I don't understand why heat transfer from hot reservoir to the system is considered reversible in this case:
$T_{reservoir}$ = $T_{system}$ + dT

but it's considered irreversible in this case:
$T_{reservoir}$ = $T_{system}$ + ΔT

Where dT is infinitesimal difference while ΔT is finite difference in temperature between reservoir and the system.

In both cases some heat is transferred from the reservoir to the system, so it should be irreversible in both cases. What understanding am I missing here?

  • $\begingroup$ An ideal reservoir does not care if the temperature difference is finite or infinitesimal but heat conduction that takes place between the heat source (reservoir) and heat sink ( the lower temperature reservoir) across a finite temperature difference is an irreversible process, it goes spontaneously by itself always from a higher to a lower temperature, never the other way around, and it always increases the entropy of the sink more than it reduces the entropy of the source. $\endgroup$ – hyportnex Apr 22 '16 at 12:48
  • $\begingroup$ Thanks for the comment but that doesn't really answer my question. I don't understand why infinitesimal temperature difference makes the process reversible while finite temperature difference makes it irreversible. It says so here: web.mit.edu/16.unified/www/FALL/thermodynamics/notes/… but I don't know why. $\endgroup$ – matori82 Apr 22 '16 at 13:37
  • $\begingroup$ i think it does answer your question because the entropy change from the reservoir giving up $\delta q$ heat at temperature $T$ is $\frac {\delta q}{T}$ so when heat flows from $T_h$ to $T_{\ell}$ then the entropy change is positive: $\delta q (\frac{1}{T_{\ell}} - \frac {1}{T_h}) > 0$ if $T_h > T_{\ell}$ $\endgroup$ – hyportnex Apr 22 '16 at 15:49
  • $\begingroup$ I agree that entropy change is positive, but for reversible process entropy change should be zero right? If it's positive then it's irreversible process right? $\endgroup$ – matori82 Apr 22 '16 at 16:09
  • $\begingroup$ exactly as you said it, and the entropy dropped across a finite temperature difference is definitely positive, not an infinitesimal. $\endgroup$ – hyportnex Apr 22 '16 at 16:22

To do it reversibly, you can heat the body from $T_1$ to $T_2$ (i.e., over a finite temperature change) using an infinite sequence of constant temperature reservoirs, in which each reservoir in turn is only dT higher in temperature than the body at any time (and also only dT higher in temperature than the reservoir before it in the sequence). Each increment in heat transfer would take place with only a differential temperature driving force between the body and the current reservoir. To reverse the process, and bring both the body and the reservoirs back to their original states, you would simply contact the body with the reservoirs in the reverse sequence, in which case the reservoirs would be dT lower in temperature than the body in each step of the process). The only difference would be with regard to the very first and very last reservoirs (which could not be returned to their original states). But this would be insignificant.

In the case where the body is heated from $T_1$ to $T_2$ by bringing it into contact with a constant temperature reservoir at $T_2$ for the entire time until the body equilibrated at $T_2$, all the heat transfer would take place with a finite temperature driving force, and there would be no way to return both the body and the reservoir to their original states without causing a significant change in something else (like using other reservoirs).

A reversible process is one in which the system is only slightly removed from being at thermodynamic equilibrium throughout the change. Thus, a reversible process can be viewed as imposing a continuous sequence of thermodynamic equilibrium states.

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    $\begingroup$ @matori82 Is there something about my answer that is confusing to you? In what way does it not answer your question? How can I clarify? $\endgroup$ – Chet Miller Apr 22 '16 at 21:37
  • $\begingroup$ Thanks for the answer! What is confusing me is the last sentence in the first paragraph: how can we consider insignificant that the first and the last reservoirs didn't return to the original state? I mean there was some net dQ heat exchange between them after the process such that the last reservoir gave dQ which at the end of the process was transferred to the first reservoir. This means that total entropy of the surrounding has been increased. ΔSsurroundings=-dQ/(T+n*dT) + dQ/T >0. So the changes in the surroundings were leftover after we returned the system to the initial state. $\endgroup$ – matori82 Apr 23 '16 at 0:01
  • $\begingroup$ A reversible process is always a limiting situation. Notice that the entropy change you calculated is infinitecimal, while along the forward path, both the system and surroundings have experienced large finite changes in entropy (the reservoirs have experienced a finite summation over infinitecimal changes, just the same thing that happens mathematically when we perform an integration of a function). So the entropy change for the combined forward and reverse paths is infinitecimal, while the individual entropy changes for both the system and surrounding over each path is large. $\endgroup$ – Chet Miller Apr 23 '16 at 0:15
  • $\begingroup$ Also note that, for an irreversible process, you would calculate a finite (non-infinitecimal) entropy change in the combination of system and surroundings. And you could not return both the system and surroundings to their original states without a finite (non-infinitecimal) change in the system and surroundings (even if the return path were reversible). $\endgroup$ – Chet Miller Apr 23 '16 at 0:45

It is reversible in the first case because it satisfies the reversibility definition. A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a system at temperature $T$ in thermal contact with a thermal reservoir at same temperature. By an infinitesimal increase $dT$ of the reservoir's temperature you get heat flowing to the body. With a further infinitesimal decrease, let us say $2dT$ you reverse the flow. The same will not happen if there was a finite difference of temperature.

  • $\begingroup$ So in your example I assume we have two reservoirs with temperatures T, T+dT, right? So if we do the things like this: 1) put system in contact with reservoir T+dT and wait for system to increase it's temperature to T+dT, 2) put system to reservoir T, and wait system to get back to original temperature T. Then we have successfully returned the system to the initial state. BUT, the surroundings is not returned to the initial state because we have transferred some heat dQ between reservoirs. Since change is left in the surroundings this means the process is actually irreversible. Is that right? $\endgroup$ – matori82 Apr 22 '16 at 15:39
  • $\begingroup$ There is no change in the surroundings. Each reservoir is characterised by its temperature, which does not change. $\endgroup$ – Diracology Apr 22 '16 at 16:33

Because the effects of heat transfer.

If the difference in temperature is large, the process can't be reversed. Heat is a form of energy that can transform into an increase of the molecules' movement in a system and if the system is water, and this water is hydrostatic, the kinetic energy will become pressure. So to reverse this, you need to do a refrigeration process which requires work and so it affects the surroundings, so the process then can't be reversed.

If heat transfer is due to infinitely small amount of temperature difference, the process can be reversed without refrigeration process and keeping the surrounding intact.

Now transferring an amount heat through an infinitesimal temperature difference would require an infinite amount of time, so in other words, rate of $Q$ transferred is infinitely small, making it possible for us to reverse the process because the temperature of the lower temperature body and higher temperature body would stay constant. So no kinetic energy and the process here is isothermal!!

So here you go

Now about about the example you shown, one process is reversible and the other is not even though they both goes through the same process and pass exactly the same state. But one is reversible internally and externally, the other would affect the surroundings.


Short answer is that there are irreversibilities associated with both the heat transfers. But the irreversibility associated with heat transfer by a finite temperature difference $ΔT$ is so much larger than the irreversibility associated with heat transfer by an infinitesimal temperature difference $dT$, that the latter process can be approximated as reversible.

Simply said the extent of irreversibility associated with the $dT$ case is so small that it can be approximated as reversible.

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