Suppose we have two bodies at different temperatures, and we let them interact thermally in such a way that the process is not quasistatic (e.g. two different metal spheres touching). Do we arrive at the same final temperature as if the same spheres had equilibrated via a quasistatic process? If so then, rigorously, why? Of course I expect the answer to be yes but I can't convince myself of why in the context of thermodynamic theory.

Edit: I think I see that since I can establish that in a final equilibrium, they have equal temperatures, then I can imagine a quasistatic process taking them to this final state? It seems thus that in the case of non-quasistatic heat transfer, ceteris paribus, we have the same entropy change as for the corresponding quasistatic heat transfer?

  • $\begingroup$ So what if they have the same entropy change for a quasi static path as for a reversible path? How does that violate the 2nd law? $\endgroup$ Commented May 27, 2023 at 23:47
  • 1
    $\begingroup$ Do you actually mean "reversible" when you write "quasistatic"? Not all quasistatic processes are reversible, for example ones with friction. $\endgroup$
    – hyportnex
    Commented May 28, 2023 at 1:09
  • $\begingroup$ @ChetMiller I am confused and/or not following you; where did I mention a violation of the Second Law? $\endgroup$
    – EE18
    Commented May 28, 2023 at 3:57
  • $\begingroup$ @hyportnex I do mean quasistatic as written. I am here asking about quasistatic heat transfer (which is in general not reversible unless the temperatures of the bodies are (virtually) identical of course). $\endgroup$
    – EE18
    Commented May 28, 2023 at 3:58
  • $\begingroup$ Let's see you analysis of the 2 sphere system you described for the quasistati path and for the sudden contact situation, as far as the entropy change and the dQ/T integrals are concerned. $\endgroup$ Commented May 28, 2023 at 9:46

4 Answers 4


Do we arrive at the same final temperature as if the same spheres had equilibrated via a quasistatic process? If so then, rigorously, why?

Of course; by conservation of energy, since nothing else exists, it doesn't matter how the spheres interact; they'll end up at the same final common temperature. (This can be shown by using $U=\int_0^{T_\text{final}} C\, dT+U_0$ as a caloric equation of state.)

As an alternative path, consider running a Carnot heat engine between the spheres; if they're identical and material properties are constant, we end up not with $T_\text{final}=\frac{T_1+T_2}{2}$ but $T_\text{}=(T_1T_2)^{1/2}$, as I discuss here, with the difference stored elsewhere as nonthermal energy (e.g., electrical energy) and no entropy production. Then, we convert the electrical energy into thermal energy via irreversible Joule heating of a resistor in each of the spheres and obtain $T_\text{final}=\frac{T_1+T_2}{2}$, with the same entropy production as if the spheres had simply been pushed into thermal contact.

  • $\begingroup$ Thank you for your answer, but I'm not sure the first sentence satisfies me. How can I know this? In the QS case, we can write $0 = \Delta U = \int_{T_{10}}^{T_f}C_1 dT + \int_{T_{20}}^{T_f}C_2 dT$ and solve for $T_f$. But in the nonQS case, we only know that there is some final temperature -- I'll call that $T_f'$. We cannot write $0 = \Delta U = \int_{T_{10}}^{T_f'}C_1 dT + \int_{T_{20}}^{T_f'}C_2 dT$ because that only applies to quasistatic processes. So how do we argue it? $\endgroup$
    – EE18
    Commented May 28, 2023 at 16:15
  • $\begingroup$ Please see my edit. $\endgroup$ Commented May 28, 2023 at 16:22
  • $\begingroup$ So you're saying because the nonquasistatic process does not change volumes or mole numbers of each subsystem by assumption, one goes ahead and uses what you've written and so obtains the same result in the end (and $T_f = T'_f$)? $\endgroup$
    – EE18
    Commented May 28, 2023 at 16:28
  • $\begingroup$ To my knowledge. What do you get? $\endgroup$ Commented May 28, 2023 at 16:33
  • $\begingroup$ I am not sure I follow, are you asking if I agree with that assessment? I guess the answer is that I think so? $U$ is a function of state so, given that a priori we only know with respect to the nonQS that, on settling to equilibrium, there will be some common final temperature, and the same volumes and mole numbers as initially. This fixes the final equilibrium state of each subsystem. We can then imagine a QS process which takes us to this same final state and so write $0 = \Delta U = \int_{T_{10}}^{T_f'}C_1 dT + \int_{T_{20}}^{T_f'}C_2 dT$. But this has the exact same structure as... $\endgroup$
    – EE18
    Commented May 28, 2023 at 17:00

