtract the reactant enthalpy from the product enthalpy to more clearly show a change in enthalpy between the products and reactants. Standard Enthalpy of Formation Reference State Standard enthalpy of formation is defined as the change in enthalpy upon forming one mole of material from the element’s most stable pure state. The most stable pure state is called the reference state. There is only one reference state for each element. In contrast, more than one standard state can exist for elements. For example for carbon, there is a standard state for graphite (also the reference state) and a standard state for diamond (not a reference state). Standard enthalpy of formation is defined for a standard pressure of 1 bar (100 kPa), and usually at a temperature of 25 C. Standard enthalpies of formation of elements at their reference state are, by definition, equal to zero. For example, the most stable state for carbon is graphite (solid, but not diamond), for hydrogen is H2, and oxygen is O2. If you look in the thermodynamic data handout, you will find that the standard enthalpy of formation for these substances is given as:
Experimentally Derived Values Standard enthalpy of formation for some substances can be determined directly from measurements. For example, the enthalpy of reaction for the combustion of carbon (graphite) with oxygen can be measured in laboratory experiments. The chemical reaction is shown below. C (graphite) + O2 (g) → CO2 (g)
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2.7.12
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
At a temperature of 25 C and pressure of 100 kPa, the standard reaction enthalpy for this reaction has been measured as ∆ 𝐻 = -393.51 kJ mol-1. This is an exothermal reaction that results in considerable heat transfer from the system to the surroundings. Let’s consider the definition of standard reaction enthalpy for the above reaction. We have ∆
𝐻 = −393.51 𝑘𝐽 𝑚𝑜𝑙 = (1)∆
𝐻
= ,
𝜈∆ 𝐻
− (1)∆
𝐻
− ,
− (1)∆
𝜈∆ 𝐻 𝐻
2.7.13
,
Since the carbon and oxygen of the reactants are at their reference states, their enthalpies of formation are zero. We then obtain the standard enthalpy of formation for CO2 as
which corresponds to the value given in the thermodynamic handout. After standard enthalpies of formation for compounds have been determined, they can be used to determine the standard enthalpy of formation for other compounds from measured data. For example in Homework Problem 2.33, the standard enthalpy of formation for carbon monoxide is computed using the ∆ 𝐻 for CO2 computed in our example. Hess’ Law Our definition of standard enthalpy of reaction follows directly from Hess’ law. Hess’ law states that the standard enthalpy of reaction can be computed using the sum of the enthalpies of individual reactions into which the overall reaction may be divided. This law follows directly from the characteristics of an exact differential where the change is independent of the path between endpoints. Application of Hess’ law is used to solve Homework Problems 2.36 and 2.37. Changes with Temperature Derivation The standard enthalpy of formation in the thermodynamic data handout can be used to compute the standard enthalpies of reaction at 25 C. You will be interested in the enthalpies at other temperatures. A schematic for this conversion is shown in Fig. 2.32. Conceptually, the change in the enthalpy with temperature is computed for each of the compounds using their molar heat capacities. After the standard enthalpy of formation has been modified to the design temperature, we can then easily compute the standard reaction enthalpy at this temperature.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.32. Changing Reaction Enthalpy with Temperature. We will start with our general definition of reaction enthalpy at standard pressure, that is, ∆ 𝐻 =
𝜈𝐻
−
𝜈𝐻
=
𝜈∆ 𝐻
−
𝜈∆ 𝐻
For changes with temperature, it useful to take the derivative with respect to temperature, which can be written as 𝑑(∆ 𝐻 ) = 𝑑𝑇
𝜈
𝑑 𝐻 𝑑𝑇
−
𝜈
𝑑 𝐻 𝑑𝑇
2.7.19
Imagine a system at a temperature corresponding to tabular value (T o = 25 C for our thermodynamic handout). Let’s consider a general infinitesimal change in standard enthalpy with temperature (for a constant pressure) using the definition of molar heat capacity at constant pressure
and therefore we obtain
where ΔrCpm is called the molar heat capacity of reaction. It is defined for a system of products and reactants. If we integrate between a temperature of known values (T o) and the temperature of interest T, or
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
∆ 𝐻 =∆ 𝐻 +
∆ 𝐶
𝑑𝑇 ≈ ∆ 𝐻 + ∆ 𝐶
2.7.22
∆𝑇
Example Problem 2.6: Standard Enthalpy of Reaction of Urea Problem Statement. For the following reaction of urea with oxygen at a constant pressure of 100 kPa: (NH2)2CO(s) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) + N2 (g) Determine:
Thermodynamic Data. From the thermodynamic data handout, we obtained for T=25 C:
25 C O2 (g): Cpm = 29.36 J K-1 mol-1; f H om = 0 kJ mol-1 25 C CO2 (g): Cpm = 37.11 J K-1 mol-1; f H om = -393.51 kJ mol-1 25 C N2 (g): Cpm = 29.13 J K-1 mol-1; f H om = 0 kJ mol-1
H2O (l): Cpm = 75.29 J K-1 mol-1;
.15 298 H = -285.83 kJ mol-1 f
Solution: Standard Enthalpy of Reaction at 25 C. Standard enthalpy of reaction is defined as ∆
𝐻 =
𝜈∆
𝐻
−
𝜈∆
,
+ (2)∆
𝐻
For the products, we have 𝜈∆ 𝐻
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= (1)∆
𝐻
2-58
𝐻
,
+ (1)∆
𝐻
,
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For the reactants = (1)∆
𝜈∆ 𝐻
𝐻
+ (3 ∕ 2)∆
,
𝐻
,
We therefore obtain the following standard enthalpy of reaction:
For a constant pressure system, we have previously shown that 𝑞 = ∆𝐻 = ∆𝑈 + 𝑃∆𝑉 If we divide both sides by the molar extent of reaction (Δξ), the molar heat transfer is simply 𝑞 =
𝑞 =∆ ∆ξ
𝐻 = −632.1 kJ mol
Solution: Molar Expansion Work and Molar Internal Energy. Expansion work is related to the gas components of the chemical reaction. We need to divide the number of moles of gases into their components, that is,
where for complete combustion no2 is zero after the reaction. Before the reaction, nco2 and nn2 are zero. We will use ideal gases where PV = nRT and use the solution obtained for a constant temperature expansion, that is, RT i=RTf = constant.. For constant pressure, PdV = d(PV) = RT dn. The expansion work for our reaction can then be evaluated as ∫ 𝑃𝑑𝑉 = 𝑅𝑇 ∫ 𝑑 𝑛
+𝑛
+𝑛
= 𝑅𝑇 (𝑛
+𝑛
+𝑛
)
where Δn = nf – ni refers to the change in the number of moles for each of the gas components. The change in number of moles for O 2 is negative. Since reaction enthalpy is
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
defined by molar extent of reaction (Δξ), we obtain an appropriate measure of the expansion work as
and therefore for this problem 𝑃 ∆𝑉 = 8.314 𝐽 𝐾 ∆ξ
𝑚𝑜𝑙
3 (298.15 𝐾) − + 1 + 1 = 1239 J mol 2
= 1.239 𝑘𝐽 𝑚𝑜𝑙
which is less than one percent of the reaction enthalpy. The expansion work per molar extent of reaction is defined as wm = - (PΔV)/ Δξ = - 1.24 kJ mol-1. The net volume of gases increased as a consequence of the reaction resulting in expansion work of the system on the surroundings. The change in internal energy per molar extent of reaction (Δ rUm) can then be defined as
∆ U = (−632.1 − 1.24) kJ mol
= −633.3 kJ mol
The change in internal energy by the chemical reaction was greater than the heat transferred from the system because of expansion work. Solution: Standard Enthalpy of Reaction at 80 C. From Part 1, we have determined the standard reaction enthalpy at 25 C as ∆
𝐻 = −632.1 𝑘𝐽 𝑚𝑜𝑙
Let’s now compute the molar heat capacity of reaction defined as: ∆ 𝐶
=
𝜈𝐶
,
−
𝜈 𝐶
,
For the products, we have
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝜈𝐶
,
= (1)(37.11) + (2)(75.29) + (1)(29.13) = 216.82 𝐽 𝑚𝑜𝑙
𝐾
For the reactants 𝜈𝐶
,
= (1)𝐶
𝜈𝐶
,
= (1)(93.14) + (3 ∕ 2)(29.36) = 137.18 𝐽 𝑚𝑜𝑙
,
+ (3 ∕ 2)𝐶
,
𝐾
We then obtain:
The standard enthalpy of reaction at To = 80 C = 353.15 K is approximately .
∆
𝐻 =∆
𝐻 +
∆ 𝐶
𝑑𝑇 ≈ ∆
𝐻 +∆ 𝐶
∆𝑇
.
∆
𝐻 ≈ −632.1 𝑘𝐽 𝑚𝑜𝑙
+ 79.64
𝐽 𝑚𝑜𝑙 𝐾
(353.15 − 298.15) 𝐾 (
𝑘𝐽 ) 1000 𝐽
The change in standard reaction enthalpy was less than 1%. The change in reaction enthalpy with temperature is generally much less than other properties (such as the change in reaction entropy given in Chapter 6). From its definition, Δ rCpm = 0 if the molar heat capacities and the stoichiometric coefficients are equal. Enthalpy of Combustion Combustion of organic compounds is often the chemical reaction of greatest interest in the design and understanding of mechanical and biological systems. Combustions are chemical reactions where organic compounds are completely oxidized by O2 (g). For common organic compounds containing C, H, and O, the products are CO2 (g) and H2O (l). For compounds that also contain N, N2 (g) is also a product. Combustion processes are important sources of energy for cells as well as mechanical engines. Ignition energy is often needed to start combustion of engine fuels. This energy allows the elemental bonds to be broken between atoms for a few organic molecules. When these atoms recombine with the oxygen atoms, additional energy is generated that releases other atoms from their molecules. The reaction will then proceed until all (effectively) of the fuel is
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
consumed (assuming adequate oxygen). If the chemical reaction is written such that the stoichiometric coefficient for the fuel is υ fuel=1, then the molar extent of reaction is equal to the number of moles of the fuel, that is,
where nfuel is the number of fuel moles in the system and, for these powerful reactions, it is equal to the actual number of fuel moles consumed (Δnfuel). Standard reaction enthalpies for complete combustion reactions (organic reactant + O 2) can be computed and stored in tables and databases for your use (you don’t need to compute them from their products and reactant components). Some examples are given in the Thermodynamic Data Handout. They are called standard enthalpy of combustion (sometimes called heat of combustion) and we will use the notation of ΔcTHm°. It is defined for a standard pressure as ∆ H =(∆ H )
=
(∆ H ) 𝑛
2.7.25
The standard enthalpy of combustion is equal to the standard reaction enthalpy for Δξ = n fuel, that is, the standard enthalpies of reactions for one mole of organic reactant. For expansion work under constant pressure, we can easily compute the heat transfer as simply q = ΔU+PΔV= nfuel ΔcTHmo. We have already analyzed the following examples of combustion in this section and the results for the combustion of carbon monoxide of Homework Problem 2.33. Once again, the numerical values of standard enthalpies of reaction and combustion are equal but ∆ 𝐻 is defined per molar extent of reaction and ∆ 𝐻 is defined by mole of organic reactant. Standard enthalpies of combustion for sugars, fats and proteins are given in the next section. 𝐶 (𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒) + O (g) → CO (g) ∆ H , = − 110.5 kJ mol (NH ) CO (s) + 3⁄2 O (g) → CO (g) + 2 H O (l) + N (g) ∆ H , = − 632 kJ mol 𝐶𝑂 + 1⁄2 O (g) → CO (g) ∆ H , = − 110.5 kJ mol Metabolic Rate and Thermoregulation Introduction Biochemical reactions are the source of the metabolic rate of humans and other animals. The energy from these reactions are defined by the standard enthalpy of reaction for the combustion of carbohydrates (including sugar), fats, proteins and other organic compounds. As shown by Homework Problem 2.26, biochemical reactions require thermoregulation
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
processes to avoid high body temperatures that can be lethal to humans (and other animals). Thermoregulation is achieved by the heat transfer of convection, conduction, and radiation and by the evaporation of perspiration. As the body temperature increase, the blood flow to the skin increases causing greater heat transfer from the body and larger evaporation rate. If the temperature of the surroundings is greater than the body temperature (and emissivities are equal), evaporation is the only process available to regulate body temperature. We are often interested in understanding and in designing systems that include humans and other animals. For conditions of constant pressure, the First Law for these system is ∆U + P∆V = ∆H
+ ∆H
+ ∆H = q
2.7.26a
where the system has all of three possible changes in its internal energy given in this chapter. Temperature changes in rooms or building are important in ventilation designs. This change can be obtained from Eq. 2.7.26a as ∆T =
∆H nC
=
q − ∆H − ∆H nC
2.7.26b
