Modern Engineering Thermodynamics

272 CHAPTER 8: Second Law Closed System Applications
When the chambers contain identical ideal gases, Eqs. (3.38) and (8.9) give
or
u 2 − u b = c v ðT 2 − T b
Þ = yc v ðT a − T b Þ
T 2 = T b + yT ð a − T b Þ (8.11)
and Eqs. (7.37) and (8.10) give
1ðS P Þ 2
= ðm a + m b Þ c p ln T 2
− R ln p
2
+ y c p ln T b
− R ln p
b
T b p b T a p a
or
and inserting Eq. (8.11) for T 2 gives
1ðS P Þ 2 = ðm a + m b Þ c p ln ½ðT 2 /T b ÞðT b /T a Þ y Š− R ln ½ðp 2 /p b Þðp b /p a
Entropy production when two identical ideal gases are mixed
1ðS P Þ 2
= ðm a + m b Þ c p ln ½ð1 + yT ð a /T b − 1ÞÞðT b /T a Þ y Š− R ln ½ðp 2 /p b Þðp b /p a Þ y Š ≥ 0 (8.12)
When the chambers contain identical incompressible liquids, Eqs. (3.33), (8.9), and (8.10) can be combined to
yield
Entropy production when two identical incompressible liquids are mixed
1ðS P Þ 2
= ðm a + m b Þc ln ½ð1 + yT ð a /T b − 1ÞÞðT b /T a Š≥ 0 (8.13)
The actual mixing process need not have the same two-chamber geometry shown in Figure 8.18, as illustrated by
the following example.
Þ y
Þ y
Š
EXAMPLE 8.15
Determine the entropy produced when 3.00 g of cream at 10.0°C are added adiabatically and without stirring to 200. g of
hot coffee at 80.0°C. Assume both the coffee and the cream have the properties of pure water.
Solution
First, draw a sketch of the system (Figure 8.19).
Cream (water)
m = 3.00 g
T cream = 10.0 °C
1 (S P ) 2 = ?
Coffee (water)
m = 200. g
T coffee = 80.0 °C
FIGURE 8.19
Example 8.15.
The unknown is the entropy produced by mixing, and the materials are coffee and cream, both modeled as liquid water.
Let a = cream and b = coffee. Then, y = 3.00/203 = 0.0148. Assuming both the coffee and the cream are incompressible
liquids with the specific heat of water, c = 4186 J/(kg·K), Eq. (8.13) gives
1ðS P
Þ 2
= ð0:203 kg
= 0:282 J/K
(
) 0:0148
Þ½4186 J/ ðkg⋅KÞŠln 1 + 0:0148
10:0 + 273:15
80:0 + 273:15 − 1 ×
80:0 + 273:15
10:0 + 273:15