3
$\begingroup$

While reading my textbook , I came across a proof which intended to verify that entropy always increases when a hot reservoir is kept in thermal contact with a colder reservoir.

The proof goes as follows :

Let the temperature of reservoir A be $T_A$ and the temperature of reservoir B be $T_B$

Two reservoirs in thermal contact

Change in entropy of Reservoir A = $\frac{-Q}{T_A}$

Change in entropy of reservoir B= $\frac{Q}{T_B}$

Total change in entropy = $\frac{-Q}{T_A}+\frac{Q}{T_B}$ =$Q(\frac{1}{T_B}-\frac{1}{T_A})$

Since, $T_A>T_B$

Therefore, total change in entropy is positive. Hence, we can say that entropy of the universe increases in this particular system.

My confusion : The total entropy change is positive which clearly implies that both the reservoirs are undergoing irreversible processes. And we know that for an irreversible process,$$dS>\frac{dQ}{T}$$ Then how can they write :

change in entropy of reservoir A = $\frac{-Q}{T_A}$.

Shouldn't they replace '=' with '>' here?

Similarly, for reservoir B they should have written

Change in entropy of reservoir B> $\frac{Q}{T_B}$

Where I am going wrong with my reasoning?

$\endgroup$
2

2 Answers 2

1
$\begingroup$

The irreversibility (entropy generation) occurs at the interface between the reservoirs, and not within the reservoirs themselves. Each of the ideal reservoirs individually experiences a reversible change.. –

$\endgroup$
0
$\begingroup$

Denote the transported entropy by $S_A$ from body $A$ at temperature $T_A$. The same entropy falls from temperature $T_A$ to temperature $T_B$ and the thermal work being dissipated is exactly $\Delta W = (T_A-T_B)S_A$ at the lower temperature $T_B$ and thereby generating in the heat conducting interface entropy in the amount of $\Delta S = S_B-S_A>0$ where $S_B$ is the amount of entropy entering body $B$ at temperature $T_B$. But the dissipated work is also $\Delta W = T_B \Delta S$ because $T_BS_B=T_AS_A$ by the formal constitutive definition of a heat conductor, an equality that holds at every cross section of a conductor $T(x)S(x) = \text{constant}=Q$ the incoming thermal energy, "heat".

If the size of each body is finite, then, of course, as entropy leaves the higher temperature side and enters lower temperature one, the difference in temperatures gets closer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.