where the reaction enthalpy is negative for metabolic processes and vaporization enthalpy is a component of thermoregulation. Animal Calorimetry Animal calorimetry is the measurement of the heat transfer between animals and their environment and is dependent on the bio-chemical reactions of metabolism. A simplified representation of the measurement approaches is given in Fig. 2.33. The direct measurement of heat transfer (with consideration of vaporization) from animals is called direct calorimetry. The simplest of direct calorimeters measures the heat transfer using principles similar to those of an adiabatic bomb calorimeter (see Homework Problem 2.32). Here heat transfer from animals is obtained by measuring changes in temperature of a substance with well-defined thermodynamic characteristics. Thermoregulation by evaporation is typically assessed with a separate set of measurements. For a rigid chamber, expansion work needs to be computed separately to define enthalpy values. Experiments using direct calorimetry need to be done carefully to account for evaporation from fountains and fecal pans.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.33. Simplified Representation of Calorimetric Methods. Indirect calorimetry uses the reaction enthalpy for the combustion of carbohydrates (including sugars), fats, protein and other organic compounds using metabolic chemical reactions to determine the equivalent heat transfer. This approach is also shown in Fig. 2.33. In contrast to direct calorimetry, the heat transfer corresponds to metabolic processes within living tissues of animals. The heat transfer leaving animals only equals this rate for negligible change in the body temperature and no tissue gain or production of products like milk or eggs. Indirect calorimetry is more widely used than direct calorimetry and is closely tied to our enthalpy of reaction relationships. Since enthalpy of reaction varies with the source, molar extents of reaction for sugar, fats and other food are needed to determine the net enthalpy of reaction (where measured changes for the reactants are negative). The measured the respiratory quotient (Rq) shown in Fig. 2.33 is useful in partitioning metabolism among the different sources of food. Table 2.3. Summary of Illustrative Metabolic Reactions. ∆
H
kJ mol-1
Source
∆
H
kJ mol-1O2
∆
H
∆
kJ mol-1CO2
H
∆
H
kJ L-1O2
kJ L-1CO2
Rq
-18.9
-18.9
1.00
-17.7
-24.8
0.71
C H O (s, glucose) + 6 O (g) → 6 CO (g) + 6 H O (l) C6H12O6 C H
-2803
-467
-467
O (l, triolein) + 80 O (g) → 57 CO (g) + 52 H O (l)
C57H104O6
-35099
-439
-616
C H N O (s, glyclphenylalanine ) + 13 O (g) → 11CO (g) + 7H O (l) + N (g) C11H14N2O3
-5645
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-434
-513
2-64
-17.5
-20.7
0.84
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
CH CH(OH)COOH (s, lactic acid ) + 3 O (g) → 3CO (g) + 3H O (l) C3H6O3
-1344
-448
-448
-18
-18
1.00
Generic Metabolic Reactions (Walsberg and Hoffmann, 2005) Carbohydrates
-473
-473
-19.1
-19.1
1.00
Lipids (fats)
-444
-624
-17.9
-25.1
0.71
Proteins
-430
-531
-17.3
-21.4
0.81
Let’s further explore the metabolic rate obtained from indirect calorimetric methods by considering the standard enthalpies of reaction for different food sources. Table 2.3 shows chemical reactions and corresponding ∆ 𝐻 ′𝑠 for glucose (sugar), triolein (fat), glyclphenylalanine (peptide used to represent proteins) and lactic acid. The stoichiometric coefficients are defined relative to a value of one for the source (organic reactant). We will begin our analysis by considering the combustion of a single food source (such as a sugar or fat) where the standard enthalpy of reaction and stoichiometric coefficients are as defined in Table 2.3. For indirect calorimetry, changes in the moles of oxygen and/or carbon dioxide are directly measured. From our definition of stoichiometric coefficients (Eq. 2.7.4), the molar extent of reaction can be defined by the measured change in the number of O 2 or CO2 moles (same result for either measurement). The reaction enthalpy of metabolism is then defined as
∆H = ∆ξ ∆ H = −∆n
∆ 𝐻 𝜈
= ∆n
∆ 𝐻 𝜈
2.7.27b
where Δn of reactants are negative values. Instead of using the molar extent of reaction, it is easier to determine the reaction enthalpy by using the direct measurement of the changes in number of moles or in the gas volume of oxygen or carbon dioxide. Alternative forms of the standard enthalpy of reaction are given in the equations below, where molar volume of V mo2 = Vmco2 = 24.8 L mol-1are defined from the ideal gas equation (i.e., RT/P) at standard ambient temperature and pressure (SATP). ∆ H ∆ H ∆ H ∆ H
∆H ∆ 𝐻 = − ∆𝑛 𝜈 ∆H ∆ 𝐻 = = ∆𝑛 𝜈 ∆H −∆n ∆ 𝐻 ∆ 𝐻 = = = −∆𝑉 −∆𝑛 (𝑅𝑇⁄𝑃) 𝑉 ∆H ∆n ∆ 𝐻 ∆ 𝐻 = = = ∆𝑉 ∆𝑛 (𝑅𝑇⁄𝑃 ) 𝑉 =
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2.7.27c 2.7.27d 2.7.27e 2.7.27f
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Values for different food sources obtained from Eqs. 2.7.27c through 2.7.27f are summarized in Table 2.3. You will compute these values as part of the assignments of Homework Problems 2.35 and 2.37. To compute ΔHr for a given food source, you only need to multiply the appropriate table value by the measured changes in number of moles or in gas volumes. Let’s consider the case where the only source of metabolism is the fat triolein. If the measured concentrations is a decrease in O2 of 20 L O2 h-1 (and therefore an increase of CO2 of 14.2 L CO2 h-1), the metabolic rate is simply (-17.7 kJ L-1 O2) (20 L O2 h-1) = -354 kJ h-1. Insight into types of consumed food sources can be obtained by examining the measured respiratory quotient (Rq) defined as R =
∆n −∆n
=
∆ξ ν ∆ξ ν
=
ν 𝜈
2.7.27g
where, once again, the measured Δno2 is negative. The measured Rqobs = Δnco2/(-Δno2) can be compared to the Rq = υco2/υo2 values in the Table 2.3 for different food sources of carbohydrates, fats and proteins. A measured R qobs = 1 corresponds to bio-chemical reactions dominated by carbohydrates; whereas a measured R qobs = 0.85 corresponds to bio-chemical reactions of more than one food source. The use of the measured respiratory quotient to determine the metabolic rate for two different food sources is illustrated in Homework Problem 2.38. This concept is also illustrated in Fig. 2.33. Here the measured R qobs is used to determine the separate molar extents of reaction for sugar and fat. The total reaction enthalpy of metabolism is then computed by adding the reaction enthalpy of each food source. A general relationship for relating metabolic rate to measured oxygen consumption and carbon dioxide production is Brouwer equation (see Nienaber et al., 2009). The most general form of this equation also includes methane production and nitrogen excretion rate. For our purposes, we will only use measured oxygen and carbon dioxide rates. The reaction enthalpy from metabolic processes is then defined from Brouwer equation as ∆ H = −(14.6 V
+ 4.53 V
2.7.28
)
where ΔrH is the effective reaction enthalpy for metabolic processes involving multiple food sources in units of kJ, Vo2 is the consumption volume of O2 in liters at SATP and Vco2 is the production volume of CO2 in liters at SATP. The use of this equation is illustrated in Homework Problem 2.39. Let’s consider the expansion work for the metabolic reactions. From Example Problem 2.5, we can use the following relationship to compute molar expansion work for the reactions in Table 2.3:
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
We are only interested in the stoichiometric coefficient for N 2 for metabolic reactions of proteins and related compounds. For glucose and lactic acid reactions where the respiratory quotient is one, the expansion work term is zero and the change in internal energy equals change in enthalpy. As shown in Homework Problem 2.35 for triolein and glyclphenylalanine, the expansion work term is less than 1% of the standard enthalpy of combustion. We will therefore not make adjustments for expansion work to estimate the change in internal energy for metabolic reactions, that is, 2.7.30
∆ U = ∆ H − (P ∆V⁄∆ξ) ≈ ∆ H Human Metabolic Rate
Values for reaction enthalpies of combustion processes obtained from calorimetric methods are often summarized in relatively simple relationships for design problems. Relationships for the design of pig and turkey facilities are given in Homework Problems 2.42 and 2.43. In this section, we will focus on relationships for human by first considering the basal human metabolic rate. The basal human metabolic rate is defined as the rate of energy expenditures for an undisturbed state in a thermally neutral environment without actively digesting food. For our purposes, an effective resting metabolic rate (corresponding to a relaxed sitting position) of Mifflin-St Jeor can be estimated by using the following basal rate equation (based on data from 498 individuals between 19 and 78)
where the ∆ 𝐻̇ is the effective metabolic rate corresponding to the heat transfer in units of kJ h-1 person-1, m is the mass of the human in kg, ht is height in m, age is person age in years, and Csex is a factor related to gender. It is equal to 5 for males and -161 for females. The metabolic rate increases with physical activities. Approximate estimates for these activities can be obtained by multiplying the metabolic rate of Eq. 2.7.31 by factors given in Table 2.4. The evaporation rate of sweat increases with body temperature. For exercising healthy young males (with a body surface area of approximately 1.85 m 2), the sweat rate increases roughly by 240 g m-2 h-1 for ΔTbody = 1 C. Estimates of the equivalent heat transfer for vaporization are shown in Table 2.3 as a percentage of the negative reaction enthalpy. Vaporization enthalpies vary substantially based on a number of factors. It increases with warmer surroundings. Table 2.4. Increase in Metabolic Rate with Activities Relative to Resting. Activity
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Metabolic Factor
% Evap
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Activity
Metabolic Factor
% Evap
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Studying Moderate Walking Moderate Labor
1.3 1.5 1.6
55% 65% 65%
Dancing Fast Walking Heavy Labor
2.5 3.0 4.0
75% 73% 70%
Example Problem 2.7: Change in Room Air Temperature Problem Statement. Consider a study room with dimensions of 15 m by 20 m with a ceiling height of 3 m. The room is completely insulated (q= 0) and has 10 males and 10 females who are happily studying thermodynamics. The men have an average height of 1.8 m, average mass of 75 kg, and an average age of 20 years. The women have an average height of 1.65 m, average mass of 60 kg, and an average age of 20 years. After studying for one-hour, what is the change in the room’s temperature? What is the total mass of water evaporated from the students? You can estimate the thermodynamic parameters using a room temperature of 25 C.
Thermodynamic Data: For this problem, we will use the following characteristics for air and water (R = 8.314 J K-1 mol-1): Air: mm = 29 g mol-1, Cpm = 29 J K-1 mol-1; ρair = 1.3 kg m-3 25C o H2O: mm = 18 g mol-1, vapHm = 44 kJ mol-1
From Table 2.4: Metabolic Studying Factor: 1.3 Evaporation fraction is the equivalent heat transfer for vaporization (q vap) relative to reaction enthalpy of metabolic processes
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Non-expansion Work Relationships: For this problem, the room is of constant volume and therefore the expansion work is zero. For q=0 and PΔV = 0, the First Law can be written as
From the ideal gas equation, we know that PdV = RT dn for vaporization (see Example Problem 2.4). We can then estimate the molar change in internal energy for Δn moles of vaporization as: ∆
H
=∆
∆
U =∆
R T ∆n =∆ U +RT ∆𝑛 1 kJ 8.314 J x 1000 J (25 + 273.5) 𝐾) = 41.5 𝑘𝐽 𝑚𝑜𝑙 H − RT = (44 − K mol U +
P ∆V =∆ ∆𝑛
U +
As previously discussed, we will neglect the expansion work term for the metabolic reactions ∆ 𝑈 ≈ ∆ 𝐻 and use
The change in internal energy of vaporization is defined from f vap. For our constant volume room, w=0 and we obtain (using the First law) ∆U
=q
+w=q
=f
∆H = −0.55 ∆H = − 0.55 ∆U
where fvap = -0.55 has been used. Solution: The number of moles of air in the room is obtained as
g 1170 kg x1000 m kg n= = = 40,344.8 mol m 29 g mol The reaction enthalpy corresponding to metabolism is defined per student as ∆H
= 1.3 ∆ Ḣ
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∆t = −1.3 0.174 (10 m + 625 ht − 5 age + C
2-69
) (1 h)
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Female students:
Male students: = − 0.226(10 (75 kg) + 625(1.8 𝑚) − 5(20 𝑦) + 5) = − 402.6
∆H
𝑘𝐽 𝑚𝑎𝑙𝑒
These rates correspond to burning an equivalent number of daily food calories of approximately 1770 and 2300 for females and males, respectively. Change in reaction enthalpy and internal energy: ∆U = ∆H = ∆ H
(# female) + ∆ H
(# male)
Change in internal energy from liquid water to vapor: ∆U
= −0.55 ∆U = −(−7126 𝑘𝐽)(0.55) = 3919 𝑘𝐽
Internal energy for related temperature: As previously shown, the 1 st Law can be rearranged as ∆U
= −∆U
− ∆U = −(−7126 𝑘𝐽) − 3919 𝑘𝐽 = 3207 𝑘𝐽
Change in Room Temperature:
∆U ∆T = 𝑛𝐶
1000 𝐽 𝑘𝐽 = (40,345 𝑚𝑜𝑙) 20.686 𝐽 𝐾 3207 𝑘𝐽
𝑚𝑜𝑙
= 3.8 𝐾 = 3.8 𝐶
Mass of water:
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
∆U 3919 𝑘𝐽 = ∆ U 41.5 𝑘𝐽 𝑚𝑜𝑙
∆𝑛
=
∆m
= ∆n
= 94.4 𝑚𝑜𝑙
m = (94.4 mol) 18
g = 1699 g = 1.7 kg mol
The change in temperature is likely too large to maintain a comfortable environment for the students. Either a window needs to be opened or ventilation system is needed for this room. The design of a ventilation system for this problem is given in Example Problem 3.5 in Chapter 3.
