In order to perform energy balances on both closed and open systems, it is necessary to be able to determine how the energy (or enthalpy) of the species in the system changes during a process. As we learned in Section 1.3, the internal energy, u, and the enthalpy, h, for a pure species are constrained by specifying two independent intensive properties.
Moreover, since u is a thermodynamic property, we can specify a hypothetical path to calculate the change in internal energy, Du. It does not have to be the path of the actual process. Likewise for Dh. While any thermodynamic properties can be used, it is often convenient to choose measured properties (T, P, or v) as the independent properties.
Temperature is almost always chosen as one of the independent properties, since it can be measured in the lab (or fi eld), and there is a direct relationship between T and u; that is, temperature is a measure of the molecular kinetic energy, which is one component of u (see Section 2.1). In fact, for an ideal gas, it is only this component that contributes to u. The other independent property is typically also a measurable property, either P or v can be chosen according to convenience.
Figure 2.3 illustrates a common hypothetical path used to calculate Du. In this case, T and v are chosen as independent properties. In step 1, we must know the tempera- ture dependence of u to calculate Du, as we go from T1 to T2 at constant volume. This information is often obtained in the form of heat capacity (or specifi c heat). Similarly, the temperature dependence of h used to fi nd Dh can be found through reported heat capacity values. Therefore, heat capacity data are crucial in this problem-solving meth- odology. In the next section, we will explore how heat capacities are experimentally determined and how they are reported.
Heat Capacity at Constant Volume, cv
To measure the heat capacity at constant volume, cv, an experimental setup as conceptu- ally shown in Figure 2.10a can be used. This closed system consists of pure species A within a rigid container. The container is connected to a heat source (in this case, a resist- ance heater) and is otherwise well insulated. The experiment is conducted as follows:
As a known amount of heat, q, is provided through the resistive heater, the tempera- ture, T, of the system is measured. As heat is supplied, the molecules of A move faster, and the temperature increases. A typical data set for pure species A is also shown in Figure 2.10a. Since we can “sense” the result of the heat input with the thermocouple, this type of energy change is labeled sensible heat. This setup is known as a constant- volume calorimeter. Since there is no work done in this process, we can apply the fi rst law [Equation (2.13b)] to this system and get:
Du5q closed system, const V (2.23) 2.6 Thermochemical Data for U and H ◄ 67
c02.indd 67
c02.indd 67 05/11/12 1:35 PM05/11/12 1:35 PM
68 ► Chapter 2. The First Law of Thermodynamics
Note that in Figure 2.10a, the amount of heat supplied is plotted on the y-axis and the temperature measured is on the x-axis. However, Equation (2.23) shows heat input is identical to Du.
We defi ne the heat capacity at constant volume, cv, as:
cv; a'u 'Tb
v
(2.24) Hence, the slope of the curve gives us the heat capacity at any temperature. In these data, the heat capacity at T1 is less than that at T2. Typically, heat capacity changes with T.
By taking the slope of this curve as a function of temperature, we can get:
cv5cv1T2
We can then fi t the data to a polynomial expression of the form:
cv5a1BT1CT21DT221ET3 (2.25)
Figure 2.10 Schematics of the experimental determination of heat capacity. (a) constant-volume calorimeter to obtain cv; (b) constant-pressure calorimeter to determine cP.
A A A A A
A A
q q
Experiment: Data:
Experiment: Data:
Pure substance, A A
A A
A A A A A
A A A A A
P = const V = const
T
T
Slope = cP at T1
Slope = cv at T1
Slope = cP at T2
Slope = cv at T2
cP ≡ ∂T∂h
P
cv ≡
∂T∂uv
(b) Constant-Pressure Calorimeter: Determination of cP (a) Constant-Volume Calorimeter: Determination of cv
T1 T2
T T1 T2
T
q = Δhq= Δu
c02.indd 68
c02.indd 68 05/11/12 1:35 PM05/11/12 1:35 PM
Parameters a, B, C, D, and E are then tabulated and can be used any time we want to know how the internal energy of species A changes with temperature at constant vol- ume. We can then fi nd Du by integration:
Du5 3
T2
T1
cv dT5 3
T2
T1
3a1BT1CT21DT224dT (2.26)
Heat capacity should only be used for temperature changes between the same phase.
When a phase change occurs, the latent heat must also be considered, as discussed shortly.
