Reversible/irreversible heat exchanges

[Meithan]

I am a TA in a class covering an introduction to fluid mechanics, thermodynamics and wave motion. We just finished the thermodynamics part of the class, but out of all the problems we solved, there is still one both me and my students are not comfortable with. Here it is:

Problem
A metal rod is put in thermal contact with a heat reservoir at a temperature of 130°C on one side and with a heat reservoir at 24.0°C on the other side. (a) Calculate the total change of entropy that is produced by the conduction of 1200 J of heat through the bar. (b) Does the entropy of the bar itself change during this process?

Here's how we solved it in class.

(a)Since the temperature of the heat reservoirs is unchanged (by definition) when they exchange the 1200 J of heat, we can calculate the entropy change by direct integration:

$$\Delta S = \int \frac{\delta Q}{T} = \frac{Q}{T}$$

(the integration of $$\delta Q$$ yields the total heat exchanged.) Then, for reservoirs H (the hot one) and C (the cold one) we have:

$$\Delta S_H = \frac{-|Q|}{T_H}$$
$$\Delta S_C = \frac{+|Q|}{T_C}$$

where the signs before the heat transferred differ because reservoir H is losing heat, while reservoir C is gaining heat. The total entropy change is then:

$$\Delta S_{total} = \Delta S_H + \Delta S_C = |Q|\left( \frac{1}{T_C}-\frac{1}{T_H} \right) > 0$$

Since $$T_C<T_H$$, we see that the total entropy change is positive, which is what we expected. Plugging in the numbers (with the temperatures converted to Kelvin), we obtain a numerical answer of

$$\Delta S_{total} \approx +1.06$$ J/K

which is correct, according to the book.

(b) In part (a) we ignored the entropy change of the rod itself. However, since there is no neat heat transfer from or into the rod during the process, we would expect its entropy change will be zero. The rod is simply conducting the heat it receives from one reservoir into the other reservoir, without keeping any of it. Therefore, for any piece of the bar, [delta]\delta Q = 0[/tex] and so the entropy integral vanishes.

This seems all nice and good. But then come the fine points.

Because the entropy of the heat-reservoirs+rod system (which is isolated from the rest of the Universe) increases in this process, we know for sure it is irreversible. However, let's look at what's happening in a little more detail.

We don't know the temperature of the rod. What would happen if we set up this experiment is that, after a transient phase, the temperature distribution of the rod will come to equilibrium by having a temperature gradient appearing over the length of the rod. The rod will be nearly at temperature $$T_H$$ on the hot side and nearly at temperature $$T_C$$ on the cold side. By nearly I mean the temperature at the ends will be only infinitesimally different from the temperature of the reservoirs. Across the rod, each small element will have a temperature only infinitesimally different from the ones of its neighbors. This must be so if the temperature gradient is a continuous function (which the heat transfer equation says it is).

So consider the heat exchanged between the hot end of the rod and the hot reservoir. Since it happens across an infinitesimal temperature difference, wouldn't it be reversible?

Then, heat is transported across the elements of the rod. Again, for each element, the heat transfer occurs across an infinitesimal temperature difference, which is reversible.

Finally, the same thing happens on the cold end.

So we have a succession of reversible processes where heat is exchanged across infinitesimal temperature differences. Shouldn't the overall process be reversible, then? Yet, we know for sure it isn't.

So what's wrong with this argument?

If you think about it in terms of entropy changes, this isn't all that different from the Carnot engine: we extract a quantity of heat isothermally from a hot reservoir, and then (after some intermediate step that doesn't provoke an entropy change) dump it into a cold reservoir, again isothermally. The two isothermal processes are reversible ones.

The main difference is that in the Carnot engine, not all heat that is extracted from the hot reservoir ends up in the cold reservoir. Some is converted into work. And the crucial thing is that exactly as much heat is converted into to work so that the Q dumped to the cold reservoir produces an entropy loss (for the engine) that exactly cancels out the entropy gain produced when it siphoned heat from the hot reservoir.

So while we could think as the Carnot engine as a machine that extracts heat from a hot reservoir and dumps it into a cold reservoir, it's not the same thing as our rod-conduction problem, because not *all* heat is conducted from one reservoir to the other.

So what's the real deal with these heat exchanges between heat reservoirs? Even if heat is extracted/dumped in reversible isothermal processes, apparently the fact that some heat is never dumped (because it is used to produce work, which means the engine is not a closed system) is what "recovers" reversibility for the Carnot engine. Why is this so?

Thanks for your comments.

 

[Andy Resnick]

Maybe I'm missing something, but there does not appear to be any major issue: the process of heat conduction by the rod is considered reversible because the rod does not heat up. The overall process is irreversible because a quantity of heat passed from the hot to cold reservoir while leaving the system in the original configuration (same temperature distribution).

I wonder if the conceptual difficulty arises from the distinction between equilibrium and steady state. The rod is in a condition of steady-state, not equilibrium.

 

[quantum123]

I guess, the rod is an open system, that is why it undergoes a reversible process.
But rod+hot bath+ cold bath is an isolated system, that is why it undergoes irreversible process.

 

[zonde]

As I understand your question is about http://en.wikipedia.org/wiki/Loschmidt%27s_paradox" .

Paradoxes usually indicate when you are taking theory too far.
I would say that symmetry of reversible processes is only very good approximation. At macro scale the deviation from perfect theoretical symmetry adds up and becomes observable.

Another objection is that definition of CLOSED system is unphysical. For example, how do you make system closed in respect to gravity?

 

[sardar]

As a scientist, it is important to understand the concept of reversible and irreversible heat exchanges. In this problem, we have a metal rod in thermal contact with two heat reservoirs at different temperatures. The total entropy change is calculated by integrating the heat transferred over the temperature of the reservoirs. However, we also need to consider the entropy change of the rod itself.

In this case, the entropy change of the rod is assumed to be zero because there is no net heat transfer into or out of the rod. However, this assumption is not entirely accurate. As mentioned in the problem, there is a transient phase where the temperature of the rod will come to equilibrium, resulting in a temperature gradient along the length of the rod.

This temperature gradient means that there is a continuous exchange of heat between infinitesimal temperature differences within the rod. Each small element of the rod experiences reversible heat transfer, but when we consider the entire rod, the process is irreversible.

The difference between this problem and the Carnot engine is that in the Carnot engine, not all heat is transferred from the hot reservoir to the cold reservoir. Some heat is converted into work, which means the system is not closed. This conversion of heat into work allows for the entropy loss in the cold reservoir to be offset by the entropy gain in the hot reservoir, making the overall process reversible.

In the case of the rod, all heat is transferred from one reservoir to the other, resulting in a net increase in entropy. This makes the process irreversible.

In conclusion, the key difference between reversible and irreversible heat exchanges is the presence of a closed system and the conversion of heat into work. In the case of the rod, the absence of these factors results in an irreversible process. It is important to understand these concepts in order to fully grasp the laws of thermodynamics and their applications.