HEAT TRANSFER BY RADIATON Basic Concepts Blackbody To gain insight into the nature of radiation, Kirchhoff conceived of a blackbody that sends out radiation at all frequencies and absorbs all radiations (hence black because it reflects none). Black bodies were used experimentally to study radiation by using soot-covered interior walls to form a blackened cavity. The hole of the cavity served as a surrogate of a black body. Black bodies are similar to ideal gases. These idealized substances provide a useful framework for deriving theoretical relationships and are often an adequate approximation to real material.
Figure 2.34. Black Body as Idealized Source of Radiation. Radiation is typically represented using radiant flux density or radiant power per unit area, defined as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
which is a function of the temperature of the body To determine the radiant power we need to integrate over the appropriate area: 𝑞̇
= ∫ 𝑅̇ 𝑑𝐴
2.8.2
where 𝑞̇ is the radiant power for any arbitrary body. To determine the radiant heat transfer (not net) for body i: ∆
𝑞
=
𝑞̇
2.8.3
𝑑𝑡
The net heat transfer requires a balance between the incoming and outgoing radiation from the body. Irradiance is as the radiant flux density incident (upon) on a surface arriving from all directions (J m-2 s-1 = W m-2). Radiant emittance is the radiant flux density leaving a surface Early Theoretical Framework By the late 19th century, there was general consensus that radiation was caused by the vibrations of molecules and atoms of solids but the proper theoretical framework was not yet developed. A simple harmonic oscillator was proposed to represent the vibration of electrons. We will introduce this topic by considering continuously vibrated springs. As discussed later in the chapter, this approach needed to be modified to account for quantum mechanics. Let’s first review the notation of a simple spring shown below. x negative
x positive
Relax State
x =0 Fx = 0
Fx negative
x
x
Fx positive
x
Figure 2.35a. Position Dependent Forces of Springs. Clearly the force is negative after the spring is extended from the relaxed and is positive if the spring is compressed. We will limit our discussion to the magnitude for a vector acting in the
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
x-direction only. Force for a spring varies with its position with x as defined by Hooke’s law, or, F
2.8.4a3
= −k x
where Fsp is the spring force acting on the block and ksp is the spring constant (units of mass per time2). We now expand our analysis for a continuously vibrating spring. For illustration purposes, we will rotate our spring so that movement is in the vertical direction. Our idealized system has no gravity or friction forces. As shown below, our continuous vibrating spring has different locations with time. We start at its relax state of z = 0. The maximum and minimum values of z correspond to z = zmax and z = -zmin. As shown in the right-sided figure, the spring will clearly return to these locations as it continuously vibrates. The time required for it to repeat a cycle is related to the frequency of the spring.
Figure 2.35b. Continuously Vibrating Spring. Since force is the product of mass and acceleration, our spring force can then be written as
where m is the oscillating mass, z is the displacement distance shown in the above figure and ksp is the spring constant. The above equation can be rearranged into the following secondorder linear ordinary differential equation with constant coefficients k d z + dt m
2.8.4c
z =0
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
A good course on ordinary differential equation will cover the general solution to this type of equation. For our purposes, we will simply suggest a solution of the form of z = a sin (2πbt), where “a” and “b” are unknown coefficients and t is time. This is certainly a reasonable suggestion given that our continuously vibrating spring periodically returns to the same vertical positions. By trying this relationship, the spring equation can be evaluated as −a (2πb) sin( 2π b t) +
k m
a sin( 2πb t) = 0
2.8.4d
We conclude that our suggested function is a valid solution if we rearrange term and define our frequency response (υw) as b=ν =
1 2π
k m
2.8.4e
When t = 1/(4υw), we obtain a sine value of one and minus one for t =3/(4υ w) We conclude that “a” is the amplitude defined as 𝒜 = zmax = -zmin. Our equation for simple harmonic motion is then defined as z=𝒜
2.8.4f
sin( 2π υ t)
Graphical solution is shown below. The period is defined as λ = 1/υ w and corresponds to the time interval for the mass returns to the relaxed state (and other locations).
Figure 2.35c. Position Locations for Vibrating Springs. From the Maxwell’s elegant electromagnetic theory, the wavelength and frequency (defined as the number of times per second the value at a fixed point repeats) of electromagnetic waves of radiation are related as 2.8.4g
c=λυ
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where c is the speed of light (299,792,458 m s-1 ≈ 300,000,000 m s-1), λ is the wave length [L], and υw is the frequency [cycles per time - Hz]. As easily shown with prisms, light is composed of electro-magnetic waves of different wavelengths. The range of wavelengths of radiation for a blackbody is shown below. The wavelength varies from a length much smaller than the diameter of a hydrogen atom to a length much larger than a football field. Wavelengths for the visible spectrum are much longer than gamma waves or X rays, but they are still quite small, approximately 1000 times smaller than the typical width of human hair.
Figure 2.35d. Range in Wavelengths for Blackbodies. Many biological and environmental processes are dependent on the energy associated with a particular range of wavelengths. For these processes, it is useful to define
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
where the first term in the brackets is loosely the number of waves per unit volume of a blackbody cavity per wavelength between frequencies of λ and λ+dλ (with adjustments to obtain flux units), and the second bracket term is the average energy per wave. In Chapter 1, we discussed the equipartition of energy corresponding to RT/2 per mole for each degree of freedom, which is equivalent to kBT/2 per molecule for each degree of freedoms. For a simple one-dimensional harmonic oscillator with two degrees of freedom corresponding to a “spring” for each direction, the average energy per wave is: ξ̅ = k T
2.8.4h
This approach closely matched experimental data for large wavelengths but poorly represented radiation for small wave lengths. This so-called ultraviolet catastrophe was successfully resolved using the concepts from quantum mechanics. These results are given later in the chapter. Spectral Flux Density Let’s now expand upon on radiation concepts introduced with vibrating springs. We have previously discussed that black-body radiation can be represented by electro-magnetic waves with wavelengths varying from very small (smaller than diameter of hydrogen atoms) to very large (longer than football fields). The radiation energy corresponding to individual wavelengths is important in understanding photosynthesis, climate change and other biological and environmental processes. Spectral flux density is used to represent the radiant flux density per wavelength. Insight into the definition of spectral flux density can be obtained by cumulating the energy for increasing wavelengths. We can start by using zero energy if there is no wavelength (i.e., λ equals zero). We can add the contribution to radiant energy by computing the energy for each Δλ as we move to larger wavelengths. We will use the symbol of 𝑅 (𝜆) for this cumulative radiant energy. This concept is shown in Figure 2.36 where the range in wavelengths varies between gamma rays to long radio waves. Let’s examine the slope of a point on the curve. This slope is the spectral flux density and is defined mathematically as
Spectral flux density, 𝑅̇ , is defined such that of 𝑅̇ (𝜆)𝑑 𝜆 is the radiant power per unit area between the wavelength of and +d. Radiant flux density is then obtained by integrating over all wavelengths.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.36. Illustration of Spectral Density Function. We are interested in developing a theory that can be used to determine the spectral flux density for a black body. A good starting point was the definition of the energy per wave previously given where the average energy per wave was initially estimated as k BT. However, Planck drastically improved the representation by assuming that energy is not continuous, but has a discrete set of values defined as ξ =𝜄ℎ 𝜐
2.8.6
𝜄 = 0, 1, 2, …
where ξp is the energy, ι is a positive integer called the quantum number, and h P is Planck’s constant (= 6.63x10-34 J s). Einstein realized that radiation is emitted or absorbed when the electron moves between different energy levels of e2 and e1, that is, hPυw = e2-e1. Photons are released by the movement of electrons between these different levels. A graphical illustration of Einstein’s concept for one form of photon generation is shown below. Also shown below is the extension of this concept within the framework of Bohr’s model of electron orbits.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Before Excited State
During
After n=3
e2
n=2 n=1 photon=hυ
Incident photon=hυ
∆e=hυ
photon=hυ
e1 Ground State
Einstein’s Approach
Bohr’s Simple Atom
Figure 2.37. Illustration of Discrete Energy Levels. For the quantum representation of energy and by using the relationship between speed of light ( c ), frequency (υw) and wavelength (λ) of c = υwλ, the average energy for each frequency is defined as
By using this relationship for the frequency energy, the spectral flux density can be defined as
𝑅̇
=
ℎ 𝑐 ⁄𝜆 ℎ 𝑐 exp −1 𝜆𝑘 𝑇
2𝜋𝑐 𝜆
=
2𝜋ℎ 𝑐 𝜆
1
2.8.8
ℎ 𝑐 exp −1 𝜆𝑘 𝑇
For hP = 6.62607x10-34 J s, c = 299,792,458 m s-1 and kB = 1.38066 x 10-23 J K-1, we can simplify as
C1 2 h P c2 3.742 x 1016 J m2s 1 3.742 x 108 W m4 m2 C2
hP c 1.4388 x 102 m K 1.4388 x 104 m K kB
2.8.9 2.8.10
which simplifies the equation as function of the wavelength and temperature as 𝑅̇
=
1 𝜆
𝐶 𝐶 exp −1 𝜆𝑇
2.8.11
Solar and Terrestrial Radiation
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
120
60
Solar (6000 K) Photosynthetically Active Radiation (0.4 to 0.7 m)
80
40
Earth (288 K)
40
20
0 0.1
Ultraviolet (< 0.4 m)
10 Wavelength (m)
1
Violet (0.40-0.43m)
Green (0.49-0.56m)
Blue (0.43-0.49m)
Orange (0.59-0.63m)
0
Earth Spectral Flux Density (W/m2/μm)
Solar Spectral Flux Density (MW/m2/μm)
The spectral flux density for the temperature of the sun (T = 6000 K) and for the temperature of the earth (T = 288 K) are shown below.
