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Solved: Unavailable Energy. The discussion of entropy
Chapter 20, Problem 57P(choose chapter or problem)
Problem 57P
Unavailable Energy. The discussion of entropy and the second law that follows Example 20.10 (Section 20.7) says that the increase in entropy in an irreversible process is associated with energy becoming less available. Consider a Carnot cycle that uses a low-temperature reservoir with Kelvin temperature \(T_{c}\). This is a true reservoir—that is, large enough not to change temperature when it accepts heat from the engine. Let the engine accept heat from an object of temperature \(\mathrm{T}^{\prime}\), where \(\mathrm{T}^{\prime}>\mathrm{T}_{\mathrm{c}}\). The object is of finite size, so it cools as heat is extracted from it. The engine continues to operate until \(T^{\prime}=T_{c}\). (a) Show that the total magnitude of heat rejected to the low-temperature reservoir is \(T_{\mathrm{c}}\left|\Delta S_{\mathrm{h}}\right|\), where \(\Delta S_{h}\) is the change in entropy of the high-temperature reservoir. (b) Apply the result of part (a) to 1.00 kg of water initially at a temperature of 373 K as the heat source for the engine and \(T_{C}=273\) K. How much total mechanical work can be performed by the engine until it stops? (c) Repeat part (b) for 2.00 kg of water at 323 K. (d) Compare the amount of work that can be obtained from the energy in the water of Example 20.10 before and after it is mixed. Discuss whether your result shows that energy has become less available.
Equation Transcription:
Text Transcription:
T_C
T’
T’>T_C
T’=T_C
T_C|deltaS_h|
deltaS_h
T_C=273
Questions & Answers
QUESTION:
Problem 57P
Unavailable Energy. The discussion of entropy and the second law that follows Example 20.10 (Section 20.7) says that the increase in entropy in an irreversible process is associated with energy becoming less available. Consider a Carnot cycle that uses a low-temperature reservoir with Kelvin temperature \(T_{c}\). This is a true reservoir—that is, large enough not to change temperature when it accepts heat from the engine. Let the engine accept heat from an object of temperature \(\mathrm{T}^{\prime}\), where \(\mathrm{T}^{\prime}>\mathrm{T}_{\mathrm{c}}\). The object is of finite size, so it cools as heat is extracted from it. The engine continues to operate until \(T^{\prime}=T_{c}\). (a) Show that the total magnitude of heat rejected to the low-temperature reservoir is \(T_{\mathrm{c}}\left|\Delta S_{\mathrm{h}}\right|\), where \(\Delta S_{h}\) is the change in entropy of the high-temperature reservoir. (b) Apply the result of part (a) to 1.00 kg of water initially at a temperature of 373 K as the heat source for the engine and \(T_{C}=273\) K. How much total mechanical work can be performed by the engine until it stops? (c) Repeat part (b) for 2.00 kg of water at 323 K. (d) Compare the amount of work that can be obtained from the energy in the water of Example 20.10 before and after it is mixed. Discuss whether your result shows that energy has become less available.
Equation Transcription:
Text Transcription:
T_C
T’
T’>T_C
T’=T_C
T_C|deltaS_h|
deltaS_h
T_C=273
ANSWER:
Solution to 57P
Step 1
Reservoir temperature =Tc
Temperature accepted from the body =T’
Heat flow when the reservoir temperature is T’=dQH
We know that for an irreversible process, the change in entropy is given by △S
△S=dS for an infinitesimal change in entropy.