I submit to you that the way you have phrased the question is actually ill-defined. You have tacitly assumed two pieces of matter whose interaction is strictly thermal energy exchange without any mechanical effect included, each is characterized by nothing else but its own temperature, while the internal homogeneity is ensured by the quasi-static assumption. So far so good, but the initial conditions are self-contradictory.

Why? Because you have stated that they be in thermal contact while they are at different temperatures. So their common interface, a mathematical surface through which thermal exchange takes place has now two temperatures. There is no such a thing in thermodynamics. In the problem, therefore, you must include a third piece with its own mathematical model of a thermal conductor to resolve the contradiction.

To make sure that you can still maintain the quasi-static nature of the process the thermal conductor connecting the pieces must have a small but finite conductivity, say, a thin metal wire. Important: This is not just empty verbiage, in fact, as the problem is stated, this model of a vanishingly thin and short thermal conductor is where the irreversible entropy generation occurs. Empirically, as long as radiation and other surface effects can be ignored, the amount of entropy that is generated is essentially independent of the size of the detailed nature of the connector if the ratio of its diameter is always much less than its length. That being the case we can declare it as a "2-port" model item that has no geometric dimensions and can simultaneously connect to two different temperatures, say, $T_1 > T_2$. When connected to $T_1$ from which it reversibly receives entropy in the amount of $S_1$ then simultaneously at the same time internally it irreversibly generates entropy $\Delta S = S_1\frac{T_1-T_2}{T_1}$ and rejects both at temperature $T_2$ so that the entropy transported/supplied quasi-statically at/to the low temperature port is $S_2=\Delta S+ S_1 = S_1\frac{T_1}{T_2}$. Now the problem becomes well-defined.

It is important always to remember that in this quasi-static model, the connector must have an arbitrarily small but non-zero conductivity, and once you ensure that the flow of entropy is of the right amount and is actually independent of said conductivity then you can safely assume the conductivity to be zero.

When instead you declare that the thermal process between the two masses is to be not only quasi-static but also reversible then this direct connection with a conductor must be excluded, and what is left is an indirect connection via a "Carnot engine" that reversibly absorbs entropy at $T_1$, lowers its temperature by reversible adiabatic work to $T_2$ at a temperature where it reversibly rejects the same entropy that it has absorbed at $T_1$; this is what @Chemomechanics is referring to in his answer below.

