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I am trying to derive the change in entropy for a reversible process.

Let's say there is a body of mass $m$ with some specific heat that is heated reversibly from from a temperature $T_i$ to $T_f$.

Now, we know that the change in energy is $dE = TdS$. We also know that $\frac{dE}{dT} = MC$. Now, if I take $\frac{dE}{dT}$, I get $ dS+T\frac{dS}{dT}$.

Usually, however, I see that the final conclusion is $\frac{dE}{dT} = T\frac{dS}{dT}$. Where did the final $dS$ disappear?

reesespieces
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1 Answers1

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Now, we know that the change in energy is $dE = TdS$. We also know that $\frac{dE}{dT} = MC$. Now, if I take $\frac{dE}{dT}$, I get $ > dS+T\frac{dS}{dT}$.

I don't see how you got $dS+T\frac{dS}{dT}$.

If

$$dE=TdS$$ and

$$\frac{dE}{dT} = MC$$

it follows that

$$TdS=MCdT$$

or

$$dS=\frac{MCdT}{T}$$

Which is the same as the entropy definition (where $\delta Q_{rev}=dE$) of

$$dS=\frac{\delta Q_{rev}}T=\frac{MCdT}{T}$$

Then, after integrating, the change in entropy between the initial and final temperatures are

$$\Delta S=\int _{Ti}^{Tf}\frac{\delta Q_{rev}}{T}=\int_{Ti}^{Tf}\frac{mMC}{T}dT=MC\ln\frac{T_f}{T_i}$$

Hope this helps.

Bob D
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