100
Infrared (> 0.76 m)
Yellow Red (560 - 590m) (0.63-0.76m)
Figure 2.38. Spectral Flux Densities for Temperatures of the Sun and Earth Most of the solar radiation corresponds to relatively short wavelengths (less than 4 m) and most of the terrestrial radiation corresponds to relatively long wavelengths (greater than 4 m). Note that the scale for the earth is considerably smaller. We are interested in determining the total energy as a function of temperature. We are also interested in the amount of solar radiation reaching the earth. In the above figure, the range of wavelengths used for photosynthesis is shown. This component is called photosynthetically active radiation (PAR). Many animals have evolved such that their eyes see those frequencies where sunlight has its maximum intensity. Radiant Flux Density Blackbody radiation We are now ready to derive the radiant flux density. From the spectral flux density, the radiant flux density can be defined by integrating over all frequencies, that is
If we use the spectral flux density relationship developed for a black body in the previous section, we obtain
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑅̇ =
𝑅̇
𝑑𝜆 = 2 𝜋 ℎ 𝑐
1 𝜆
1 ℎ 𝑐 exp −1 𝜆𝑘 𝑇
2.8.15
𝑑𝜆
where the subscript “b” has been used to indicate radiant flux density for a black body. The integration is easier if we define an integration variable of y = h Pc/(λkBT) and dλ = [hPc/(kBTy2)] dy. We then obtain 𝑅̇ =
𝑅̇
𝑑𝜆 =
2𝜋𝑘 𝑇 ℎ 𝑐
𝑦 2𝜋 𝑘 𝑑𝜆 = 𝑒 −1 15 ℎ 𝑐
2.8.16
T
where the integral can be evaluated as (π4/15). Since all of the terms between parentheses are constants, we will lump them together into the Stefan-Boltzmann constant of σ = 2π5kB4/(15hP3c2) = 5.6704 x 10-8 J m-2 K-4 s-1 to define our radiant flux density as
where 𝑅̇ is the radiant flux density of a black body and T is the absolute-scale temperature. The emittance from the Sun corresponds to a temperature of approximately 6000 K. The radiant density flux for a black body at the temperature of the sun is defined as 𝑅̇ (𝑇 = 6000 𝐾) = 𝜎 T = 5.6704x10
J (6000 K) = 7.35x10 J m s m K s
= 7.35x10 W m and for the Earth at a temperature of approximately 288 K 𝑅̇ (𝑇 = 288 𝐾) = 𝜎 T = 5.6704x10
J (288 K) = 390 J m s m K s
= 390 W m
Gray bodies The radiant flux density for a non-black body is done using the emissivity defined as
and therefore the radiant flux density is defined as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
𝑅̇ = 𝜀 𝑅̇ = 𝜀 𝜎 𝑇
2.8.19
In general, emissivity is a function of temperature, emission angle and wavelength. Dependence on emission angle and wavelengths is discussed in greater detail in Chapter 3. Gray bodies are defined for substances where the emissivity is constant for all wavelengths. This is the assumption used in this chapter. Reported values for emissivity are given in Table 2.5. Table 2.5. Total Emissivity for Selected Substances. Surface Clay Concrete Wood Polished Stainless Steel Polished Silver Plaster Aluminum Paints
ε 0.91 0.94 0.85 0.13 0.05 0.91 0.52
Surface Ice Liquid water Snow Clouds Dry soil Typical plant values Typical clear skies
ε 0.97 0.96 0.82 0.98 0.90– 0.95 0.90 - 0.99 0.60 - 0.75
Radiation Properties Measured Solar and Terrestrial Radiation Radiation data have been measured by satellites at the edge of the atmosphere and by instruments located at the ground surface. An example of these measurements is shown below. Let’s first discuss the spectral flux density measured at the top of the atmosphere for the irradiance from the Sun. These values are well approximated by the spectral flux densities corresponding to a T= 5900 K, adjusted for the distance between the Sun and Earth. Also shown in the figure is the solar radiation measured at the ground surface. These values are smaller because of (1) the reflection of solar radiation back into space and (2) the absorption of radiation by atmospheric gases as it propagates to the Earth’s surface. Additional details of the absorption of gases are shown in the lower figure. Different gases absorb radiation at different wavelengths. For example, O 2 and O3 (ozone) absorb radiation at wavelengths corresponding to solar radiation. These gases reduce the energy from potentially dangerous X-rays and other small-wavelength radiation. Carbon dioxide and water vapor absorb radiation at wavelengths more common for terrestrial radiation. This impact can be seen by the observed radiation leaving the top of the atmosphere. As shown by the right-sided graph, the spectral flux density for a temperature corresponding to Earth is well approximated by the observed values for wavelengths with little absorptivity by CO2 or H2O.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Observed Incoming Solar
Spectral Flux Density (W m-2 μm-1)
2500
40
Atm Window
Predicted, T = 5900 K
2000
30 Predicted T = 294 K
1500
Observed Ground 20
1000 10
500
0
0 PAR
Absorptivity (%)
Spectral Flux Density (W m-2 μm-1)
Bruce N. Wilson
1
Wavelength (μm)
10
Observed Outgoing100 Terrestrial
100 50 0
O2
O3
O2
H2 O
CO2
H2 O
CO2
H2 O
Figure 2.42. Observed and Predicted Solar and Terrestrial Radiation Based on these observations, we will discuss the following topics related to radiation:
Interactions with Incident Radiation Let’s consider the response between radiation and matter as shown below. We have already discussed the emittance of radiation using Stefan-Boltzmann radiation laws. The radiation incident on the surface can either by reflected, absorbed or transmitted. These properties are shown in Fig. 2.43. In this chapter, we will consider the total reflectivity, absorptivity and transmissivity for all wavelengths. Our more general formulation given in Chapter 3 includes wavelength and directional dependence on reflectivity, absorptivity, and transmissivity.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Absorptivity [fa]: Fraction of the incident flux density absorbed by the material. It results in heat transfer to the material. The absorptivity over the entire range of wavelengths can be written as
Transmissivity [ft]: Fraction of the incident flux density transmitted by the material. The transmissivity over the entire range of wavelengths can be written as 𝑓 =
𝑅̇ 𝑅̇
2.8.21
Reflectivity [fr]: Fraction of the incident flux density reflected by the material. The fraction of solar radiation reflected from a material (especially at the Earth’s surface) is also called albedo. The reflectivity over the entire range of wavelengths can be written as 𝑓 =
𝑅̇ 𝑅̇
2.8.22
More Discussion Since the incident radiation has only three possible pathways, we conclude that 2.8.23
𝑓 +𝑓 +𝑓 =1
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Values for absorptivity and reflectivity are commonly available in tables and charts. The transmissivity value can then be determined as
For a black body, all of the incident radiation is absorbed, and therefore
Kirchhoff’s law Kirchhoff’s law states that for a system in thermodynamic equilibrium, the emissivity equals the absorptivity, that is,
which, of course, for a black body both the emissivity and absorptivity equal one. Insight into this law can be obtained by considering thermodynamic equilibrium in a medium of uniform temperature and where the incident radiation is from a black body. At equilibrium, the emitted radiation equals that absorbed. We then have 𝜀 𝜎𝑇 = 𝑓 𝑅̇
2.8.27
=𝑓 (𝜎𝑇 )
For uniform temperatures, the emittance temperature (T e) equals the temperature of the incoming radiation (Ti). For our conditions, we then conclude that ε = f a. Although Kirchoff’s law is derived for equilibrium conditions, it widely used as an approximation for other conditions. More rigorous discussion of Kirchhoff’s law is given in most textbooks on radiation including Chapter 5 of Thomas and Stammes (1999). Absorptivity values Absorptivity is tied to the energy levels of molecules as determined by their chemical structure. These energy levels for solids or liquids allow absorption (and emissions) of radiation as a nearly continuous process with wavelengths. Absorptivity of gases is tied more closely to wavelengths. Gas absorptivities (and emissions) can be relatively large for some
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Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
wavelengths and non-existent for others. In general, polyatomic gases have vibrational and rotational energy states and therefore have more interesting absorption characteristics. Absorptivity of gases has been shown in our previous figure. It is discussed in greater detail in Chapter 3. Reflectivity values Some measured values of reflectivity for terrestrial surface are shown below. Table 2.6. Reflectivity for Different Surfaces. Cover Clean snow Black soil, dry Black soil, moist Grey soil, dry Grey soil, moist Sand Tops of oak
fr 0.90 0.14 0.08 0.26 0.11 0.35 0.19
Cover Alfalfa Green grass Cotton Water bodies (0-30 lat) – year Water bodies (30-60 lat) Dec Water bodies (30-60 lat) Jun Water bodies (30-60 lat) year
fr 0.23 0.26 0.21 0.05 0.17 0.05 0.08
The reflectivity of water varies with wave activities, turbidity, and other water body characteristics. One of the most important factors is the angle of the rays striking the water. For an angle of 5 degrees, the reflectivity is as high as 90%; whereas the reflectivity is as low as 2% when the angle is 60 degrees. The values in the table are from Cogley (1979). For latitudes closer to the equator, the monthly variability in f r is small. Reflectivity values and variability with month are greater for the larger latitudes.
RADIATATIVE BALANCE FOR THERMAL EQUILIBRIUM Earth’s Orbit around the Sun The most important source of energy for us is the radiation from the Sun. The average solar radiant flux density at the edge of the Earth’s atmosphere is commonly called the solar constant. Determining the solar constant requires knowledge of the Earth’s orbit. A schematic of this orbit is shown below. The Earth’s elliptical orbit is nearly circular with an eccentricity of 0.017. Along the main axis, the distance the Earth is closest to the sun is a distance of approximately 147 million km and the farthest distance is 152 million km. The dates corresponding to these distances vary slightly with year.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Eccentricity 1
( Minor ) 2 0.017 (Major ) 2
147 Million km
152 Million km
Aphelion ~ July 3
Sun
Perihelion ~ January 3
Earth
Figure 2.44. Orbit of Earth around the Sun. The Earth is tilted at an angle of 23.45 degrees. This tilt plays an important role in the distribution of solar radiation on the surface of the Earth. We will first explore the implications of the radii of the Earth and Sun and between them on the amount of solar energy. The distribution of solar radiation on Earth is given in Appendix 2-A. Radiation and Thermal Equilibrium Relationships Extraterrestrial Radiation A schematic illustrating key concepts used to develop relationships for establishing thermal equilibrium using radiative heat transfer for Earth are shown below.
Figure 2.45. Key Concepts for Thermal Equilibrium.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Let’s start by computing the total solar radiant power (energy per time) using the radius of the Sun. Since the surface area of a sphere at that radius is A s,sun = 4π rs2, we obtain 𝑞̇
,
= 𝑅̇
𝐴
,
= (𝜀 𝜎𝑇 )(4 𝜋 𝑟 )
2.9.1
The solar constant is the solar radiant flux density to reach earth. It can be computed using the total radiant power over a surface area corresponding to a radius equal to the mean distance between the Earth and the Sun (As,s-e = 4πr2s-e). The solar constant for the mean orbit distance is then obtained as
Let’s now consider the net radiant flux density absorbed by the Earth (and atmosphere). Net radiation is equal to the incoming radiation (solar constant) minus the radiation reflected back into the atmosphere. The net radiant power absorbed by the Earth corresponds to the product of the net radiant flux density and the projected area of the Earth. The projected area is the area of a circle for the Earth’s radius. The net radiant power (𝑞̇ , ) is then obtained as 𝑞̇
,
= 𝑆̅ (1 − 𝑓 )(𝜋 𝑟 )
2.9.3
The global radiant flux density is defined as the average solar radiant energy per area of the Earth’s surface. This form is useful in determining the solar energy available for use by plants, evaporation and melting, and solar heating systems. To determine global flux density, we need to divide the net radiant power by the surface area of Earth (As,earth = 4πre2). We then obtain 𝑅̇ where 𝑅̇
=
𝑞̇ , 𝑆̅ (1 − 𝑓 )(𝜋 𝑟 ) 𝑆̅ (1 − 𝑓 ) = = 4𝜋𝑟 4πr 4
2.9.4
is the global radiant flux density.
Thermal Equilibrium for Non-absorbing Atmospheric Gases After a sufficiently long time, the Earth should reach a thermal equilibrium with the radiation from the sun. At thermal equilibrium, the heat transfer entering the Earth by solar radiation is balanced by the radiation heat transfer leaving by corresponding to the Earth’s (including its atmosphere) emittance. From the First Law’s perspective, we have
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Where the net rate of change in internal energy is zero, and since the pressure at the top of the atmosphere is zero, the rate of expansion work is also zero. The heat transfer is divided into the radiant power entering the Earth from the Sun (𝑞̇ , as defined by Eq. 2.9.3), radiant power leaving from Earth’s emittance (𝑞̇ , = 𝑅̇ 𝐴, ) and other sources of heat transfer (𝑞̇ ) that are equal to zero for thermal equilibrium. As shown in Homework Problem 2.52, 𝑞̇ ≈ 0 is a valid condition of Venus, Mars, and Earth, but not for that of Jupiter. Thermal equilibrium conditions then correspond to
For non-absorbing atmospheric gases, the radiant power from Earth’s emittance is defined by 𝑞̇ , = 𝑅̇ 𝐴, . By also using Eq. 2.9.3, we obtain (𝜀 𝜎 𝑇 )(4 𝜋 𝑟 ) = 𝑆̅ (1 − 𝑓 )(𝜋 𝑟 ) = 𝜀 𝜎𝑇
𝑟 𝑟
(1 − 𝑓 )(𝜋 𝑟 )
2.9.7
where Te is the temperature of the Earth’s surface. We can then solve for this temperature at thermal equilibrium as
T =T
r 2r
ε (1 − f ) ε
2.9.8
Example Problem 2.9: Thermal Equilibrium and Earth’s Temperature Problem Statement. For approximately spherical bodies of Earth and the sun, estimate the (1) solar constant, (2) the temperature of the earth at thermal equilibrium, and (3) the global radiant flux density for heating the earth and atmosphere. We will use the following characteristics: Sun radius = rs = 6.96 x 108 m
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Distance between Sun and Earth = rs-e = 1.496 x 1011 m Sun temperature = Ts = 5778 K Emissivity of sun = εs = 1.0 (excellent approximation) Emissivity of earth = εe = 1.0 (good approximation for wavelengths in infrared range) Reflectivity of the atmosphere is approximately 21% and Earth’s surface is 9% Total reflectivity of atmosphere and Earth’s surface: fr = 0.3
Earth
Sun
εe =1 fr = 0.3 Pi T = 5778 K εs = 1
T? Mean Distance = 1.496
x1011
m
re = 6.37x106 m
rs = 6.96x108 m
Solution for Extraterrestrial Radiation. The solar constant for the mean orbit distance is
S = (1)
(6.96 𝑥 10 𝑚) 𝑊 (5778 𝐾) ( ) (1.496 𝑥 10 𝑚) 𝑚 𝐾 = 4.92 MJ m h
5.6704 𝑥 10
S = 1368 W m
We then obtain the following estimate of the global radiant flux density for the atmosphere and surface of fr = 0.3, as 𝑅̇
=
𝑆̅ (1 − 𝑓 ) 1368 (1 − 0.3) = = 240 W m 4 4
The irradiance global radiant flux density is simply (1368/4) = 342 W m -2. Solution: Average temperature of Earth. The average temperature of Earth for thermal equilibrium using radiative heat transfer is
T =T
r 2r
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ε (1 − f ) ε
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
T = (5778 K)
6.96 𝑥 10 𝑚 2 (1.496 x 10 𝑚)
(1) (1 − f ) 1
Discussion The measured temperature of the earth is approximately 288 K (15 C). A temperature of -18 C computed in the example problem is too cold to support our most important biotic communities. The absorption of terrestrial radiation by the so-called greenhouse gas are why our temperature is warmer than the equilibrium temperature. The temperature of the Earth surface needs to increase to 288 K to achieve an outgoing radiation equal to that of incoming global radiant flux density of 240 W m-2 because of the role of greenhouse gases. Thermal equilibrium conditions are shown in Fig. 2.46 using the simple thermal equilibrium of this problem (without considering absorptivity of atmospheric gases) and with absorptivity of these gases. Equilibrium conditions of the left-sided figure correspond to those values of the example problem. Thirty percent of the solar radiation is reflected. The transmitted radiation is balanced by the emitted terrestrial radiation at a temperature of 254 K. The actual thermal equilibrium conditions with absorbing atmospheric gases are shown in the right-sided figure. Conditions are more complicated with these gases, but the incoming and outgoing radiation are still equal to each other (342 W m -2). Only 50% of the solar radiation is transmitted to the Earth’s surface, that is, 20% of the incoming solar radiation is absorbed by the atmospheric gases. For an Earth’s temperature of 288 K, the emitted radiation is 390 W m-2. Only 5% (approximately) of this radiation is transmitted to outer space, the rest in absorbed by the atmospheric gases. An additional 103 W m -2 of energy is absorbed by these gas from convection and phase change processes. The total emitted radiation from the atmospheric gases (and clouds) is 542 W m-2. A significant fraction of this radiation is absorbed by Earth. Understanding the impact of increasing the greenhouse gases requires greater knowledge of the structure of the atmosphere and the selective absorptivity of wavelengths for these gases. These concepts are discussed in Chapter 3.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Figure 2.46. Thermal Equilibrium Without and With Absorptivity of Atmospheric Gases.
