Modern Engineering Thermodynamics

62 CHAPTER 3: Thermodynamic Properties
EXAMPLE 3.2 (Continued )
and multiplying both sides of this equation by the mass m of the block gives the total volume V as
V 2 = mv 2 = V 1 fexp½βðT 2 − T 1 ÞŠg
It can be seen from Table 3.1 that β for copper is not constant in the temperature range of 250. to 800. K. To come up with
a reasonable value for an average β, we must see how β varies with temperature. Figure 3.2 shows the data for β vs. T for
copper taken from Table 3.1.
80
70
β × 10 6 (K −1 )
60
50
40
30
β vs. T for copper (data from Table 3.1)
FIGURE 3.2
Example 3.2, solution.
20
0 200 400 600 800 1000 1200
Temperature (K)
This figure shows that β varies linearly with T in the range of 250. to 800. K. The average value of β in this temperature range is
easily found to be
Now we can calculate the final volume as
β avg = 60:7 × 10−6 + 48:0 × 10 −6
2
= 54:4 × 10 −6 K −1
V 2 = mv 2 = V 1 fexp ½βðT 2 − T 1 ÞŠg
= ð1:00 cm 3 Þ exp½ð54:4 × 10 −6 K −1 Þð800: − 250: KÞŠ
= 1:03 cm 3
Note that we could also fit a straight line to the β vs. T data between 250. and 800. K and come up with a formula of the
form β = C 1 T + C 2 . Inserting this formula into Eq. (3.7) and integrating it (with dp = 0) yields a different (but equally valid)
relation among v 1 , v 2 , T 1 , and T 2 . This is left as an exercise at the end of this chapter.
Exercises
3. Use Tables 3.1 and 3.2 to find values for the isobaric coefficient of volume expansion β and the isothermal coefficient of
compressibility κ for (a) copper at 1200. K, (b) benzene at 68°F, and (c) mercury at 20.°C. Answers: (a) β = 38.7 × 10 –6
R –1 = 69.7 × 10 –6 K –1 and κ = 49.30 10 –11 ft 2 /lbf = 1.030 × 10 –11 m 2 /N; (b) β = 0.689 × 10 –6 R –1 = 1.240 × 10 –6 K –1
and κ = 4550 × 10 –11 ft 2 /lbf = 95.0 × 10 –11 m 2 /N; (c) β = 0.101 × 10 –6 R –1 = 0.182 × 10 – K –1 and κ = 192 × 10 –11 ft 2 /
lbf = 4.02 × 10 –11 m 2 /N.
4. Rework Example 3.2 for a 1.00 cm 3 block of solid platinum, whose average isobaric coefficient of volume expansion
over the temperature range from 250. K to 800. K is 3.00 × 10 –5 K –1 . Answer: V = 1.017 cm 3 .
5. Liquid water at 68°F is isothermally compressed from 14.7 psia to 3000. psia. Determine the percent change in the
volume of the water, [(v 1 − v 2 )/v 1 ] × 100. Answer: 0.94%.
3.5 SYSTEM ENERGY
For historical reasons, the total energy of a system that has no magnetic, electric, surface, or other effects is
divided into three parts. Classical physicists recognized two easily observable forms of energy: (1) the total
kinetic energy KE = mV 2 /2g c , and (2) the total potential energy PE = mgZ/g c . The remaining unobservable part of
the total energy is simply called the total internal energy U. Thus, the total energy E of a system is written as
or
E = U + mV 2 /2g c + mgZ/g c (3.9)
E = U + KE + PE (3.10)