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Consider 3 objects, each with different heat capacities ($C_1$, $C_2$, $C_3$) and initial temperatures $T_1$, $T_2$, and $T_3$. Heat engines (HE1 and HE2) are placed between the objects to generate work, which powers a heat pump (HP) to transfer heat from object 2 to object 4. Are the final temperatures of objects 1, 2 and 3 all the same? Or is object 2 colder than objects 1 and 3?

I think the final temperatures of objects 1-3 should be the same, since the heat engines will keep running until there are no temperature differences. HE1 will run until $T_1 = T_2$, and HE2 will run until $T_2=T_3$. By the third law of thermodynamics, we therefore have $T_1=T_3$ at this final stage. I don't think the heat pump should influence this result.

Intuitively, however, it seems like object 2 should be colder in the final stage, since it is the only one rejecting heat to a heat pump.

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  • $\begingroup$ Have you tried analyzing this problem quantatively (I.e., modeling the problem)? $\endgroup$ – Chet Miller Sep 18 '18 at 0:28
  • $\begingroup$ @Chester_Miller yes, using $d U = \delta Q + \delta W$ on each system is only useful if I know the final temperature of each system. So I take objects 1 and 2 and realize the heat engine runs until $T_1=T_2$. Well, the same happens with the heat engine between objects 2 and 3. Third law therefore says that $T_1=T_3$. That's really the only analysis I know how to do for this problem. I'm not trying to solve for the final temperature, I just want to determine if the 3rd law applies between objects 1, 2 and 3. $\endgroup$ – Drew Sep 18 '18 at 0:39
  • $\begingroup$ Suppose you assume that the engines and heat pump operate in instantaneous reversible Carnot cycles, based on the current temperatures. I think you can do something quantitative with that. $\endgroup$ – Chet Miller Sep 18 '18 at 0:46
  • $\begingroup$ Heat reservoirs are normally considered to be "infinite". This means that T3>T2>T1 for the "normal" thermodynamic case. It also means that the three temperatures are constant. $\endgroup$ – David White Sep 18 '18 at 1:38
  • $\begingroup$ @David_White, these heat "reservoirs" are actually of finite heat capacity (i.e., normal objects that can change temperature). I made this problem up, so the problem itself may not even make sense or be solvable with the given parameters. $\endgroup$ – Drew Sep 18 '18 at 1:45
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I describe a similar example here in the context of three reservoirs, in which the associated heat engine runs one way (and only needs to run one way). The temperatures of two reservoirs are made equal using a reversible heat engine, and then the extracted energy is used to cool down both reservoirs using a heat pump to maximize the temperature of a third reservoir:

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Your four-component system (for which I'll assume that initially T3 > T2 > T1) introduces an additional factor: can heat engine 1 run in reverse? If not, it's possible that the energy provided by heat engine 2 allows the heat pump to cool T2 below T1. In this case, T2 = T3, which is lower than T1, and T4 is not made as hot as it could be, since a temperature difference still exists among the first three components.

Even if heat engine 1 can operate in reverse (if necessary) but you restrict the order of operation (e.g., heat engine 1 must run before heat engine 2), then you may end up with an inefficient result in which T2 < T1 with no way to exploit this temperature difference.

However, if heat engine 1 can operate in reverse and you allow both heat engines to run as long as a temperature difference remains, then all temperature differences can be exploited among T1, T2, and T3 to make T4 as hot as possible. In this case, as you state, T1 = T2 = T3 ultimately.

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