THERMODYNAMIC PROCESSES OF PLANT CANOPIES Photosynthesis and Transpiration Photosynthesis and transpiration within plant canopies are vital processes for life. Both are largely driven by solar radiation. Photosynthesis is the process that converts solar radiation and inorganic raw materials into organic compounds. It provide a mechanism for the conversion of solar energy into forms available for metabolic activities of animals. Transpiration is the process by which water is evaporated from the air spaces in plant leaves. It is an important component of the water cycle. Approximately 70% of U.S. precipitation is returned to the hydrologic cycle by evaporation from lakes, rivers and soil and by transpiration from plants. It is critically important in the design of irrigation systems, in the analysis of water supply for municipalities and in maintaining flow rates and aquatic communities in streams and rivers. The simplest form of photosynthesis can be written as
where C6H12O6 is simple sugar (glucose or fructose). The standard enthalpy of reaction for the photosynthesis of glucose can be written as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
∆
H = ν
∆ 𝐻 , − ν ∆ 𝐻
,
+ν ∆ 𝐻 +ν ∆ 𝐻 ,
,
2.10.1
= 2803 kJ mol
which is equal to that previously given for the oxidation of glucose (see Table 2.3) for a reversal of the products and reactants compounds. The standard enthalpy for this photosynthetic reactions is therefore equal to a positive 2803 kJ mol -1 (heat transfer into a constant pressure system). The critical role of solar radiation in the amount of biomass in plants is illustrated in Fig. 2.47. Cumulative biomass for eleven different agronomic plants is shown to vary linearly with cumulative photosynthetic radiation. These results are for crops that has sufficient water and nutrients, are not limited by competition from weeds or by other pests, and have temperatures suitable for plant growth. For these conditions, the daily biomass before plant maturity can be estimated as ∆B
,
= 2.83 f
𝑞
,
2.10.2
,
where ΔBm,A is the daily production of dry mass per area for the canopy in units of g m -2, qsun,in,A is the solar radiation per unit area reaching the leaves within the canopy in units of MJ m-2, and fpar is the fraction of the solar radiation (absorbed by plants)that is available for photosynthesis. For canopy where the surface area of leaves (one sided) is three times greater that the soil area, fpar is approximately 0.45.
Figure 2.47. Relationships between Biomass and Photosynthetic Radiation. Additional insight into photosynthesis and transpiration can be obtained by considering the processes within the leaves in greater detail. A simplified schematic of plant leaves is shown in Fig. 2.47. The epidermis is the outer, protective cell which secretes a waxy cutin. Most of the space between the upper and lower epidermis consists of thin-walled cells called mesophyll that are full of chloroplasts and are the primary site of photosynthesis. Water and nutrients are delivered from the soil to the plant leaves by conducting tissue called xylem.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Scattered over the epidermis are many small openings called stomata. There approximately 5000 stomata per cm2 for a corn leaf. These openings are essential for movement of water vapor (transpiration), oxygen, and carbon dioxide. The size of the openings is controlled by guard cells. The guard cells expand or contract based on turgor pressure resulting in smaller openings for periods of no photosynthesis (at night), and when the plant experiences water stress because of inadequate supply of water delivered from the roots. The evaporation of water within plant leaves and the movement of water vapor through the stomata is called transpiration. This process is fundamentally related to the First Law. O2, H2O
CO2
Upper Epidermis
Xylem (H2O)
Stomata
CO2
Mesophyll (Chloroplasts)
Lower Epidermis
O2, H2O CO2
CO2
Stomata O2, H2O
Transpiration (Vaporization)
Figure 2.48. Components of a Plant Leaf.
Figure 2.49. Energy Components for Transpiration Processes. Transpiration and the First Law
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
System Definition The First Law will be applied to the plant canopy shown in Fig. 2.49. The atmospheric pressure is effectively constant during transpiration. Total heat transfer (q t) is divided into a component from convection and conduction from the canopy to the atmosphere and soil (q) and radiation (qrad) from solar and terrestrial sources. The First Law of thermodynamics can then be written as
where PΔV is expansion work (constant pressure) and wo = 0 is the other non-expansion work. We can rearrange this equation using the definition of enthalpy for constant pressure as ∆U + P∆V = ∆H = ∆H
+ ∆H
+ ∆H = q + q
2.10.4
where ΔH is the total change in enthalpy, and it generally includes changes as the result of a change in temperature (ΔHtpr), change in phase transitions (ΔHvap) and the change from chemical reactions (ΔHr). We can rewrite the First Law as
Use of Bowen Ratio We will use the simple Bowen ratio to assist us in our calculations. The Bowen ratio is defined as the ratio of heat transfer by conduction and convection (q) out of the system (to the surroundings of air) and the energy released by vaporization (ΔHvap):
Bowen ratio is dependent on many factors. In the Example Problem 2.9 (and the homework problem), its value will be given to you. By using a known Bowen ratio, the convective/conductive heat transfer can be written as 2.10.7
𝑞 = −𝛽 ∆𝐻 and the First Law can then be evaluated as
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
We can then solve for the enthalpy used to evaporate (vaporize) in the plant canopy as ∆H
=
q
− ∆H − ∆𝐻 1+β
2.10.9
which has units of energy (kJ). Example Problem 2.10: Transpiration Depth from Plant Canopies Problem Statement We are interested in the transpiration from a well-watered alfalfa field where the plant canopy represents the system, and the air and the soil represents the surroundings. The field is located in Waseca, MN. Let’s consider our system for a clear day on July 15 for a particular year where the average air temperature during transpiration is 25 C and the corresponding average canopy temperature is 28 C. Heat transfer to the soil and radiant heat transfer during non-daylight periods are assumed to be negligible. We wish to compute (1) the net daily radiation energy entering the system, and (2) the daily transpiration depth. Selection of parameter values Key parameters are summarized in the figure below. They are: 28vapCHom = 43.7 kJ mol-1 at 28 C, mm = 18 g mol-1 Stefan-Boltzmann Constant = 5.67 x10-8 J m-2 K-4 s-1, Density (liquid water) = 996 kg m-3 Emissivity of alfalfa (plants): εp = 0.95, Emissivity of air: εa = 0.75 Reflectivity of sky: fra = 0.10, Absorptivity of sky: fa= 0.15 Reflectivity of alfalfa (plants): frp = 0.20, Bowen ratio for well-watered plants: β = 0.30 Representative area in the middle of canopy: Ap = 100 m2 Canopy height = 1.25 m, Atmospheric pressure = 101 kPa Molar heat capacity at constant pressure for air: Cpm,a = 29.15 J K-1 mol-1 Change in temperature of air in canopy = 4 C Fraction of solar radiation in plant canopy available for photosynthesis: f par = 0.45 Photosynthesis represented by glucose reaction shown below: 6CO + 6 H O ⎯⎯⎯⎯⎯⎯
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C 𝐻 𝑂 + 6𝑂 ; ∆
2-95
H = 2803 𝑘𝐽 𝑚𝑜𝑙
; 𝑚 = 180.16 𝑔 𝑚𝑜𝑙
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Results from Appendix 2-A In Appendix 2-A, we have computed the day length for July 15 for Waseca, Minnesota as ∆t = 15.01 h and the solar irradiance at the top of the atmosphere for the day of July 15 at the latitude of Waseca as q
,
= 4.194x 10 J m
= 41941 kJ m
Net solar radiation Transmissivity can be defined using the reflectivity of the atmosphere (f ra) and the absorptivity of the atmosphere (faa) as
and therefore the solar radiation on the system is q
,
,
=f q
= (0.75)41941 = 31456 kJ m
,
The solar energy entering the system of a plant canopy can be computed using the reflectivity of the plant (frp) as q
, ,
= 1−f
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q
,
,
= (1 − 0.2)31456 = 25165 kJ m
2-96
= 25.16 MJ m
Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Net terrestrial radiation Net radiant flux density from the plant canopy and the surrounding air (terrestrial) can be computed from the emissivities and temperatures. This corresponds to the heat transfer by radiation, or
For the values of this problem Ta = 25 + 273.15 = 298.15 K and Tp = 28 + 273.15 = 301.15 K, Ṙ
= 5.67 x 10
,
J [ 0.75(298.15) − 0.95(301.15) ] = −107 J m s m K s
To estimate the net terrestrial energy during the daylight hours, we use the number of daylight hours computed in the previous section. We then obtain q
,
,
= Ṙ
,
∆t
= −107 (15.01 h)
3600 s h
kJ = −5782 kJ m 1000 J
where the radiant heat transfer during the non-daylight periods is neglected. Net Solar and Terrestrial Radiation The net energy flow into the system is equal to the sum of the solar and terrestrial energies. This is the heat transfer by radiation. We obtain
Change in Enthalpy with Temperature Within the plant canopy, temperature of the air and the plants (mostly water by mass) are typically different. In the morning, the air is typically cooler than the plants but in the late afternoon the air temperature is often warmer than that of the plants. For our purposes, the volume of the liquid water in plants and the change in its temperature during the transpiration are negligible. We will then focus on the change in enthalpy of the canopy air. The number of moles of this air can be estimated from the ideal gas equation. For a temperature of T = 28
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
C = 301.15 K, a pressure of P = 101 kPa, and a volume of V = 100 m 2 (1.25 m) = 125 m3, we obtain
By using ΔT = 4 C = 4 K, the change in enthalpy of the canopy by temperature change is ∆H
∆T = (5042 mol) 29.15
≈nC
J 1 kJ (4 K) = 587.9 kJ mol K 1000 J
Change in Enthalpy with Photosynthetic Reactions The biomass obtained by photosynthesis can be estimated using solar radiation reaching the plants within the canopy using
which corresponds to an equivalent number of moles of glucose of ∆n
=
(3207 g m ) 100 m = 17.8 mol 180.16 𝑔 𝑚𝑜𝑙
Since the stoichiometric coefficient for glucose, we conclude that ∆𝜉 = ∆𝑛 23.7 𝑚𝑜𝑙. By standard enthalpy for this photosynthetic reaction, we obtain
⁄𝜈
=
Change in Enthalpy of Vaporization of Plant Canopy As previously derived, we can determine the enthalpy for transpiration as ∆H
=
q
− ∆H − ∆𝐻 1938300 − 588 − 49,850 = = 1,452,000 kJ 1+β 1.3
We conclude from the First Law of thermodynamics that the 1,439,000 kJ of energy is used to evaporate water (and provide for expansion work) by transpiration from the plant canopy of 100 m2. The change in enthalpy by temperature difference is clearly a small component of the
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
energy balance. If there is insufficient water for this evaporation, then the Bowen ratio would be larger. Mass of Evaporated Water and Transpiration Depth To compute the mass of vaporization, we need to determine the specific enthalpy of vaporization at 28 C. Specific enthalpy can be obtained by dividing the molar enthalpy by the molar mass. Therefore we obtain
where the standard enthalpy of vaporization is used for our pressure of 101 kPa. The mass of water vaporized per unit area is defined as ∆m
=
∆H ∆
H
=
1452000 kJ = 598,000 g = 598 kg 2.428 kJ g
The transpiration volume is then obtained as V
=
∆m ρ
=
598 kg 996 kg m
= 0.600 m
The transpiration depth is equal to the volume per unit area, or
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
PROBLEM ASSIGMENTS Format All homework assignments must be done in a neat and organized manner. Clearly identify the solution for those problems with several mathematical steps by underlining or boxing in your answer. Points will be deducted from problems that are sloppy and difficult to follow. Problems 2.1.
