When a homogeneous chemical system is heated with no phase changes or chemical reactions tak- ing place, its temperature increases: the bigger the Q, the bigger the increase in T:
therefore, T Q = (const)
.
QWe let (1/C) be the constant of proportionality between T and Q, where C is known as the heat capacity of the system.
Thus, T 1
C Q
=
and it follows directly that C Q
T
Heat Capacity
Defined . . . [8]
There are different ways to represent the heat capacity of a substance:
(i) We use an upper case “C ” to represent the total heat capacity of the substance; C has the units of J K–1.
1 Eqn [7] holds both for processes involving chemical reactions and for those in which no chemical reactions take
(ii) A lower case “c” with a macron–– c ––is used to designate the heat capacity per unit mass of the substance, and is called the specific heat capacity; c has the units of J kg–1 K–1.
(iii) A lower case “c” is used for the molar heat capacity, which is the heat capacity of one mole of the substance; c has the units of J mol–1 K–1.
If we know the heat capacity of a system, the heat is easily calculated by measuring the change in temperature of the system when we heat it:
Thus, Q = C T = mc T = nc T . . . [9]
where m and n are, respectively, the mass and the number of moles of the substance being heated or cooled.
There are two common ways to heat a system:
(1) Constant Volume Heating––in which the system is enclosed in a rigid container at constant volume ( Vsyst= 0). For a constant volume heating process the pressure of the system is allowed to change. Thus, CV, cV, and cV are, respectively, the total heat capacity at constant volume, the spe-
cific heat capacity at constant volume, and the molar heat capacity at constant volume.
(2) Constant Pressure Heating––in which the pressure is kept constant and the volume of the system is allowed to change. In this case we use the symbols CP, cP, and cP to represent the total
heat capacity at constant pressure, the specific heat capacity at constant pressure, and the molar heat capacity at constant pressure. For solids and liquids it is found that the heat capacity at con-
stant volume is approximately the same as the heat capacity at constant pressure. (This is a conse- quence of the fact that the volume of a solid or liquid doesn’t change appreciably on heating, so that negligible PV-work is done when a solid or liquid is heated at constant pressure.)
From thermodynamics it can be shown that for an ideal gas,
cP = cV + R Ideal Gas . . . [10] where R is the gas constant. Eqn [10] holds for diatomic and polyatomic ideal gases as well as for monatomic ideal gases. For most real gases, too, the relationship is approximately valid, and can be used in the absence of more detailed information.
We showed that for any constant volume process in a closed system in which no work of any kind is done, UV= QV. Furthermore, if the process is merely the heating or cooling of a homogeneous chemical system (i.e., no phase changes or chemical reactions take place), then we also can say that QV= CV T. Combining the two equations gives
UV = QV = CV T W = 0; Closed System,No Chemical Reactions
or Phase Changes . . . [11]
Thus, for a closed system, when W = 0, UV = QV always; but UV = CV T only when the con-
or chemical reactions occurring. When heating or cooling m kilograms, of course, UV = mcV T, and when heating or cooling n moles, UV= ncV T.
Example 7-1
A candle is burning in a sealed insulated room––the “system”––releasing 100 J s–1 of thermal en-
ergy into the room, which is at one atm pressure. The total volume of the room is 100 m3, and the
initial room temperature is 20.0°C. For air, cV = 20.47 J mol–1 K–1. The molar mass of air is
28.97 g mol–1.
(a) Neglecting any compositional changes in the air resulting from the combustion products of the burning candle, and assuming uniform mixing, what is the temperature of the air in the room after one hour? The heat capacity of the walls can be neglected.
(b) What is Qsystas a result of the candle burning for one hour? (c) What is Usystas a result of this process?
Solution
(a) Let the final temperature of the air be ƒ°C.
The heating of the air is a constant volume process, for which QV= ncV T.
Number of moles of air in the room: n = PV
RT =
( )( )
( . )( . )
101 325 100
8 314 20 273 15+ = 4157 mol The heat released by the burning candle in one hour is (100 J s–1)(3600 s) = 3.60 × 105 J.
This heat is transferred to the air in the room, which becomes warmer. Thus,
(QV)air = ncV T
3.60 × 105 = (4157)(20.47) T
T = 4.23 K = ƒ – 20
ƒ = 20 + 4.23 = 24.23°C
Ans: 24.2°C
(b) The system is the room itself. Since the room is insulated, no heat can cross its boundaries; therefore Qsyst = 0. There is, of course, heat exchange between the subsystems that make up the contents of the room; namely, the candle and the air. The heat evolved by the candle (Qcandle = –3.60 × 105 J) is equal to the heat absorbed by the air (Q
air = +3.60 × 10
5 J).
Ans: Qsyst = 0 (c) Similarly, no work crosses the boundary of the room; therefore, Wsyst = 0. Thus:
Usyst = Qsyst + Wsyst = 0 + 0 = 0
Since nothing enters or leaves the room, its internal energy must remain the same; i.e., U = 0. All that happens inside the room is that one form of concentrated energy (the chemical energy con- tained in the wax of the candle) is converted into another, lower grade, form of energy (dispersed thermal energy of the air molecules).