Consider an ideal gas. Equation (2.24) shows us that if we increase T, we increase u. The amount the internal energy increases with temperature is quantifi ed by the heat capacity according to Equation (2.26). The more energy it gains, the larger cv. On a molecular level, we may want to know how this increase in energy manifests itself. We can associate the increase in molecular kinetic energy and, therefore, in u with temperature to three possible modes in which the molecules can obtain kinetic energy. The fi rst mode is related to the center-of-mass motion of the molecules through space. In Chapter 1, we saw that the Maxwell–Boltzmann distribution characterizes the velocities of the mol- ecules at a given temperature. This translational energy contributes kT/2 per molecule (or RT/2 per mole) to the kinetic energy in each direction that the molecule moves.
Since molecules translate through space in three directions, the translational motion contributes 3RT/2 per mole to the internal energy of the molecules. The contribution of translational motion to cv, given by the derivative of the internal energy with respect to temperature, is 3R/2. In fact, monatomic gases have heat capacities given by this value.9
Diatomic and polyatomic molecules can manifest kinetic energy in rotational and vibrational modes as well. Except at extremely low temperature, the additional kinetic energy due to rotational motion for linear and nonlinear molecules is RT and 3RT/2 per mole, respectively. The kinetic energy due to vibration is much more interesting. It is related to the specifi c quantized energy levels of the molecule. The distribution of these levels depends on temperature. To account for the vibrational modes of kinetic energy, we would need to resort to quantum mechanics. We will not formally address that prob- lem here; however, it is useful to realize that the temperature dependence of the heat capacity indicated by Equation (2.25) manifests itself in the vibrational mode. At low temperature, the vibrational contribution goes to zero and the heat capacity is given by the translational and rotational modes only. At high temperature, where the vibrational motion is fully active, the contribution is R per mole. In summary, cv can be attributed to molecular structure and the ways in which each species exhibits translational, rota- tional, and vibrational kinetic energy. Heat capacities for gases, liquids, and solids can be intepreted in a similar way.
Heat Capacity at Constand Pressure, cp
The heat capacity at constant pressure, cP, is measured in a similar manner, only gas A is no longer held within a rigid container, but in a system that can expand as it is heated so as to keep the pressure constant. A conceptual representation of the experimental setup to measure cP is shown in Figure 2.10b. While the actual apparatus may look dif- ferent, this depiction is in terms of the piston–cylinder assembly that we have previously examined.
9 Except at very high temperature when electrons occupy excited states.
2.6 Thermochemical Data for U and H ◄ 69
c02.indd 69
c02.indd 69 05/11/12 1:35 PM05/11/12 1:35 PM
70 ► Chapter 2. The First Law of Thermodynamics
Since the system is now doing Pv work as it expands, the energy balance contains a term for work:
Du5q1w5q2PDv (2.27)
So Equation (2.27) can be rewritten as:
Du1 D1Pv2 5q since at constant pressure, DP50, and:
D1Pv25PDv1vDP5PDv Applying the defi nition for enthalpy [Equation (2.18)], we get:
Dh5q closed system, const P (2.28) Hence, in this case, an energy balance tells us the heat supplied at constant pressure is just equal to the change in the thermodynamic property, enthalpy. Therefore, we defi ne the heat capacity at constant pressure as:
cP; a'h 'Tb
P
(2.29)
Again, typical data for species A are presented in Figure 2.10b and can be fi t to the polynomial form:
cP5A1BT1CT21DT221ET3 (2.30) The parameters A, B, C, D, and E are reported for some ideal gases in Appendix A.2.
Heat capacity parameters at constant pressure of some liquids and solids are reported in this appendix.
Enthalpy—A Second Common Use
Recall that we “constructed” the property enthalpy to account for both the internal energy and fl ow work for streams fl owing into and out of open systems. However, inspec- tion of Equation (2.28) suggests a second common use of enthalpy. This equation holds, in general, for closed systems at constant P. In this case, it accounts for both the change in internal energy and the Pv work as the system boundary changes in order to keep pres- sure constant. In both cases, the property h couples internal energy and work. Therefore, experiments that are conveniently done in closed systems at constant P are reported using the thermodynamic property enthalpy. For example, the energetics of a chemical reac- tion, the so-called enthalpy of reaction, is reported in terms of the property Dhrxn. In this way, the experimentally measured heat can be related directly to a thermodynamic property.
