Minimum energy - Maximum entropy equivalence

Question

I'm going through Callen's Thermodynamics book. After demonstrating from a physical point of view the maximum entropy implies minimum energy, the inverse argument is left as an exercise:

Show that if the entropy were not maximum at constant energy then the energy could not be minimum at constant entropy.

Hint: First show that the permissible increase in entropy in the system can be exploited to extract heat from a reversible heat source (initially at the same temperature) and to deposit it in a reversible work source. The reversible heat source is thereby cooled. Continue the argument.

The minimum energy principle states that for a given entropy, the system will reach equilibrium at minimum energy. Now if we look at the whole system, including the heat and work sources, transferring heat from a reversible source will neither change the energy of the system, nor the entropy (since $ds=\frac{dQ}{dT}$, and $T_1=T_2$). Storing the energy in the work will yet cause no change in the total energy, and since it's reversible neither in the entropy as well. The heat source will be cooled by $\frac{dQ}{C}$. Yet, I can't see where is it taking us.

Answer 1

I think I figured it out. We attach to our system this reversible heat source and consider it as a unit. Transferring $dQ$ indeed cause no change of entropy nor total energy in the system. We than takes this energy out of the system by storing it in a reversible work source. Since it's reversible, this cause no change of entropy as well. However, we bottomed up with losing energy from the system, staying with the same entropy and being at equilibrium, contrary to our assumption that the equilibrium is reached for minimum energy given an entropy. Thus, our original system was originally at maximum entropy, thus couldn't extract heat from the heat source.

Answer 2

About your answer, have you found it to be true? I would say that there is a problem with your reasoning because the hint is that you can increase the entropy.

My answer by following the same thought process as the book:

Since your system has minimum energy and the entropy is not a maximum then you can increase it's entropy by exchanging heat dQ from the heat source with the same temperature T as the system. This revesible isothermal process increases the energy of the system by an amount dU and it would also increase it's entropy since dS=dQ/T. Since the heat source is loosing energy dU it cools.

Now we can use a reversible work source to extract the same amount of energy dU (the amount received from the heat souce) from the system without changing it's entropy. This energy is then transfered to the heat source in the form of heat dQ=dU which then should make it restore it's original temperature T.

At this point the system has an increased amount of entropy dS and the same internal energy as in the initial state, which is suposed to be the minimum for the original equilibrium state (this is the important part).

Now we can have the system give back to the reversible heat source the heat amount dQ wich mean it will decrease it's entropy by the same amount it originally gained (dS) but the heat transfer also means it will decrease it's energy by the amount dU as stated before.

At this point we now have the system with the same entropy as in the initial state of equilibrium but its internal energy is lower than the internal energy which was suposed to be a minimun for that given amount of entropy!

This proves that if the energy is a minimum at constant entropy then the entropy must be maximum at minimum energy.

Both answers arrive at the same conclusion but i'm not sure that your thought process is correct.