Example
This example of an isolated composite thermal system (Fig. 8.6) has been
discussed in Chaps. 3 and 5: blocks X and Y [X is aluminum
c pX ¼ 0:900 kJ/kg Á K
À
Á
with m X = 0.5 kg, Y is copper c pY ¼
À
0:386 kJ/kg Á
KÞ with m Y = 1 kg] initially at 100 and 500 °C, respectively (state A), are
brought together undergoing a spontaneous heat transfer process to a final
state B of 557.84 K (see Problem 3.8) with corresponding entropy change
0.0549 kJ/K (see Problem 5.4).
Now consider a Carnot heat engine operating between X and Y producing
work while the system undergoes a reversible change to an isentropic temperature at internal thermal equilibrium with corresponding reversible work
output (lower left part of Fig. 8.6). This is essentially the reversible event
considered in Fig. 8.5. But, instead of using a special heat reservoir for the
spontaneous event as it was done in Fig. 8.5, here the spontaneous event is
under isolated condition (see the top of Fig. 8.6), involving no reservoir. For
the comparison of the two events, one needs in the reversible event to bring
the system back from state B Isen (522.35 K, after the first isentropic step) to
the final end state B—calling it the second step of the reversible events.
A moment of reflection shows three possible reversible events (of an infinite
number of possibilities) associated with three different heat reservoir temperatures (see Fig. 8.6).
Show: (a) For the first step of the reversible event, the isentropic temperature to be 522.35 K and determine the corresponding reversible work.
(b) Describe the second step of the reversible event for a heat reservoir at
557.85 K and the corresponding additional work output of the second step
(confirming that the overall work output is equal to the prediction of [129]).
(c) Describe the second step of the reversible event for a heat reservoir at
522.35 K and the corresponding heat pump work input of the second step
(confirming that the overall work output is equal to the prediction of [129]).
(d) Consider a second step of the reversible event for a heat reservoir at a
particular given temperature so that according to (129), the overall work
output of the reversible event (for both steps) is the same as the isentropic
work. That is, the second step requires no (nor produces) additional work.
Determine the reservoir temperature.
Solution
(a)
0 ¼ m X c pX ln
T Isen
373:15
þ m Y c pY ln
T Isen
773:15
¼ 0:5 Â 0:9 Â ln
T Isen
373:15
þ 1 Â 0:386 Â ln
T Isen
773:15
T Isen ¼ 522:3521 K
208
8 The Second Law: The Entropy Growth Potential Principle …
This example of an isolated composite thermal system (Fig. 8.6) has been
discussed in Chaps. 3 and 5: blocks X and Y [X is aluminum
c pX ¼ 0:900 kJ/kg Á K
À
Á
with m X = 0.5 kg, Y is copper c pY ¼
À
0:386 kJ/kg Á
KÞ with m Y = 1 kg] initially at 100 and 500 °C, respectively (state A), are
brought together undergoing a spontaneous heat transfer process to a final
state B of 557.84 K (see Problem 3.8) with corresponding entropy change
0.0549 kJ/K (see Problem 5.4).
Now consider a Carnot heat engine operating between X and Y producing
work while the system undergoes a reversible change to an isentropic temperature at internal thermal equilibrium with corresponding reversible work
output (lower left part of Fig. 8.6). This is essentially the reversible event
considered in Fig. 8.5. But, instead of using a special heat reservoir for the
spontaneous event as it was done in Fig. 8.5, here the spontaneous event is
under isolated condition (see the top of Fig. 8.6), involving no reservoir. For
the comparison of the two events, one needs in the reversible event to bring
the system back from state B Isen (522.35 K, after the first isentropic step) to
the final end state B—calling it the second step of the reversible events.
A moment of reflection shows three possible reversible events (of an infinite
number of possibilities) associated with three different heat reservoir temperatures (see Fig. 8.6).
Show: (a) For the first step of the reversible event, the isentropic temperature to be 522.35 K and determine the corresponding reversible work.
(b) Describe the second step of the reversible event for a heat reservoir at
557.85 K and the corresponding additional work output of the second step
(confirming that the overall work output is equal to the prediction of [129]).
(c) Describe the second step of the reversible event for a heat reservoir at
522.35 K and the corresponding heat pump work input of the second step
(confirming that the overall work output is equal to the prediction of [129]).
(d) Consider a second step of the reversible event for a heat reservoir at a
particular given temperature so that according to (129), the overall work
output of the reversible event (for both steps) is the same as the isentropic
work. That is, the second step requires no (nor produces) additional work.
Determine the reservoir temperature.
Solution
(a)
0 ¼ m X c pX ln
T Isen
373:15
þ m Y c pY ln
T Isen
773:15
¼ 0:5 Â 0:9 Â ln
T Isen
373:15
þ 1 Â 0:386 Â ln
T Isen
773:15
T Isen ¼ 522:3521 K
208
8 The Second Law: The Entropy Growth Potential Principle …