Relations between cP and cV
By comparing Figure 2.10a to Figure 2.10b, we can estimate the difference incvand cP for the different phases of matter. If species A is in the liquid or solid phase, its vol- ume expansion upon heating should be relatively small; that is, the molar volumes of liquids and solids do not change much with temperature. Hence, the piston depicted
c02.indd 70
c02.indd 70 05/11/12 1:35 PM05/11/12 1:35 PM
in Figure 2.10b will not move signifi cantly and the value of work in Equation (2.27) will be small compared to q. Thus Equation (2.29) and Equation (2.24) are approximately equivalent and, consequently:10
cP<cv liquids and solids
On the other hand, the volume expansion of a gas will be signifi cant. For the case of an ideal gas, we can fi gure out the relationship between cP and cv by applying the ideal gas law to the defi nition for cP as follows:
cP; a'h 'Tb
P5 c'1u1Pv2 'T d
P5 a'u 'Tb
P1 a'RT 'T b
P5 a'u 'Tb
P1R (2.31) since Pv5RT for an ideal gas. However, for an ideal gas, the internal energy depends on temperature only; that is, the only change in molecular energy is in molecular kinetic energy. Therefore,
a'u 'Tb
P5 du dT 5 a'u
'Tb
v
(2.32)
Plugging Equation (2.32) into (2.31) gives:
cP5cv1R for ideal gases
Values of heat capacity for gases are almost always reported for the ideal gas state. Thus, when doing calculations using these data, you must choose a hypothetical path where the change in temperature occurs when the gas behaves ideally.
Mean Heat Capacity
For many gases, heat capacity data are often reported in terms of the mean heat capacity, cP. Use of cP eliminates the need for integration and can make the mechanics of problem solving easier. As its name suggests, the mean heat capacity is the average of cP between two temperatures. It is usually reported between 298 K and a given temperature, T.
Hence, the enthalpy change becomes:
Dh5cP1T22982 Solving for cP gives:
cP5 eT
298
cPdT
T2298 (2.33)
Note that Equation (2.33) is also, by defi nition, the mathematical average of the continu- ous function cP over temperature between 298 K and T.
10 We will revisit the relation between cP and cv of liquids in Chapter 5, Problems 5.20 and 5.21.
2.6 Thermochemical Data for U and H ◄ 71
c02.indd 71
c02.indd 71 05/11/12 1:35 PM05/11/12 1:35 PM
72 ► Chapter 2. The First Law of Thermodynamics
Consider heating 2 moles of steam from 200ºC and 1 MPa to 500ºC and 1 MPa. Calculate the heat input required using the following sources for data:
(a) Heat capacity (b) Steam tables
SOLUTION (a) Since this process occurs at constant pressure, the system will expand as T increases. In accordance with the discussion above, enthalpy is the appropriate property to calculate the heat input. The extensive version of Equation (2.28) can be written as:
Q5nDh
If we assume water is an ideal gas, we use the values of heat capacity given in Appendix A.2 to calculate Dh:
cP ideal gas
R 5A1BT1DT2253.47011.45031023T10.1213105 T2 Using the defi nition of heat capacity, we get the following integral expression:
Dh5 3
T2
T1
cPdT5R3
773
473
3A1BT1DT224 dT
Integrating,
Dh5RcAT1B 2T22D
Td
473 773
Substituting in values:
Dh58.31433.4701300210.7253102317732247322 20.1213105a 1
7732 1
473bR510,991 c J mold and, Q5nDh521,981 3J4
(b) From the steam tables, Appendix B.2:
h^11at 1 MPa, 200°C2 52827.9 3kJ/kg4 h^21at 1 MPa, 500°C2 53478.4 3kJ/kg4 So, Dh^ 5h^22h^15650.5 3kJ/kg4
Since the steam tables give us the specifi c enthalpy, we must multiply by mass. Thus, we must use the molecular weight for water, MWH2O50.018 3kg/mol4:
Q5mDh^ 5 12 3mol4 2 10.018 3kg/mol4 2 1650.5 3kJ/kg4 2 11033J/kJ4 2523,418 3J4
The answer for part (b) is approximately 6% higher than for part (a). At 1 MPa, water is not an ideal gas but rather has attractive intermolecular interactions. The extra energy needed in part (b) results from that needed to pull the water molecules apart. We will learn more of these things in Chapter 4.
EXAMPLE 2.7 Heat Input Calculations Using Different Data Sources
c02.indd 72
c02.indd 72 05/11/12 1:35 PM05/11/12 1:35 PM
Use the data available in Appendix A.2 to calculate the mean heat capacity, cP, for air between T15298 K and T25300 to 1000 K, in intervals of 100 K.
SOLUTION Using the defi nition of cP from Equation (2.33):
cP5 Dh
1T22982 5
3
T
298
cPdT
T2298 (E2.8A)
The heat capacity can be integrated with respect to temperature using the parameters in Appendix A.2:
3
T2
T1
cPdT5R3
T
298
3A1BT1DT224dT5RBAT1B
2T22D TR
298 T
(E2.8B) where for air:
A53.355, B50.57531023, and D5 20.0163105 (E2.8C) The solution to Equation (E2.8A), using (E2.8B) and (E2.8C), is presented at intervals of 100 K in Table E2.8.
