Proving Entropy statement is equivalent to Clausius statement

In summary, the conversation discusses the application of the change in entropy equation for solids, where the temperature for each solid is not constant during heat exchange. It is mentioned that this is done by only allowing a small amount of heat transfer to prevent significant temperature changes before the bodies are separated. After separation, each body is allowed to re-equilibrate. The conversation also mentions a more precise way to write the equation for this process.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
##\Delta S = \frac{Q}{T}##
For this,
1680464760625.png

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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  • #2
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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  • #3
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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