  • $\begingroup$ I am not sure I follow your calculation for $\Delta S$, but perhaps that's because I don't know enough thermodynamics yet (I think this has to do with the Clausius inequality?). I will have to reread this answer a bit later, once I've read deeper into Callen, I fear. $\endgroup$
    – EE18
    Commented May 28, 2023 at 16:11
  • $\begingroup$ In pure "heat conduction", one that conducts and generates entropy down a temperature gradient, the irreversibly generated entropy $\delta S$ is exactly equal the ratio of the dissipated energy $\delta Q$ and the temperature $T$ at which this dissipation occurs. as a result the process will conserve the quantity called "heat". In other words, the quantity $TS$ is conserved down a homogeneous linear conductor at every of its cross section. This is really energy conservation combined with the empirical Fourier's law, and you might call it the mathematical definition of heat conduction here. $\endgroup$
    – hyportnex
    Commented May 28, 2023 at 16:58
  • $\begingroup$ If $TS$ is a constant then $d (TS) = T dS + S dT =0$ that is $dS = -S \frac{dT}{T}$, meaning that if entropy $S$ is at one cross section at temperature $T$, then at a cross section where the temperature is $T+dT$ the entropy is $S+dS = S-S \frac{dT}{T}$. In this expression I used the differential $d$ as is conventional in mathematical analysis. In thermodynamics it is more conventional to use the negative of this differential signifying that entropy $S$ moves down and not up on a gradient, then with $\delta T = -dT >0$ you get the formula $\delta S = S \frac{\delta T}{T} >0$. $\endgroup$
    – hyportnex
    Commented May 28, 2023 at 17:08
  • $\begingroup$ Does Callen discuss what you are discussing in your first comment? I've never seen/heard of that? $\endgroup$
    – EE18
    Commented May 28, 2023 at 18:03
  • $\begingroup$ Callen does not spell this out this way explicitly, but it is just energy conservation in a steady state, no more, no less. Imagine two reservoirs at temperatures $T_1$ and $T_2 = T_1 -delta T$ connected via a conductor. In the steady state and in a unit time the higher temp emits $\delta Q$ thermal energy and the same amount of energy is absorbed by the lower temp reservoir. This means that at the high temp $S_1=\frac{\delta Q}{T_1}$ entropy was emitted... $\endgroup$
    – hyportnex
    Commented May 28, 2023 at 19:07

If you equilibrate the two spheres using a quasi-static path or an irreversible path, in both cases you end up with the same temperature, $(T_H+T_C)/2$. So the entropy changes of each of the two spheres must be the same for both paths: $$\Delta S_H=MC\ln{\left(\frac{T_H+T_C}{2T_H}\right)}$$$$\Delta S_C=MC\ln{\left(\frac{T_H+T_C}{2T_C}\right)}$$If you add these entropy changes together, their sum is greater than zero.

To carry out the quasi-static reversible path, you first separate the two sphere from one another, and then place each of them into contact with a continuous sequence of constant temperature reservoirs running gradually from their initial temperature to their final temperature. This reversible process path thus requires the use of a new set of surroundings (the sequence of reservoirs) for each of the spheres.

The irreversible path requires simply placing the two spheres into direct contact and allowing them to equilibrate. During this direct contact, the temperature in the contact area (through which all the heat Q flows is the average temperature $(T_H+T_C)/2$. So checking the Clausius inequality for each of the spheres separately, we have $$\Delta S_H=MC\ln{\left(\frac{T_H+T_C}{2T_H}\right)}\geq-MC\frac{T_H-T_C}{T_H+T_C}$$$$\Delta S_C=MC\ln{\left(\frac{T_H+T_C}{2T_C}\right)}\geq+MC\frac{T_H-T_C}{T_H+T_C}$$As long as $T_H\geq T_C$, both these inequality are satisfied.

  • $\begingroup$ Thank you for your answer. I'm afraid I'm not deep enough into Callen at the moment to know about the Clausius inequality but, when I do encounter it, I will come back to this answer. $\endgroup$
    – EE18
    Commented May 28, 2023 at 15:50
  • $\begingroup$ I am still thinking about the first sentence in your answer. How do we prove that we get the same final temperature even in the fully nonquasistatic case? Suppose we have a process in which mole numbers and volume of the two subsystems are fixed. If the process is QS I can write $0 = \Delta U = \int_{T_{10}}^{T_f}C_1 dT + \int_{T_{20}}^{T_f}C_2 dT$ (and solve for $T_f$). But in the nonQS, as per your "recipe" article on Physics Forums, I need to deduce a corresponding QS process which takes me to the same final state in order to compute properties. But how do I know what that final state is? $\endgroup$
    – EE18
    Commented May 28, 2023 at 16:09
  • $\begingroup$ The final state is the same for the reversible path as for the irreversible path, since $\Delta S$ is a function only of the two end states. Even for an irreversible process, $\Delta U=Q-W=0$ $$\Delta U$ is also a state function and depends only on the two end states. $\endgroup$ Commented May 29, 2023 at 1:10

Because any thermodynamical equilibrium state is accessible via reversible processes by any other equilibrium state, and that the initial and final states are equilibrium states, so it is trivially true that there must be reversible processes connecting the initial states to the final state.