TABLE E2.8 Calculated Values for Mean Heat Capacity of Air at Different Temperatures.
T
[K] Dh
3J/mol4 T2298
cP
3J/mol K4
300 58.35 2 29.17
400 3003.93 102 29.45
500 6001.75 202 29.71
600 9049.59 302 29.97
700 12146.51 402 30.22
800 15292.02 502 30.46
900 18485.87 602 30.71
1000 21727.89 702 30.95
EXAMPLE 2.8 Determination of Mean Heat Capacity for Air
2.6 Thermochemical Data for U and H ◄ 73
You need to preheat a stream of air fl owing steadily at 10 mol/min from 600 K to 900 K.
Determine the heat rate required using the mean heat capacity data from Example 2.8.
SOLUTION This process occurs at steady-state with one stream in and one stream out; hence Equation (2.19) can be written as follows:
05n#
1ah1MW V2 2
S
1MWgzb
12n#
2ah1MW V2 2
S
1MWgzb
21Q# 1W#
s
EXAMPLE 2.9 Heat Calculation Using Mean Heat Capacity for Air
0 0 0 0 0
(Continued)
c02.indd 73
c02.indd 73 05/11/12 1:35 PM05/11/12 1:35 PM
74 ► Chapter 2. The First Law of Thermodynamics
where we have set the bulk kinetic and potential energies and shaft work to zero. A mole balance yields:
n#
15n#
25n#
air
so the fi rst law balance simplifi es to:
Q# 5n#
air1h9002h6002 5n#
air3 1h9002h29822 1h6002h2982 4 (E2.9A)
where h900 is the enthalpy of air at 900 K, h600 is the enthalpy at 600 K, and h298 is the enthalpy at 298 K. Equation (E2.9A) was rewritten to use the defi nition for mean heat capacity given by Equation (2.33):
1h9002h2982 5cP,900190022982 (E2.9B)
1h6002h2982 5cP,600160022982 (E2.9C)
Substituting Equations (E2.9B) and (E2.9C) into (E2.9A) and using the values from Table E2.8, we get:
Q#
5n#Dh510330.71190022982229.97160022982 4594.363 3J/min4
Air is contained within a piston–cylinder assembly, as shown in Figure E2.10A. The cross- sectional area of the piston is 0.01 m2. Initially the piston is at 1 bar and 25ºC, 10 cm above the base of the cylinder. In this state, the spring exerts no force on the piston. The system is then reversibly heated to 100ºC. As the spring is compressed, it exerts a force on the piston according to:
F5 2kx
where k550,000 3N/m4 and x is the displacement length from its uncompressed position.
(a) Determine the work done.
(b) Determine the heat transferred.
SOLUTION (a) Since the process is reversible, the system pressure is always balanced by the external pressure and the work done is given by:
W5 23
v2
v1
PdV (E2.10A)
We can draw a free-body diagram to determine how all the forces acting on the piston balance, as shown in Figure E2.10A.
The displacement of the spring, x, can be written in terms of the change in volume:
x5V2V1
A 5DV A EXAMPLE 2.10
Spring-Assembled Piston–Cylinder
c02.indd 74
c02.indd 74 05/11/12 1:35 PM05/11/12 1:35 PM
Air 10 cm Pext = 1 bar F = −kx
P1 = 1 bar A = 0.01 m2
T1 = 25°C
Piston
FAir = PAirA
Fext = PextA
|Fspring|= kx
where DV5V2V1. We can then equate the force per area acting on each side of the piston to get:
P5Pext1kx
A 5Pext1kDV
A2 (E2.10B)
Plugging Equation (E2.10B) into (E2.10A) and integrating gives:
W5 23
V2
V1
PextdV2 3
DV51V22V12
0
kDV
A2 d1DV2 5 2Pext1V22V12 2k1V22V122
2A2 (E2.10.C) To solve Equation (E2.10C), we must fi nd V2. Applying the ideal gas law gives:
P1V1
T1 5P2V2
T2 5V2
T2¢Pext1k1V22V12 A2 ≤ Solving this quadratic equation for V2 gives:
V250.00116 3m34 which can be plugged back into Equation (E2.10 C) to get:
W5 2166 3J4
(b) To fi nd the heat transferred during this process, we can apply the fi rst law to this closed system:
DU5Q1W (E2.10D)
Figure E2.10A Piston–cylinder assembly with a spring attached to the piston. The initial state of the system for Example 2.10 is shown.
