A reversible transformation, but there is finite temperature heat transfer?

Question

Suppose to have 3 heat reservoirs, the first at temperature $T_1$, the second at temperature $T_1+dT$ and the third at temperature $T_2>T_1$

Then, consider a system, which volume is constant, in thermal equilibrium with the colder reservoir, $T_{system}=T_1$, and do the following steps (order matters):

1. put the system in contact with the hotter reservoir, and wait until $T_{system}=T_2$
2. put the system in contact with the reservoir at temperature $T_1+dT$ and wait until $T_{system}=T_1+dT$

Note that, the thermodynamic equilibrium state of the system, at the initial condition point, is infinitesimally close to that of point, 2. So, considering those two equilibrium states, we can write :

$dE=TdS-PdV=Q+W$

However, since $dV=0$ by assumption, and $W=0$:

$dE=TdS=Q$, that means $dS=Q/T$.

Since the last relation holds only for reversible processes, the transformation is reversible. This seams paradoxical, because in the transformation there is heat transfer with finite temperature difference.

EDIT:

Why don't we get rid of the reservoir at $T_1+dT$ and just bring the system back into contact with the first reservoir?

The reservoir at $T_1+dT$ is a ploy to write down differentials and obtain $Q/T = dS$. Clausius inequality states $Q/T \leq dS$, where the equality hold for reversible transformation. So, since, here $Q/T = dS$, the transformation seems to be reversible.

EDIT:

Thanks you, i figure out that i was wrong, the transformation is not irreversible. I was confused about the expression $\delta Q/T$, that, many times, is written in an ambiguous way. I think the following recap about $\delta Q/T$ could be useful.

• $dS =\delta Q_{rev} /T$ is the thermodynamic definition of entropy. Note that, for a heat exchange to be reversible, temperature differences must be infinitesimal, so, $T=T_{surr}$, and, $dS =\delta Q_{rev} /T$ is equivalent to $dS =\delta Q_{rev} /T_{surr}$

• $dS \ge \delta Q_{real}/T_{surr}$ is the Clausius inequality, where the equality holds only for reversible transformation. So, if $dS=\delta Q/T_{surr}$, the transformation is reversible, that means $T=T_{surr}$, that means $dS=\delta Q/T_{surr}$

• It's important to note that $dS=\delta Q/T$ doesn't imply reversibility, because there are transformations in which $T \neq T_{surr}$, as in the example above. But, $dS=\delta Q/T_{surr}$ imply reversibility, and, in this case, $dS=\delta Q/T$, is also true

Answer 1

The equation dU=TdS-PdV holds for any pair of closely neighboring thermodynamic equilibrium states, even if they represent the starting and end points for an irreversible path. If you had devised a reversible path between the same two end points, you would have obtained the same entropy change, which is essentially what you did.

Answer 2

Since the last relation holds only for reversible processes, the transformation is reversible.

No, that process involving $T_2$ is not reversible, because it involves irreversible transfer of heat from body of temperature $T_2$ to body of different temperature $T_1$.

The relation

$$ dS = \frac{dQ}{T}, $$ where $S$ is entropy of the system studied and $T$ is its temperature, holds true also when the system partakes in some irreversible processes.

E.g. consider a process involving slow enough heat acceptance from a hotter reservoir, so that temperature of the system changes only slowly and remains uniform throughout the system, so it always has single temperature $T$. System's internal energy is function of entropy and volume $U(S,V)$, and the process can be such that all variables $U,S,T,P$ are defined at every instant (quasi-static process) while the system accepts heat from the hotter body. Then $dS$ and $dV$ define $dU = \frac{\partial U}{\partial S}dS - \frac{\partial U}{\partial V}dV$. If no change of volume and no work is allowed, $dU$ is equal to $dQ$ and thus $dQ$ is determined by $dS$ to be $TdS$; it can't be chosen arbitrarily by choosing some different path in the space of equilibrium states, because the path is fixed by the condition $dV = 0$.

You are probably confused by the fact there is also the general relation (assuming single source of heat)

$$ dS \geq \frac{dQ}{T_r} $$ where $T_r$ is temperature of the source of heat (reservoir). This can be sharpened/replaced by equality for any part of the process that is reversible (and then $T_r=T$). For irreversible transformations, the equality may not be valid, that's why we write defensively the unsharp inequality; but the equality is also not forbidden to be valid, especially if we change the formula by putting in $T$ instead of $T_r$.

Answer 3

For a simpler example of the same thing, get rid of the reservoir at $T_1+dT$ and just bring the system back into contact with the first reservoir. You now have a cyclic process, where the final state is the same as the initial state, and so must have the same entropy (and so $dS=dQ/T=0$), but if the process of irreversible then entropy must have increased. The solution is that we have neglected the entropy of the reservoir at $T_2$, which will have increased during the various heat transfer processes