Proving Entropy statement is equivalent to Clausius statement

AI Thread Summary
The discussion addresses the application of the change in entropy equation for solids during heat exchange before reaching thermal equilibrium. It clarifies that a small amount of heat transfer is allowed, preventing significant temperature changes in the solids until they are separated. After separation, each solid can re-equilibrate independently. The correct equations for final temperatures and entropy change are provided, emphasizing the importance of not allowing the combined system to reach thermal equilibrium during the process. The conversation highlights the nuances of applying thermodynamic principles in specific scenarios.
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Homework Statement
Please see below
Relevant Equations
##\Delta S = \frac{Q}{T}##
For this,
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I don't understand how we can apply the change in entropy equation for each solid since the ##\frac{dT}{dt}## for each solid will be non-zero until the solids reach thermal equilibrium. My textbook says that the ##\Delta S## for a system undergoing a reversible process at constant temperature is given by
##\Delta S = \frac{Q}{T}##, however, the temperature of the each solid is not constant while the heat is getting exchanged.

Dose anybody please know what allows them to do that?

Many thanks!
 
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They are allowing only a small amount of heat to transfer, so that the temperatures of the two bodies do not significantly change before they are separated. They are not allowing the combined system to reach thermal equilibrium. After the bodies are separated, each body is allowed to re-equilibrate by itself.

To be more precise, they should write $$T_{1f}=T_1-\frac{Q}{mC}$$ $$T_{2f}=T_2+\frac{Q}{mC}$$and $$\Delta S=mC\ln{(T_{1f}/T1)}+mC\ln{(T_{2f}/T2)}$$
 
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Likes member 731016
Chestermiller said:
They are allowing only a small amount of heat to transfer, so that the temperatures of the two bodies do not significantly change before they are separated. They are not allowing the combined system to reach thermal equilibrium. After the bodies are separated, each body is allowed to re-equilibrate by itself.

To be more precise, they should write $$T_{1f}=T_1-\frac{Q}{mC}$$ $$T_{2f}=T_2+\frac{Q}{mC}$$and $$\Delta S=mC\ln{(T_{1f}/T1)}+mC\ln{(T_{2f}/T2)}$$
Thank you for your help @Chestermiller ! That is very helpful!
 
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