Assume two identical solid blocks of copper with molar heat capacity at constant volume $C_{V,m}=24.5 \text{ J$\cdot$K$^{-1}$mol$^{-1}$}$ . Assume the heat capacity to be constant in the temperature range of the experiment. Block 1 has a temperature $T_1$ and block 2 has a temperature $T_2=4T_1$. The copper blocks are thermally isolated from the surrounding. Only energy exchange between the two blocks is possible.

a) What is the final temperature $T_{final}$, both blocks will eventually reach, when they are in thermal contact?

b) Calculate the entropy change $\Delta S$ for each of the blocks. What is the total change in entropy? Would it have been possible to predict the sign (minus or plus) of the entropy change (give arguments!).

c) Now, instead of direct thermal contact, a reversible thermodynamic machine is placed between both copper blocks at their initial temperatures $T_1$ and $T_2$. What is the total entropy change now (give arguments)?

d) Calculate the entropy change $\Delta S$ for each of the blocks. Use the result to show, that for the case of the reversible thermodynamic machine $T_{final}=2T_1$.

e) Show, that the work done by the reversible thermodynamic machine is: $w= C_{V,m}T_1$

I tried to answer this question as part of an exam preparation (no homework) and got stuck at part C. Part A and B are quite trivial. You can show that for part A, $T_{final}=2.5T_1$. For part B we see that

$$\Delta S = \int_{T_1}^{T_{final}}\frac{C_{V,m}\mathrm{d}T}{T}+\int_{T_2}^{T_{final}}\frac{C_{V,m}\mathrm{d}T}{T}=C_{V,m}\ln \frac{25}{16}\approx 10.9 \text{ J$\cdot$K$^{-1}$mol$^{-1}$}$$ Now I do not quite get C and D (E follows very easily from the first law and the answer found at part D, i.e. $T_{final}=2T_1$). What is exactly meant by a reversible thermodynamic machine, and moreover, how do I calculate part C and D? I can see that the total entropy change will be lower when some of the thermal energy is used to do work, but I don't know how to quantitatively calculate the new entropy change.


Some hints:

  • In C, the "reversible thermodynamic machine" is meant to be some gadget that transfers heat between the blocks reversibly. So what's $\Delta S$ for the universe (both of the blocks together) if the process is thermodynamically reversible? (What would it be for any thermodynamically reversible process?)

  • In D, you can solve $\Delta S_{universe} = \Delta S_1 + \Delta S_2$ for $T_f$, using the $\Delta S_{universe}$ you obtained in C and the entropy changes for the individual blocks. You should see that $T_f = 2 T_1$, as the problem states.

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  • $\begingroup$ Hmm, of course. $\Delta S_{system}=0$ for a reversible process as seen from the Clausius inequality. The fancy machinery must perform some kind work to be able to keep $ S_{system}$ unchanged as direct flow of heat to a cooler area is irreversible. The internal working of the machine is unimportant as long as it uses a cycle with no entropy change (like the Carnot cycle). Am I right? Thanks a lot! $\endgroup$ – Jori Jan 20 '15 at 17:47
  • $\begingroup$ You're welcome! But be careful about "system" and "universe". When you compute the entropy change of just one of the blocks, you're calling that the system and the other block the surroundings. The universe is both blocks together, system + surroundings, and THAT is zero, not the entropy change of the individual blocks. $\endgroup$ – Fred Senese Jan 20 '15 at 18:14

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