This is obviously a trivial and useless point, which is why hyportnex wrote about it being ill-defined.

Let us treat the situation in higher detail. We speak of extensive quantities, so that $U=U_H+U_C,$ $V=V_H+V_C,$ $N=N_H+N_C,$ say. What is important for us to realise, however, is that if you want to study this non-quasistatic transfer of heat, then $S_\text{final}\gneq S_H+S_C$ because there will be internal generation of entropy. So, if you were originally intending to ask if there is a quasistatic process connecting the initial states and final state, assuming that the total entropy is the addition of the two initial states, then that is not possible.

Really, this is why we should be considering thermodynamics as a subject by focusing upon the study of entropy as the thing to scrutinise, not energy. What we are supposed to do, is to take the entropy formula $$S=S(U,V,N)\qquad\rightarrow\qquad S_\text{final}=S(U_H+U_C,V_H+V_C,N_H+N_C)$$ in order to find the actual final equilibrium state. We will find that $S_\text{final}-(S_H+S_C)\gneq0$, such that if we plot the constant entropy hypersurfaces of the combined system as the space within which we look for reversible processes connecting the initial and final states, we will immediately see that this is futile. Note that this is an intuitive, but really unphysical additional restriction in the space of reversible processes that we are imposing, and it is no wonder that it does not work.

Another way to see that this is the case, is to find the reversible processes that will actually take the initial states to the final state. The theoretically clean way to study this is to consider Carnot-like processes. Here, it is smarter to not study the metal spheres, but rather study ideal gases. In the above, we found the final entropy, and since all the relevant extensive parameters of the system has been found, we thus know the exact thermodynamical state of the entire system. Which means we also know the temperature, pressure, chemical potential, etc.

This thus allows us to work backwards. Knowledge of $T_f$ means that we can first isentropically compress the colder subsystem to $T_f$ and then isothermally expand it back to $V_C.$ We could have also chosen to isothermally compress and isentropically expand. There is thus already one ambiguous choice here that would alter the details of the calculation.

We can make a similar choice on the hot side. Because of the ambiguity of having choices on both the hot and cold sides of this calculation (and actually, since we can break up the processes into many sub-processes, the actual details can vary somewhat continuously between the extremes), we cannot actually speak exactly of what particular amounts of the transfer of heat and work happened in the reversible processes. However, what we can observe, is that, since $U=U_1+U_2,$ this means that there is no net transfer of energy. But we also know that, if we have reversible processes in all these, then there is a net transfer of entropy $\Delta S=S_\text{final}-(S_H+S_C)$ from the reservoir into the system, which means also that there is a net work done from the system to the outside. This is the part that is lost, due to the irreversibility of the non-quasistatic transfer of heat.

Note even further that, because of the ambiguity, we have a net transfer of heat or work, equivalent in magnitude to $T_f\Delta S$ in one specific case, and may be some combination of $T_C$ and $T_H$ thereof, or any value in-between.

I leave the detailed calculations to you. It has been tedious even getting to this point.

Right before we leave, I want to re-emphasise: If you use reversible processes, you can reach any temperature as the final state as you want. If you stay on the smaller entropy hypersurface, you should reach a smaller temperature, by doing some net work and having a smaller final energy. If you stay on the same energy hypersurface, then you should absorb some entropy and convert that to work. There is thus ambiguity in the phrasing of the question.


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