Figure E2.10B Schematic of the forces acting on the piston in Figure E2.10A as the gas expands.
2.6 Thermochemical Data for U and H ◄ 75
(Continued)
c02.indd 75
c02.indd 75 05/11/12 1:35 PM05/11/12 1:35 PM
76 ► Chapter 2. The First Law of Thermodynamics
The change in internal energy is given by:
Du5 3
T2
T1
cvdT5 3
T2
T1
1cP2R2dT5R3
T2
T1
3 1A212 1BT1DT224dT
5RB1A212T1B 2T22D
TR
T1
T2
The heat capacity parameters for air can be found in Appendix A.2:
A53.355, B50.57531023, and D5 20.0163105 Thus,
Du51,580 3J/mol4
and, DU5nDu5 ¢P1V1
RT1≤Du5638 3J4 We can now solve for the heat transfer from Equation (E2.10D):
Q5 DU2W5803 3J4
Latent Heats
When a substance undergoes a phase change, there is a substantial change in internal energy associated with it (see Section 2.1). We need to be able to determine a value for this energy change if we want to apply the fi rst law to a process involving a phase change. Like heat capacities, the energetics characteristic of a given phase change are reported based on accessible measured data.
For example, consider the vaporization of a liquid. Liquids are held together by attractive forces between the molecules. To vaporize a liquid, we must supply enough energy to overcome the forces of attraction. A typical experimental setup is schematically shown in Figure 2.11. A given amount of liquid substance A is placed in a well-insulated closed system at constant pressure. A measured amount of heat is supplied until A becomes all vapor. We chose a system at constant pressure, as depicted in Figure 2.11, so that the temperature will stay constant during the phase change. A schematic for the data acquired in this experiment is presented on the right of Figure 2.11. The temperature of the subcooled liquid and the superheated vapor increases as energy is supplied via heat. It is only the heat input at constant temperature during the phase transition that is reported as the enthalpy of vaporization.
Examination of Equation (2.28) shows that if we measure the heat absorbed as A changes phase; it is equal to the enthalpy of the vapor minus the enthalpy of the liquid. We term this dif- ference the enthalpy of vaporization:
Dhvap5hv2hl
If we need to calculate the energetics of a vapor condensing to a liquid, we simply use the negative of Dhvap. Similarly, the change in enthalpy from the liquid phase to the solid phase is reported as the enthalpy of fusion, Dhfus:
Dhfus5hs2hl
c02.indd 76
c02.indd 76 05/11/12 1:35 PM05/11/12 1:35 PM
And the change from the solid phase to the vapor phase is the enthalpy of sublimation, Dhsub: Dhsub5hv2hs
How would you fi nd the internal energy of vaporization, Duvap, given Dhvap?
The term to describe the change of enthalpy during a phase transition at constant pressure is latent heat. Latent means “hidden,” and it is called “latent” because we cannot “sense” the heat input by detecting the temperature change, as is the case with “sensible” heat, described previously.
The latent heat changes with temperature. However, we usually only know its value at one state. For example, the enthalpy of vaporization is typically reported at 1 bar, the so-called normal boiling point, Tb. To fi nd Dhvap at another pressure (and therefore another temperature), we need to construct an appropriate hypothetical path (see Section 2.2). Figure 2.12 illustrates a path for the calculation of Dhvap,T at any T based on the measured value to which we have access at Tb. It consists of three steps. In step 1, we calculate the change in enthalpy of the liquid from T to Tb, using heat capacity data. In step 2, we vaporize the liquid at the normal boiling point, since this is the value we have for Dhvap. In step 3, we calculate the change in enthalpy of the vapor from the normal boiling point to T. Adding together the three steps, we get:
Dhvap, T5 3
Tb
T
cPldT1 Dhvap, Tb1 3
T
Tb
cPvdT5 Dhvap, Tb1 3
T
Tb
DcPvldT Figure 2.11 Schematic of the experimental determination of enthalpy of vaporization.
Experiment:
AAAAA
A A AAA
P = const
Liquid State 1
q q
Data:
A A
A A
A
A A
A A A A State 2
Vapor
T
T T
Tvap Δhvap
q =Δh
Step 2
Temperature
Phase
Liquid Vapor
Step 3
Step 1 ∫
Δhvap,Tb
Δhvap,Tb
Tb c| pdT T ∫ TTb cvpdT
Tb
T
Figure 2.12 Hypothetical path to calculate Dhvap at temperature T from data available at Tb and heat capacity data.
2.6 Thermochemical Data for U and H ◄ 77
c02.indd 77
c02.indd 77 05/11/12 1:35 PM05/11/12 1:35 PM