Thermodynamics – How a Reversible Transformation Allows Finite Temperature Heat Transfer

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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

Best Answer

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.

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