Consider an elevator of mass of 2300 kg that is at rest at a position 5 m above the base of the elevator shaft. It is lifted to a height 80 m above the base of the shaft. The cable holding it breaks, and it falls freely to the base of the shaft (where it is brought to rest by a strong spring). Assume the entire process to be frictionless. Calculate (1) the gravitational potential energy of the elevator at its initial and highest positions, relative to the base of the shaft, (2) the work done in lifting the elevator to its highest position, and (3) the kinetic energy and velocity of the elevator just before it strikes the spring.
2.2.
A billiard ball of mass = 0.15 kg is dropped from a four story building, corresponding to a height of 12 m. If mechanical energy is conserved, what are the (1) kinetic energy and (2) the velocity of the ball just before it strikes the ground surface? Assume that the ball remains intact after impact and that the collision is elastic (kinetic energy is conserved). By using the definitions given for the kinetic theory of gases in Chapter 1, what is the impulse? If the collision duration is 0.01 s, what is the average force of the collision?
2.3.
The heat transfer rate and the temperature difference have been measured for different types of mineral-fiber insulation materials by Tye et al. (1980). For fiberglass insulation, they measured a heat transfer rate of 2.5 J s -1 over an area of 0.19 m2 and a thickness of insulation of 0.105 m (Δx = - 0.105 m). The temperature difference between the two sides of the material was determined to be 27.7 K. Assuming heat transfer by conduction, what is the thermal conductivity in units of J s -1m-1K-1 and the Rvalue in units of K m2 s J-1? R-values for insulation in the United States are reported using units of F ft2 h BTU-1. By using the conversion that 1 K m2s J-1 = 5.68 F ft2 h BTU-1, what R-value would you use to label this fiberglass material?
2.4.
Consider a spring with a spring constant of 50 N m -1 and a mass of 0.05 kg. What is the natural frequency for the spring in a simple harmonic motion? For an amplitude of 0.1 m, compute and plot the location of the spring with time between 0 and 0.5 seconds using 0.01 seconds increments. Identify on the graph the wavelength for the spring motion. In addition for each time increment, compute the work done, the elastic potential energy, the spring velocity, the kinetic energy, and mechanical energy.
2.5.
Consider a human lung as your system. A typical volume of a human lung is 5 x 10 -4 m3 (0.5 L) and a typical air pressure is 101 kPa. How much pressure work is done by the muscles (of the surroundings) to exhale air from a typical human lung? Convert this work into an equivalent number of textbooks raised to a height of Δz = 1 m. Use an average mass of textbooks of 1 kg.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
2.6
Measured pressures and volumes are given below for a cylinder-piston system where gas expanded slowly under quasi-equilibrium conditions. The first value in the table corresponds to the initial state and the last value to the final state. Plot the observed pressure as a function of volume. What is the work done by the gas during the expansion? P bar 12 10 8 6 4 2
2.7
V m3 0.045 0.052 0.063 0.079 0.109 0.189
Consider the following relationship between pressure and volume for a cylinder-piston system under quasi-equilibrium conditions between initial state of P i = 50 kPa and Vi = 0.5 m3 and a final state of Pf = 200 kPa and Vf = 0.1 m3. 𝑃 =𝑎+𝑏𝑒
(
)
= 𝑎 + 𝑏 exp 𝑐(𝑉 − 𝑥 )
You wish to analysis two possible paths between the endpoint states. For Path 1, the coefficients are defined as a = -72.4 kPa, b = 122.4 kPa, c = -2 m -3 and xo = 0.5 m3. For Path 2, the coefficients are defined as a = 322.4 kPa, b = -122.4 kPa, c = 2 m -3 and xo = 0.1 m3. Part I: You are required to compute the work for each path using only the conditions of the initial and final states. Is work defined by an exact or inexact differential? Part II: You are also required to plot the pressure for Path #1 and Path #2 for volumes between 0.5 m3 and 0.1 m3. You can compute the pressure for both paths using ΔV = 0.02 m3, that is, find the pressure for V=0.5 m3, V=0.48 m3, V=0.46 m3, ... , V=0.1 m3. Part III. For a system with n = 10 moles of nitrogen gas, compute gas temperature for each volume obtained using ΔV = 0.02 m3 for the two different paths. Also compute the work done between each of the consecutive volume for both Paths #1 and #2. Compare the total work done for the two paths obtained from Part III with the solution obtained for Part I. 2.8
You have determined the following linear relationship between pressure and temperature for a specially designed cylinder-piston system: 𝑃 = 200 + 10 𝑇 where the coefficient “200” has units of Pa and the coefficient of “10” has units of
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Pa K-1. For an initial temperature of 200 K and a final temperature of 350 K, what is the work done per mole of ideal gas? 2.9
A colleague in your consulting firm has studied the relationship between pressure and temperature for a particular cylinder-piston system with 50 moles of O 2. She obtained the following equation: 𝑃 = 10 √𝑇 where the coefficient “10” has units of kPa K-1/2. For an initial temperature of 200 K and a final temperature of 400 K, how much work is done?
2.10.
What is the kinetic energy of a system of 1 kg of water just prior to landing in a pool after falling 150 m over a waterfall? What would be the increase in temperature of the system if all of the kinetic energy is converted into internal energy? You can assume that water is an incompressible substance. You can treat the 1 kg of water as an isolated system once it enters the pool at the bottom of the waterfall.
2.11.
As discussed in the chapter, the molar heat capacity at constant pressure is sometimes related to temperature using the equation given below. Coefficients for helium (monatomic gas), nitrogen (diatomic gas) and carbon dioxide (polyatomic gas) are also shown below. Compute and plot the molar heat capacity at constant pressure for temperature between 10 and 150 C in 10 C increments. What is the largest percent error in Cpm between the equation values and Thermodynamic Data Handout value (given for 25 C) for each of these gases? For this problem, use the C pm obtained from the equation below as the standard for computing percent error (% error = 100% x (Cpm-Cpm,25 C)/Cpm).
Cpm a b T c T 2 d T3
Gas He N2 CO2 2.12.
a
-1
J K mol 20.786 28.883 22.243
-1
b
-2
-1
J K mol 4.851 x 10-13 -0.157 x 10-2 5.977 x 10-2
c
-3
-1
J K mol -1.583 x 10-16 0.808 x 10-5 -3.499 x 10-5
d
-4
J K mol-1 1.525 x 10-20 -2.871 x 10-9 7.464 x 10-9
We are interested in the error obtained in change in the molar enthalpy by assuming a constant Cpm. Consider the helium, nitrogen and carbon dioxide gases of Problem 2.10 for temperature between (1) 15 C to 35 C, (2) 15 C and 60 C, and (3) 50 C and 100 C. For each of these temperature ranges and for each of the gases, compute the molar enthalpy using a (a) constant Cpm from the Thermodynamic Data Handout (corresponding to 25 C) and (b) using Cpm defined by the relationship of
Cpm a b T c T 2 d T3
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
What are the percent errors obtained by using the simpler constant C pm to compute change in molar enthalpy for each of the temperature intervals and each of the gases? For this problem, use the more accurate solution obtained from the relationship (Method b) as the standard for computing the percent error. 2.13.
In Problem 1.20 of Chapter 1, you computed the theoretical change in internal energies per mole between 10 C and 40 C for a helium (He, monatomic gas) using U m=3RT/2, nitrogen gas (N2, diatomic gas) using Um=5RT/2 and carbon dioxide (CO2, polyatomic gas) using Um=3RT. Re-compute the changes in molar internal energy using the Thermodynamic Data Handout that is based on measured values. You can assume a constant heat capacity between the two temperatures. What are the percent errors in the internal energies obtained using the theoretical values of Chapter 1 and that obtained using the thermodynamic data obtained from experimental measurements?
2.14.
Consider 1 kg of air compressed slowly in a piston-cylinder assembly from 150 kPa to a final pressure of 900 kPa at constant temperature of 38 C. Determine (a) the change in the internal energy of the gas, (b) the work interaction in kJ, and (c) the quantity of heat transfer in J kg-1. You can use a molar mass of air of 29 g mol -1.
2.15.
A piston-cylinder assembly is filled with 2 kg of oxygen gas that has an initial temperature of 30 C. Under conditions of constant pressure, heat transfer into the cylinder between initial and final states has been determined to be 20 kJ. What is the temperature of the oxygen gas at the final equilibrium state? How much expansion work was done in the process?
2.16.
A vertical piston-cylinder assembly with a volume of 28 L is filled with 45 g of nitrogen gas. The piston is weighed so that the pressure on the nitrogen is always maintained at 135 kPa. Heat transfer is allowed to take place until the volume is 80% of its initial value. Determine (a) the energy transfer by heating, (b) final temperature of the nitrogen at equilibrium, (c) the change in internal energy and (d) the expansion work done in the process.
2.17.
Consider a rigid tank with 7 kg of methane gas. The gas is heated from 7 to 174 C. By assuming an ideal gas, determine (1) the work done by the gas and (2) the heat transferred. Also compute the change in enthalpy for the temperature difference (i.e. ΔHtpr) from a constant Cpm over the range of temperatures. Why is the heat transfer not equal to this change in enthalpy?
2.18.
A rigid insulated tank is divided into equal volumes by a partition, 450 g of carbon dioxide is introduced into one side of the partitioned tank, and the other side remains evacuated. For this initial equilibrium state, the pressure and temperature of the gas are measured and recorded as 105 kPa and 345 C. The partition is then pulled out, and the gas is allowed to expand into the entire tank. Expansion in this type of system is called Joule expansion. Determine the final pressure and temperature at a new equilibrium state for an ideal gas.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
2.19.
Many students find the outcome of Joule expansion (Problem 2.18) to be counterintuitive. This outcome (examined experimentally by Joule) played an important role in defining ideal gases. Ideal gases are defined such that “no change in temperature occurs when air is allowed to expand in such a manner as not to develop mechanical power” (i.e., w=0). Let’s consider the change in temperature for the system of Problem 2.18 for a non-ideal gas using van der Waals equation shown below (and given in Chapter 1 for the complete set of lecture notes). The translational molar internal energy is then defined as 3 na U = RT − 2 V where for carbon dioxide, the parameter “a” has been determined to be a = 0.000365 kPa m6 mol-2. You can use an initial volume of 0.5 m3. What is the change in temperature for the conditions of Problem 2.18 using the more rigorous non-ideal gas equation?
2.20
A pressure-release valve has been installed on a 0.5 m 3 rigid tank. The valve releases gas so that the tank pressure doesn’t exceed 300 kPa. Initially, the tank is filled with nitrogen gas to a pressure of 200 kPa and a temperature of 50 C. Consider a heat transfer into the tank of 200 kJ. How many moles of nitrogen is released from the tank? What is the temperature of nitrogen in the tank? You can neglect possible work with the release of nitrogen.
2.21.
Let’s revisit the 10 moles of nitrogen gas placed in a cylinder of Problem 2.7. For each step of changing volumes, compute (1) change in enthalpy, (2) change in internal energy, (3) heat transfer and (4) change in entropy estimated as 𝑞/𝑇. You can use the temperatures and work done computed in Problem 2.7. Perform these calculations for both Paths #1 and #2. Assume constant molar heat capacity. For a given path, is the net heat transfer equal to the total change in enthalpy? Which of the thermodynamic characteristics appear to be defined by exact differentials and which ones by inexact differentials?
2.22.