Ans: Usyst = 0
Example 7-2
An insulated container made of a certain metal has a mass of 3.6 kg and contains 14.0 kg of water. Both the container and the water are initially at a temperature of 16.0°C. A 10.0 kg piece of the same type of metal initially at a temperature of 50°C is dropped into the water. After reaching ther- mal equilibrium, the entire system is at 18.0°C. What is the specific heat capacity of the metal? The specific heat capacity of water is (c )P W = 4.18 kJ K–1 kg–1. The process takes place at constant
pressure. Solution
Let the specific heat capacity of the metal be cP J K–1 kg–1.
The heat gained by the container is Q1 = m1cP T1
= (3.6) cP(18.0 – 16.0) = +7.2 cP J
The heat gained by the water is Q2 = m2( cP)w T2
= (14.0)(4180)(18.0 – 16.0) = +117 040 J
The heat lost by the piece of metal is Q3 = m3cP T3
= (10.0) cP(18.0 – 50.0) = –320.0 cP J
Since the system is insulated, no heat escapes, so that for the overall system Qsyst = 0. Thus the sum of the various subsystem internal heat transfers must equal zero:
Q1 + Q2 + Q3 = 0
+7.2 cP + 117 040 – 320.0 cP = 0
312.8 cP = 117 040 from which cP = 374.2 J K–1 kg–1 = 0.3742 kJ K–1 kg–1
Exercise 7-2
A 75 g piece of copper metal is heated to 312°C and dropped into an insulated glass beaker con- taining 220 g of water at 12°C. What is the final temperature after thermal equilibrium has been at- tained? The heat capacity of the glass beaker is Cb = 188 J K–1. The specific heat capacities of wa-
ter and copper are 4.18 kJ K–1 kg–1 and 386 J K–1 kg–1, respectively.
Example 7-3 2
The molar heat capacity at constant pressure for water vapor varies with temperature according to
cP = 30.54 + 0.0103T . . . [a]
where cP is in J mol–1 K–1 and T is the temperature in kelvin.
Calculate: (a) W (b) Q (c) U when one mole of water vapor is heated at constant volume from 25°C to 200°C. Assume ideal gas behavior.
Solution
(a) Since this is a constant volume process, Vsyst = 0; therefore, the PV-work, defined as
WPV= –Pext Vsyst, also must be zero.
Ans: W = 0
(b) QV, the heat at constant volume, is given by QV= ncV T where n is the number of moles of
gas being heated, cV is the molar heat capacity at constant volume, and T = T2 – T1 is the tem- perature difference through which the gas is heated.
We are given cP, the molar heat capacity at constant pressure, but we need to know cV, the molar heat capacity at constant volume, because this is a constant volume process.
For an ideal gas, cP= cV + R;
therefore, cV = cP – R
= (30.54 + 0.0103T) – 8.314 = 22.226 + 0.0103T
We can use the formula QV = n cV T only when cV is constant over the temperature range
T1 T2. In this problem, however, Eqn [a] tells us that cV has a different value at each tempera- ture. However, over an infinitesimally small temperature range from T to T + dT the value of cV will be constant at the value for T; therefore, we can calculate the incremental bit of heat Q over this small temperature range dT using the expression3 Q
V= cVdT and integrate (add up) all
2 This is an important type of calculation often encountered. If you’re deficient in integration, read Appendix 8. 3 In this problem, n = 1 so that QV = cVdT.
the small increments of Q to obtain the total heat Q over the complete temperature range T. Thus: QV = c dTV T T 1 2 = ( T) dT 298.15 473.15 22.226 0.0103+ = 22 226 298 15 473 15 . . . T + 0 0103 2 298 15 473 15 . . . T2 = (22.226)(473.15 – 298.15) + (0.00515) 473 15
(
. 2 298 15. 2)
= 3889.5 + 695.1 = 4584.2 J = +4.58 kJ Ans: QV = + 4.58 kJ Three important points can be made here:First, when calculating a T it doesn’t matter whether you use °C or kelvin, since the difference between the two numbers will always be the same. However, when carrying out integrations we often end up with expressions involving T 2 or even higher powers of T, as well as with the loga-
rithm of T. For these cases you must use the absolute temperature, otherwise you will get the wrong answer. For this reason, when in doubt, it is always safe to express the temperature as the absolute temperature.
Second, be careful to remember that T22 T 1
2
(
)
is not the same as T(
2 T1)
2. Third, note that although ( lna lnb ) = ln ab , ln a
ln b does not equal ln a b .
(c) For a constant volume process UV= QV, therefore U = +4.58 kJ.
Ans: U = + 4.58 kJ
Exercise 7-3
When water is shaken, the mechanical kinetic energy input by each shake dissipates 100% through friction to thermal energy. How long would it take to raise the temperature of 500 mL of water ini- tially at 20°C to its normal boiling point by putting the water into a completely insulated vacuum flask and shaking it? Assume that you can do 40 shakes per minute, and that for each shake the water falls through a distance of 20 cm. The heat capacity of water is 4.18 kJ K–1 kg–1 and the ac-