- #1
Meithan
- 16
- 1
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.
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.