You are interested in ΔUtpr, ΔHtpr and ∫δq/T for the system of Problem 2.21; however for this problem, the pressure-volume path between the endpoint states is unknown. Since the actual path is unknown, a friend suggested that you take the simplest possible path defined by assuming a linear relationship between pressure and volume. By using a linear relationship between endpoint pressure and volume conditions, repeat the calculations done in Problem 2.21 (and Problem 2.7). Since the path is the simplest, and not the actual path, you are not interested in the work done and heat transfer, but these characteristics are needed to determine ∫δq/T. Compare the computed ΔU tpr, ΔHtpr and ∫δq/T obtained using the linear relationship with the results of Problem 2.21. Could have you determine these changes obtained for the actual Paths #1 or #2 using your simple linear relationship (yes or no)?
2.23.
A particular automobile tire has an approximately constant volume of 0.05 m 3. Assume that the heat transfer into the tire is 0.025 kJ mile-1. For an initial temperature of 20 C and an initial pressure of 210 kPa, how many miles need to be driven to increase the tire
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
pressure to 235 kPa? You can use a molar heat capacity at constant pressure of air of Cpm = 28.25 J K-1 mol-1. 2.25.
What is the heat transfer required (1) to melt 200 g of ice at 0 C, (2) to increase its temperature from 0 C to 100 C and (3) to evaporate it at 100 C? You can assume constant heat capacities with temperatures and that water is only evaporated at T=100 C. How much more heat transfer is needed to evaporate the 200 g of water than to raise its temperature from 0 to 100 C?
2.26.
Recommended caloric intake to support metabolic processes is approximately 2400 calories per day for a typical college student (definition of calories for food is actually kilocalories in thermodynamics). By using 1 food calorie = 4.1868 kJ, convert these calories into a change of the internal energy from metabolic biochemical reactions (-ΔUr) in units of MJ per day. If the human body of mass 65 kg is an isolated system that has the heat capacity of liquid water, what would be the change in temperature in the human body as a consequence of its metabolic processes?
2.27.
An important mechanism for controlling the temperature of the human body is perspiration and subsequent evaporation. Convert the standard enthalpy of vaporization given in the lecture notes of 40.7 kJ mol-1 at 100 C to 37 C (body temperature). Assuming that the body generates –ΔUr =10 MJ d-1 from metabolic processes (see Problem 2.26), how much water in kg needs to be evaporated per day to maintain the body at constant temperature of 37 C? What is the percent difference in mass if you had computed the evaporated mass using the standard enthalpy of vaporization at 100 C? Convert the evaporated mass of liquid water to liters for ∆ 𝐻 . How many one-halfliter glasses of water needs to be drunk per day to meet this evaporative demand? The recommend intake of water by the Institute of Medicine is approximately 3 liters for men and 2.2 liters for women. As shown by Example Problem 2.1, the human body also has heat transfer by radiation, convection and conduction.
2.28.
Methane has a similar molar mass (mm = 16 g mol-1) to water, but has standard enthalpy of vaporization of only 8.2 kJ mol-1. If ∆ 𝐻 of water was equal to that of methane (still use water properties for all other characteristics), how many additional one-halfliter glasses of water per day is needed to meet the evaporative demand of Problem 2.27? You don’t need to adjust 8.2 kJ mol -1 to 37 C.
2.29.
The change in internal energy and heat transfer resulting from the condensation of water vapor is an important process in the development of thunderstorms and hurricanes. Consider a depth of rainfall of 25.4 mm (1 inch) for Minneapolis, MN. Minneapolis has an area of 151 km2 within its city limits. If the rain falls uniformly over the land area of Minneapolis, what is the total mass of precipitation in kg? Using the standard enthalpy of condensation at 25 C, how much energy is available for heat transfer with the condensation of this mass of water from vapor to rain droplets (liquid phase)? The Monticello, MN nuclear plant generates approximately 600 megawatts of electrical
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
power. How many days must the Monticello plant generate power to obtain the energy corresponding to the phase transition of our 1-inch storm? 2.30.
Consider the phase transition of vaporization and fusion as consequence of the following reactions: H O (l) → H O (g) and H O (s) → H O (l) For vaporization, compute the standard enthalpy of reaction at 25 C using the Thermodynamic Data in the Handout. How does this value compare to the standard enthalpy of vaporization given in the lecture notes? For fusion, show that the molar heat capacity of reaction (i.e. ΔrCpm) can be used to compute the standard enthalpy of fusion at different temperature using the equation of ∆
2.31.
H ≈∆
H +
C
− C
(T − T )
Let’s consider an alternative solution to the determination of the standard enthalpy of reaction for urea (see Lecture Example Problem 2.5) at T= 80 C. The reaction of interest is: (NH2)2CO(s) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l) + N2 (g) Instead of using the molar heat capacity of reaction as done in lecture, you need to adjust the standard enthalpy of formation for change in temperature for each of the compounds separately, using their individual molar heat capacities. For example, the standard enthalpy of formation for urea would be evaluated as
80f C H mo ,urea 25f C H mo ,urea C pm,urea T After adjusting each of the reactants’ and products’ standard enthalpies of formation to T= 80 C, compute the standard enthalpy of reaction at T= 80 C? How well does it compare to the value computed in Lecture Example Problem 2.5? 2.32.
The chemical reaction for the combustion of ethanol is shown below. You wish to determine the standard enthalpy of reaction from measured data and from our definition using standard enthalpy of formation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l) Part I. The measured data for the reaction can be obtained using an adiabatic bomb calorimeter. A schematic is shown below. Adiabatic bomb calorimeters are constructed from rigid and insulated containers that have an inner chamber where combustion occurs. Heat transfer occurs between the inner chamber and a liquid surrounding the chamber with a heat capacity of 3570 J K-1. The inner chamber is also rigid so that the heat transfer from the reaction to the surrounding liquid is the change in the internal energy (no expansion work) from the ethanol reaction.
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
For our experiment, the combustion of 0.7 g of ethanol (C2H5OH) caused the temperature of the liquid to increase from 25 to 30.814 C. What is the molar internal energy for the combustion of ethanol? To determine the enthalpy of reaction, you need to compute the expansion work that would have occurred under constant pressure for the ethanol reaction at a temperature of 25 C. The molar expansion work is determined using (as derived in Example Problem 2.5) P ∆V w =− = −R T −ν + ν ∆ξ Based on these experimental results, what value would you report for the standard enthalpy of reaction for the combustion of ethanol? Part II: Also compute the standard enthalpy of the ethanol reaction (∆ 𝐻 ) at 25 C using the ∆ 𝐻 values given in the thermodynamic handout. Is this value in good agreement with that obtained experimentally in Part I? Compare your calculated value of ∆ 𝐻 with the reported value of the standard enthalpy of combustion (∆ 𝐻 ) for liquid ethanol given in the thermodynamic handout. Are the values in reasonably close agreement? 2.33.
The reaction for the combustion of carbon monoxide is shown below: CO (g) + (1/2) O2 (g) → CO2 (g) For an experiment at a temperature of 25 C and standard pressure, the standard reaction enthalpy for this reaction has been measured as ∆ 𝐻 = - 282.98 kJ mol-1. Determine the standard enthalpy of formation for CO. Is your computed value in good agreement with that given in the thermodynamic data handout?
2.34.
From the table of standard enthalpies of formation given below, calculate the standard reaction enthalpy for the following chemical reaction: Ca3P2 (s) + 6 HCl (g) → 3 CaCl2 (s) + 2 PH3 (g) Compound Ca3P2 (s) PH3 (g)
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ΔfTHo (kJ mol-1) - 504 +9
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
HCl (g) CaCl2 (s) 2.35.
- 92 - 794
Metabolism is largely dependent on the combustion of sugar, fats, and proteins. For this problems, we are interested in the standard enthalpies of combustion of fats (using triolein) and proteins (using the polypeptide of glyclphenylalanine as a surrogate). The chemical reaction for the combustion of these compounds are shown below. NIST reports a standard enthalpy of formation for triolein of ∆ 𝐻 = -2193.7 kJ mol-1 and an equivalent ∆ 𝐻 for glyclphenylalanine of ∆ 𝐻 = - 684.3 kJ mol-1. C H O (l, triolein) + 80 O (g) → 57 CO (g) + 52 H O (l) C H N O (s, glyclphenylalanine ) + 13 O (g) → 11CO (g) + 7H O (l) + N (g) To illustrate the reported range in thermodynamic properties for complex organic compounds, NIST also reports ∆ 𝐻 = -1592.2 kJ mol-1 for triolein (you don’t need to do anything with this alternative value). Part I. For both reactions, what are the respiratory quotients and the standard enthalpy of reactions (i. e. , ∆ 𝐻 )? You can use a temperature of 25 C. Also for both reactions, compute the change in molar internal energies (i. e. , ∆ 𝑈 ). For this calculation, you will first need to compute the molar work. Is the ratio of ∆ 𝑈 to ∆ 𝐻 nearly one (yes or no)? Is the error likely small if we use ∆ 𝐻 instead of ∆ 𝑈 in the thermodynamic analysis of rooms or buildings of constant volume (yes or no)? Part II: Metabolic rate of animals is often estimated by measuring the amount of oxygen consumed and the amount of carbon dioxide produced. To make these calculations easier, convert the ∆ 𝐻 values of Part I to kJ per mole of O2 and kJ per mole of CO2 (i. e. , ∆ 𝐻 𝑎𝑛𝑑 ∆ 𝐻 in lecture notes). As shown by Homework Problem 1.11, the volume of 1 mole of O2 or CO2 at SATP is 24.8 L. What are the standard enthalpies of combustion in units of kJ L-1 O2 and kJ L-1 CO2 (i.e., ∆ 𝐻 𝑎𝑛𝑑 ∆ 𝐻 in lecture notes)?
2.36.
We wish to investigate whether the change in enthalpy of chemical reactions is independent of path (as done for changes in enthalpy with temperature and phase transitions) by considering the production of methane gas from coal (approximated as graphite) at a temperature of T= 220 C. The sequence of three reactions given below for Path #1 corresponds more closely to the actual path. Path #2 is the chemical reaction more commonly used to represent this process. If you consider the moles of reactants as negative and those of the products as positive, the sum of the moles for reaction sequence of Path #1 results in the chemical reaction given for Path #2. Path #1: Sequence of three separate reactions whose sum equals the reaction of Path #2. 2 C (s) + 2 H2O (g) → 2 CO (g) + 2 H2 (g) CO (g) + H2O (g) → CO2 (g) + H2 (g) CO (g) + 3 H2 (g) → CH4 (g) + H2O (g)
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Path #2: Single chemical reaction from coal to methane: 2C (s) + 2H2O (g) → CH4 (g) + CO2 (g) For Path #1, you are required to compute the standard reaction enthalpy at 220 C for all three reactions using the Thermodynamic Data Handout. Sum these three standard enthalpy of reactions to determine the total change in enthalpy from carbon to methane gas by Path #1. You are also required to compute the standard reaction enthalpy at 220 C for Path #2. For this example, is the total change in reaction enthalpy independent of path? These results support Hess’ law. 2.37.
The energy derived from sugar is an important chemical reaction for biological cells. For cells with an abundance of oxygen, glucose is oxidized as C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) However during exercise, oxygen levels in muscle cells may be insufficient to complete the above reaction. In this case, glucose is converted in lactic acid (C 3H6O3) by a process called glycolysis. This reaction is shown below. C6H12O6 (s) → 2 CH3CH(OH)COOH (s) Part I. Compute the standard enthalpy of reaction (∆ 𝐻 ) for the first glucose reaction using the standard enthalpies of formation given in the Thermodynamic Data Handout. Compare your computed ∆ 𝐻 to the standard enthalpy of combustion (∆ 𝐻 ) for glucose (also given in the Thermodynamic Data Handout) by using the ratio of ∆ 𝐻 to ∆ 𝐻 . Is this ratio nearly one (yes or no)? Part II: The standard enthalpy of formation for lactic acid is not available in the Thermodynamic Data Handout. However, Atkins and de Paula (2006) have reported the molar enthalpy of reaction for following reaction of lactic acid as Δ rTHm° = -1344 kJ mol-1. CH3CH(OH)COOH (s) + 3 O2 (g) → 3 CO2 (g) + 3 H2O (l) From the results of Problem 2.36, you have shown that the standard enthalpy of reaction can be computed by solving the reaction for three separate reaction. Apply this principle to determine the standard enthalpy of reaction for glycolysis using the following sequence of reactions C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) 6 CO2 (g) + 6 H2O (l) → 2 CH3CH(OH)COOH (s) + 6 O2 (g) What is the ratio of the standard enthalpy of reaction of the oxidation of glucose to that obtained by glycolysis? Which of the two reactions provide more energy for other biological activities?
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
Part III. Convert the ∆ 𝐻 obtained in Part I to enthalpy of reaction per mole of O 2 (i. e. , ∆ 𝐻 in lecture notes). Compute the enthalpy of reaction per L of consumed O2 at SATP (i.e., ∆ 𝐻 in lecture notes). 2.38.
