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

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

• The $T$ in the Clausius inequality is not equal to the temperature of the system. It's the temperature of the immediate environment of the system (or the boundary of the system in an analysis where you're using a control volume). Commented Feb 15, 2023 at 16:15
• The last edit is almost fine, except I would not write the "rev" in $\delta Q_{rev}$. The formula $dS = dQ/T$ is sometimes valid even when $dQ$ is exchanged during a process that is overall (when looking at the whole supersystem) irreversible. The decisive factor is, is the system going only through equilibrium states (then the formula holds) or not (then it need not hold, especially when there is no single temperature $T$). Commented Feb 15, 2023 at 19:34
• @JánLalinský Consider an irreversible transformation in which $dS=\delta Q_{irr}/T$ holds, as the one in the example above. Even in this case, i could device a reversible transformation, that connects the same initial and final equilibrium states, and write, $dS= \delta Q_{rev}/T$. In this case, $dS=\delta Q_{irr}/T=\delta Q_{rev}/T$ Commented Feb 16, 2023 at 8:32
• Yes, that's possible. Commented Feb 16, 2023 at 15:12

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.

• sorry, i think the question was not clear, i edited the post enlightening that, the point of my question, is that $dS=Q/T$ only for reversible trasformations Commented Feb 15, 2023 at 12:44
• For the irreversible process you considered, $$\Delta S>\frac{Q}{T_2}-\frac{Q}{T_1}$$where $Q=mc(T_2-T_1)$ which satisfies the Clausius Inequality Commented Feb 15, 2023 at 12:55
• ok, but from the differentials in my question, it seems that $dS=Q/T$ Commented Feb 15, 2023 at 12:59
• are you suggesting that $dS=Q/T$ can hold also for non reversible transformations? Commented Feb 15, 2023 at 13:04
• The Q/T you wrote is not the integral of $\int{dQ/T_{boundaary}}$ for the irreversible path you defined. Commented Feb 15, 2023 at 13:09

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 $$T_r$$.

• Please provide an example of an irreversible process of a closed system for which dS=dQ/T, and, in an irreversible process involving spatial temperature variations within the system, which T value would you use? Commented Feb 15, 2023 at 15:59
• @ChetMiller Let there be a hot metal cube 2 and a cold metal cube 1, both connected with each other by a long wooden beam of negligible heat capacity, otherwise all objects are thermally insulated. Heat will pass from the hotter cube to the colder one, but very slowly, so both cubes will maintain uniform temperature (independent of position within the cube). For the cold cube we have $dS_1 = dQ/T_1$, for the hot cube $dS_2 = -dQ/T_2$, and for the cold cube we can also write the famous inequality as $dS_1 > dQ/T_2$ ($T_2$ being temperature of the source of heat). Commented Feb 15, 2023 at 16:26
• If there are variations of temperature within the system, such as there are in the whole combined system cube+beam+cube, then there is no single $T$. If there was another source of heat with temperature $T_3$, and the combined system accepted heat $Q'$ from it, we could write the famous inequality as $dS_{cube,rod,cube} \geq dQ'/T_3$. Commented Feb 15, 2023 at 16:30
• @JánLalinský thanks for your answer, it cleared up many doubts for me. I've edited the question, to recap the main concepts, about $\delta Q /T$, that i figured out form this answer. Can you check if they are correct? Commented Feb 15, 2023 at 17:18
• In the 2-cube and wooden beam example you gave, each of the cubes can be treated as a closed system, each experiencing individually an essentially reversible process with $\Delta S=Q_{cube}/T_{cube}$. All the irreversibility takes place within the wooden beam, which receives entropy from the hot cube and discharges a greater amount of entropy to the cold cube. Commented Feb 15, 2023 at 17:28

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

• I think you meant that the entropy of T2 reservoir decreases by less than the entropy of T1 reservoir increases. Commented Feb 15, 2023 at 12:46
• thanks for the answer, i think the question was not clear, so i edited it. Indeed, yes, i see that it's not reversible, this is why it is strange. But the core of the question is that $dS=Q/T$ only for reversible transformations, so, according to Clausius inequelity, it should be reversible Commented Feb 15, 2023 at 12:47