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## Main Question or Discussion Point

In what follows I refer to the ideas of "Thermodynamic: Foundations and Applications" by Gyftoploulos and Beretta. The abbreviated form of my question is: In a reversible weight process,

Ω

When the above equation is true, why does the book occasionally (problem 6.4, example 6.4 on p. 94) calculates the energy change of the reservoir as E

Because, according to the equation above, E

My proof of E

Given two states, A

α: A

β: A

The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω

α: Ω

β: Ω

Comparing both results, one gets that all states of the reservoir have the same energy: E

So: Ω

The last equation corresponds to equation 6.18, p. 99.

Ω

_{1}^{R}-Ω_{2}^{R}= E_{1}- E_{2}(see eqn. 6.18, p. 99) is transferred out of the composite of a system A and a reservoir R to the weight, where Ω^{R}is the available energy of a system A with respect to reservoir R and E is the energy of system A.When the above equation is true, why does the book occasionally (problem 6.4, example 6.4 on p. 94) calculates the energy change of the reservoir as E

_{1}^{R}- E_{2}^{R}≠ 0 for the case of reversible weight processes?Because, according to the equation above, E

_{1}^{R}= E_{2}^{R}must be true (see below). Also, (DE_{12}^{R})_{rev}^{A}should always be 0, p. 108/9. However, this quantity is used extensively in the discussion of entropy as if it was non-zero.My proof of E

_{1}^{R}= E_{2}^{R}:Given two states, A

_{1}and A_{2}, of a system A and a reservoir R. Consider two reversible weight processes, α und β, consisting of two sub-processes, where A_{0}R_{0}is the state where A and R are in mutual stable equilibrium. The only difference is the final state of the reservoir (R_{1}in process β instead of R_{2}in process α).α: A

_{1}R_{1}→ A_{0}R_{0}→ A_{2}R_{2}β: A

_{1}R_{1}→ A_{0}R_{0}→ A_{2}R_{1}The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω

_{1}^{R}. The second sub-process in α and the second sub-process in β both transfer the available energy Ω_{2}^{R}to the weight, because the stable equilibrium state is unique (see proofs on adiabatic availability on p. 76 and regard the adiabatic availability of A+R as the available energy von A) and Ω_{2}^{R}is independent of the initial (or end state in this case) of the reservoir (statement 3 on p. 89). The energy balances for both processes are:α: Ω

_{1}^{R}-Ω_{2}^{R}= E_{1}- E_{2}+ E_{1}^{R}- E_{2}^{R}β: Ω

_{1}^{R}-Ω_{2}^{R}= E_{1}- E_{2}Comparing both results, one gets that all states of the reservoir have the same energy: E

_{1}^{R}= E_{2}^{R}So: Ω

_{1}^{R}-Ω_{2}^{R}= E_{1}- E_{2}The last equation corresponds to equation 6.18, p. 99.