Consider a hypothetical animal whose metabolism can be well approximated by the combustion of glucose and triolein. These chemical reactions have been analyzed in Homework Problems 2.35 and 2.37, and thermodynamic results are summarized in Table 2.3. For our animal, the consumption rate of oxygen has been measured to be 20 L O2 h-1 , the production rate of CO2 has been measured to be 16 L CO2 h-1 and the production rate of water vapor by vaporization has been measured to be 90 L H 2O h-1. All of these measurements correspond to standard ambient temperature and pressure of T=25 C and Po = 100 kPa for ideal gases. Part I. Compute the metabolic rate in units of MJ d-1 and the molar extents of reaction rate (mol h-1) for both glucose and triolein? Part II. Compute the fraction of the metabolic rate obtained in Part I that is used to evaporate water. You can use the standard enthalpy of vaporization corresponding to 39 C (animal temperature) of ∆ 𝐻 = 43.2 𝑘𝐽 𝑚𝑜𝑙 . Part III. Compare the metabolic rate of Part I to that obtained by Brouwer’s relationship (i.e., Eq. 2.7.28). What is the percent error that you obtained using Brouwer’s relationship?
2.39.
For four piglets, Professor Larry Jacobson at the University of Minnesota measured a consumption rate of 30.83 L O2 h-1 and a production rate of CO2 of 20.08 L CO2 h-1 defined using the STP of a temperature of T = 0 C and a pressure of P = 101.325 kPa (1 atm). What are the consumption and production rates in units of moles of O 2 and moles of CO2 per hour? in units of L O2 and L CO2 per hour adjusted for standard ambient temperature and pressures SATP of T= 25 C and P o=100 kPa? What is the observed respiratory quotient? By using Brouwer’s relationship (see Eq. 2.7.28), what is the metabolic rate in units of MJ d-1 pig-1?
2.40.
Consider an experiment for an animal that is consuming carbohydrates, fats, and proteins. You can use the metabolic reactions given in Table 2.3 for glucose, triolein and glyclphenylalanine to determine those for carbohydrates, fats, and proteins, respectively. The experiment measured a change in the number of moles of oxygen, carbon dioxide and nitrogen of Δno2 = - 0.6 mol O2 h-1, Δnco2 =0.5 mol CO2 h-1, and Δnn2 = 0.01 mol N2 h-1. What is the metabolic rate for the animal in units of MJ d -1? What are the molar extents of reaction rate (per mol h-1) for carbohydrates, fats and proteins?
2.41.
You wish to determine the metabolic rate of animal that consumes carbohydrates, fats, and proteins. Previous studies have shown that the calorie consumption from fats for this animal is approximately twice that of proteins (that is, ΔHr,fat = 2 ΔHr,prot). You can use the general metabolic reactions and respiratory quotient given in Table 2.3 for our food sources. You have experimentally measured a change in the number of moles of oxygen and carbon dioxide of Δno2 = - 0.6 mol O2 h-1 and Δnco2 =0.55 mol CO2 h-1 when the animal is resting and Δno2 = - 1.0 mol O2 h-1 and Δnco2 =0.8 mol CO2 h-1 when the
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
animal is exercising. What are the resting and exercising metabolic rate for the animal in units of MJ d-1? What fractions of these total metabolic rates are from carbohydrates, from fats and from proteins? Compute these fractions for both resting and exercising conditions. 2.42
You have a pig barn that has no heat transfer with the surroundings. The barn contains 50 pigs and has a width of 10 m, a length of 18 m and a height of 6 m. The average mass of pigs is 15 kg. The reaction enthalpy rate of metabolism per pig, for masses between 10 and 100 kg, can be approximated as ∆ Ḣ
.
= −14.95 m
where m is the mass of the pig in kg and ∆ Ḣ is the reaction enthalpy rate with units -1 of J s per pig (the coefficient “14.95” has units of J s -1 pig-1 kg-0.6). The percentage of (negative) reaction enthalpy used for vaporization is approximately 50%. What is the change in the air temperature in the barn after 90 minutes? What is the total mass of water evaporated from the pigs? For air within the barn, you can use m m = 29 g mol-1, Cpm = 29 J K-1 mol-1; and ρair = 1.3 kg m-3. You use standard enthalpy of vaporization corresponding to 39 C (pig temperature) of ∆ 𝐻 = 43.2 𝑘𝐽 𝑚𝑜𝑙 . You can neglect expansion work component of the reaction enthalpy. 2.43.
You have a turkey barn that has a constant heat transfer rate into the building of q = 50,000 kJ h-1. The barn contains 10,000 birds with an average mass of 5 kg. The reaction enthalpy of metabolism per bird can be estimated using ∆ Ḣ
= −35.5 m
.
where m is the mass of the bird in kg and ∆ Ḣ is the reaction enthalpy rate with -1 units of kJ h per bird (the coefficient “35.5” has units of kJ h-1 bird-1 kg-0.77). Forty percent of the (negative) reaction enthalpy is used to evaporate water from turkeys. What air flow rate is needed (units of kmol of air per hour) from ventilation fans to prevent the temperature of the building to increase by 5 C? What is the rate of water evaporated from the birds (kmol h-1)? You can use mm = 29 g mol-1 and Cpm = 29 J K-1 mol-1 for air and a standard enthalpy of vaporization at the bird temperature of ∆ 𝐻 = 43.2 𝑘𝐽 𝑚𝑜𝑙 . Consider the barn to be a rigid system. You can neglect the expansion work component of the reaction enthalpy. 2.44.
Propane gas tanks are widely used to supply fuel for outdoor grills. The combustion of propane for grilling is done under conditions of nearly constant atmospheric pressure. Grill gas tanks are typically filled with 9 kg (20 lbs) of propane. What is the possible heat transfer in MJ for the complete combustion of propane in one of these tanks? This heat transfer can be used for cooking, and it will also increase the air temperature of the surroundings.
2.45.
You are planning on taking food in a picnic basket on a hot summer day and wish to avoid food spoilage. You have determined that there is no spoilage if the temperature in the basket at the end of the day is 20 C. A friend (who has completed a heat transfer
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
course) has estimated that the heat transfer of 22.5 kJ will occur during the day given the temperature difference between the outside air and inside the picnic basket. You are required to determine the mass of ice to place into the basket. Your picnic basket is not rigid system and can expand with changes in temperature. It has a representative volume of 0.25 m3. You can assume a constant pressure process and perform your computations without any food in the basket. You can further assume that the initial temperature of the air (and the ice) is 0 C. You can neglect evaporation of liquid water, use a constant air pressure of 101 kPa, and a molar heat capacity at constant pressure for air of 29.15 J K-1 mol-1. 2.47.
For an electromagnetic wave of wavelength of 625 nm, what is the frequency of this wave? What region of the electromagnetic spectrum is it found in? What is its molar photon energy?
2.48.
Our eyes can perceive green light (λ = 5.5 x 10-7 m) wave length when the rate of incidence of radiation is 2 x 10-16 J s-1. Calculate the number of photons of green light that must fall on the retina per second to produce vision.
2.49.
Compute the energy of one quantum of red light (λ = 700 nm). How many joules of energy are in a mole of these photons? How many red light quanta are in one joule of energy?
2.50.
Photosynthetically active radiation (PAR) wavelengths lie between 400 and 700 nm. Compute the radiant flux density for PAR by numerically integrating the area under the spectral flux density over these wavelengths for the temperature of the Sun of 5778 K (neglecting the impact of absorptivity of atmospheric gases). First divide the range in PAR wavelengths into a minimum 10 intervals. For each interval, compute the spectral flux density using the midpoint wavelength. You can approximate the area of each interval as rectangles. Also compute the total radiant flux density corresponding to all wavelengths for T = 5778 K. What fraction of the total radiant flux density lies within the wave lengths of PAR?
2.51.
The energy content of the proven reserves of coal, oil and gas has been estimated to be 36 x 1021 J. The real reserve of fossil fuels (all fossil fuels under the earth’s surface) is often estimated as five times greater than the proven reserves. Part I: Let’s obtain insight into the magnitude of solar energy by using a solar constant of 1368 W m-2 and the radius of the Earth’s land surface of 6.37 x 106 m. Assume that the net transmissivity of the atmosphere is 0.5, that 1% of the projected Earth’s surface is available for capturing solar radiation, and that this surface receives sunlight corresponding to the solar constant for an equivalent 6 hours per day. How many years would it take for solar radiation landing on this surface to obtain the energy equivalence of the proven reserves of fossil fuels? of real reserves of fossil fuels? Part II. Hall and Rao (1999) has estimated that the global production of biomass by photosynthesis as 2 x 1011 tonnes of organic matter per year, which is equivalent to 4 x
Last Modified: March, 2018
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
1021 J y-1. Annual food intake by humans has been estimated as an equivalent of 16 x 1018 J y-1. For proven reserves of coal, oil and gas, how many years of net photosynthesis – adjusted for the biomass necessary to feed humans - would it take to generate the energy equivalent of these reserves? How many years of net photosynthesis – again adjusted for the biomass necessary to feed humans - would it take to generate the energy equivalent of all fossil fuels under the Earth’s surface? 2.52.
Characteristics of Earth, Venus, Mars and Jupiter are given below. The subscript “o” refers to characteristics on the surface of the planet, and rs-p and rp are the distance from the sun to the planet and the radius of the planet, respectively. The symbol T p,meas is the approximate temperature of the planet (surface and atmosphere) and T o,meas is the approximate temperature of the planet surface. You can assume that the emissivities of the sun and planets are one. For Venus, Mars, and Jupiter, compute the solar constant and the global radiant flux density for a sun temperature of T=5778 K. Also compute the equilibrium temperatures for these planets. For which planets (include Earth based on Example Problem 2.8) are the equilibrium temperatures in good agreement with Tp,meas? with To,meas? Does the atmosphere play an important role on the surface temperatures? rs-p km
rp km
fr
Planet
g m s-2
Po kPa
ρo kg m3
Tp,meas K
To,meas K
Venus
8.9
1.0820 x 108
6052
0.77
9200
65
230
750
Mars
3.7
2.0660 x 108
3396
0.15
0.636
0.02
220
218
Jupiter
23.1
7.7860 x 108
71492
0.58
NA
NA
130
NA
Earth
9.8
1.4960 x 108
6378
0.30
101
1.217
250
280
2.53.
Compute the solar irradiance at the top of the atmosphere for Crookston, MN (latitude = 47.75 degree N, longitude = 96.5 degree W) for July 1.
2.54.
For the solar irradiance determined in Problem 2.53, compute the transpiration depth for a well-watered alfalfa field located near Crookston, MN for a clear-sky day of July 1, where the average air temperature during transpiration is 25 C and the average difference between the system (plants) and the surrounding (air) is 5 C. You can assume emissivity, reflectivities, absorptivities, and Bowen ratio are equal to those values used in the example problem solved in lecture and perform your calculations on a representative area of the canopy of 100 m2.
2.55.
For a location in southern Minnesota with a day length of 15 hours, the solar irradiance at the top of the atmosphere has been determined to be 50000 kJ m -2. You are interested in the transpiration depth for a well-watered alfalfa field for a clear-sky day of July 1, where the average air temperature during transpiration is 26 C for the plant canopy and 20 C for the surrounding air. Use a transmissivity of atmosphere of f ta = 0.7; emissivities of εp = 0.9 for alfalfa and εa = 0.7 for air; reflectivity of frp = 0.22 for alfalfa (plant), and a Bowen ratio of β = 0.25. The temperature change of the air within the
Last Modified: March, 2018
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Lecture Handout
Chapter 2: First Law of Thermodynamics
Bruce N. Wilson
canopy is 3 C. Perform your calculations on a representative area of the canopy of 100 m2, a canopy height of 1.1 m, air pressure of 100 kPa, and molar heat capacity at constant pressure of air of 29.15 J K-1 mol-1. You can assume that the fpar =0.4 and represent the reaction enthalpy using the photosynthesis of glucose. What is the transpiration depth? What is the depth if you neglect ΔH tpr? What is the depth if you neglect ΔHtpr and ΔHr? 2.56. Transpiration from plant canopies is often estimated using measured evaporation depths from a standard Class A pan. This type of pan is a cylinder with a surface area of 1.14 m 2 and is typically filled with water to a depth of 0.2 m. For one week (7 days) in southern Minnesota, the evaporation depth of 56.4 mm was measured from one of these pans. The incoming solar radiation (at the pan) for that week was measured to be 181,000 kJ m -2. The reflectivity of the pan and water is approximately 0.4. Use an average air temperature and pan water temperature during the week of 30 C and 18 C, respectively. Assume the emissivity of air of εa = 0.8 and the emissivity of the evaporation pan (with water) of εw = 0.9. Use a change in the temperature of the water in the evaporation pan of 10 C during the evaporation process. Determine the heat transfer by into (or out-of) the evaporation pan by conduction and convection processes. What is the Bowen ratio for the evaporation pan?
Last Modified: March, 2018
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Lecture Handout
