Temperature of two subsystems at equilibrium

[f3sicA_A]

Homework Statement

Prove that the temperatures of two subsystems at equilibrium separated by an immovable impermeable wall which only allows for the transfer of heat are equal.

Relevant Equations

At equilibrium $dS=0$, $(\partial U/\partial S)_{V, N}=T$.

I was going through the textbook Thermodynamics and an Introduction to Thermostatistics by Herbert B. Callen, and in Chapter 2.4 of the book, the author proves that given two subsystems separated by an impermeable and immovable wall which only allows for the transfer of heat, the temperature of the two subsystems are equal at equilibrium. The proof goes as follows:

At Equilibrium $dS=0$, and the energy conservation condition states:

$$U^{(1)}+U^{(2)}=\text{const.},$$

where $U^{(1)}$ and $U^{(2)}$ are the internal energies of subsystem 1 and subsystem 2 respectively. Since entropy is defined as a function which is additive over the subsystems:

$$S=S^{(1)}(U^{(1)},V^{(1)},N^{(1)})+S^{(2)}(U^{(2)},V^{(2)},N^{(2)})$$

$$\implies dS=\left(\frac{\partial S^{(1)}}{\partial U^{(1)}}\right)_{V^{(1)},N^{(1)}}dS^{(1)}+\left(\frac{\partial S^{(2)}}{\partial U^{(2)}}\right)_{V^{(2)},N^{(2)}}dS^{(2)}.$$

Since at equilibrium $dS=0$, and $(\partial S/\partial U)_{V,N}=1/T$, we have:

$$\frac{1}{T^{(1)}}dU^{(1)}+\frac{1}{T^{(2)}}dU^{(2)}=0$$

Now, by the energy conservation condition, we have $dU^{(1)}=-dU^{(2)}$, so:
$$\left(\frac{1}{T^{(1)}}-\frac{1}{T^{(2)}}\right)dU^{(1)}=0.$$

Over here the author claims that since the above statement must be true for any arbitrary $dU^{(1)}$, we have that $1/T^{(1)}=1/T^{(2)}$ that is $T^{(1)}=T^{(2)}$. However, my problem here is, given that the system is at equilibrium, that is the two subsystems are also at equilibrium, shouldn't $dU^{(1)}=dU^{(2)}=0$? And given $dU^{(1)}=0$, there would be no conclusive way to say that $1/T^{(1)}-1/T^{(2)}=0$ as well, at least from this derivation.

Is my idea of $dU$ being 0 at equilibrium wrong?

 

[Lnewqban]

Is my idea of $dU$ being 0 at equilibrium wrong?

I would risk stating so.

You could have fluid contained inside each subsystem that has tremendous internal energy, but similar temperature to the fluid contained next door (steam, for example).

Replacing that fluid with ice-water mix, would reduce the magnitude of the internal energy, but again, the temperature of both subsystems would reach the same value once thermal equilibrium through the separating wall is achieved.

Please, see:
https://en.wikipedia.org/wiki/Internal_energy

 

[Vanadium 50]

However, my problem here is, given that the system is at equilibrium, that is the two subsystems are also at equilibrium,

Is that necessarily true?

 

[robphy]

In the paragraph after
$$U^{(1)}+U^{(2)}=\text{const.},$$
Callen writes (emphasis mine)

According to the fundamental postulate, the values of $U^{(1})$ and $U^{(2})$ are
such as to maximize the entropy. Therefore, by the usual mathematical
condition for an extremum, it follows that in the equilibrium state a
virtual infinitesimal transfer of energy from system 1 to system 2 will
produce no change in the entropy of the whole system.

suggests (to me) that $dU^{(1)}$ is not zero. (As you quote, there is an allowed transfer of heat and (since the wall is immovable) no work is done.)

 

[f3sicA_A]

Is that necessarily true?

I guess not. Since we have that:

$$dS=dS^{(1)}+dS^{(2)},$$

given $dS=0$, we can only say $dS^{(1)}=-dS^{(2)}$ but we cannot claim that both are 0 as well (the necessary condition for an equilibria of the two subsystems). Is the idea here to say that whilst the entropy over the entire system might be maximized, the entropy of the individual subsystems is not necessarily maximized, and to maximize the entropy over a particular subsystem could lead to the fall in the entropy of the other subsystem such that the overall entropy of the entire system goes down (and hence not maximized)?

 

[f3sicA_A]

suggests (to me) that $dU^{(1)}$ is not zero.

I see, it seems I have made an unmotivated assumption. As long as $dU$ is infinitesimal there would be no change in entropy owing to $dS=0$, and so at equilibrium such infinitesimal interchange of internal energy between the two subsystems is not disallowed (since that does not lead to a change in the entropy of the eqm. state). Thank you for the clarification.

 

[f3sicA_A]

You could have fluid contained inside each subsystem that has tremendous internal energy, but similar temperature to the fluid contained next door (steam, for example).

I realize that two systems of the same temperature could have vastly different internal energies; however, this in itself does not lead me to believe that there should be an exchange of internal energies between two such systems at equilibrium. What does convince me is Callen's argument (as @robphy pointed out) that at least for infinitesimal energy changed given that $dS=0$, there would be no change in the entropy, and so infinitesimal exchange in internal energy does not seem to be disallowed.

 

[robphy]

@f3sicA_A It might be helpful to read the next section 2-5 and section 2-7 (which generalizes to include mechanical equilibrium).