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Entropy change of the surroundings during irreversible process

  1. Sep 29, 2012 #1
    According to my textbook, during an irreversible process, the entropy change of the surroundings is given by [itex] \frac{q}{T} [/itex] where q is the heat transferred to the surroundings during the process. Why are we allowed to use this equation, considering that this equation only holds for reversible processes?

  2. jcsd
  3. Sep 29, 2012 #2
    And what is T?
  4. Sep 30, 2012 #3

    Andrew Mason

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    The change in entropy between states A and B is defined as [itex]\int_A^B dQ/T[/itex] over a reversible path between states A and B.

    If the temperature of the surroundings does not change from A to B while heat flow q occurs, then a reversible path between states A and B would be a reversible isothermal path in which heat flow q occurs. So the change in entropy would be

    [tex]\Delta S = \int_A^B \frac{dQ_{rev}}{T} = \frac{1}{T}\int_A^B dQ_{rev} = q/T[/tex]

  5. Sep 30, 2012 #4
    Sorry but according to my textbook, that is incorrect. The surrounding's entropy does not require the calculation of heat along a reversible path. According to my textbook, ANY path can be used for calculating heat for the change in entropy of the surroundings.

    So if there are say, two isothermal paths from A to B, let's say path P which is reversible, and path Q which is irreversible. Then if the system were to move from state A to state B along the path Q, the entropy change of the surroundings would be the heat along path Q devided by T. Path P (the reversible path) would be irrelevant in the calculation of the sorrounding's entropy change.

    This is what I'm trying to understand and though my textbook states this, it does a poor job of explaining it.

  6. Sep 30, 2012 #5
    You still haven't answered my question....?

    You told us what q is so why not T?
  7. Sep 30, 2012 #6
    Oh sorry Studiot I didn't see your post.
    T refers to the temperature of the surroundings.

    To prevent confusion, let's suppose the process is infinitesimal, or
    [tex] dS = \frac{dq_{rev}}{T} [/tex] which holds true for all processes, whether or not they are isothermal.

  8. Sep 30, 2012 #7
    That's good.
    I expect that in the situation under consideration the surroundings are so massive that an addition of heat q will not (significantly) change the temperature of the surroundings.

    That is the usual assumption.

    For example pouring a kettle of hot water into a swimming pool will not significantly change the pool temperature. The kettle water will take on the swimming pool temp and loose heat equal its specific heat time mass times temp above pool temp.

    We can then say that whatever the process in the system, the heat is transferred to the surroundings at constant surroundings temperature.

    So we can apply the second law in a straightforward manner.

    Total entropy change in surroundings = Total heat transferred / constant surroundings temp = q/T

    It's as simple as that.
  9. Oct 1, 2012 #8

    Andrew Mason

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    While dS = dQirrev/T may work for an isothermal transfer of heat to a reservoir with an arbitrarily large heat capacity, in general dS = dQrev/T.

    But state B is the state of the reservoir (surroundings) that is the same as in A except that it has received heat flow Q. So all paths between A and B must have the same heat flow, Q. Since all paths must also have the same T, it doesn't matter whether you pick the reversible or irreversible one in this case. But in general it does.

    The difference between the reversible and irreversible processes in this case has to do with the temperature of the body that the reservoir is in contact with. In the reversible case, the temperature of that body is kept infinitessimally lower than the reservoir while in the irreversible case, there is a finite temperature difference. But the heat flow to the reservoir and the reservoir temperature are the same in both cases.

  10. Oct 2, 2012 #9
    I think that what Studoit and Andrew Mason are saying is that, as far as the surroundings are concerned, if its temperature doesn't change (significantly) during the transfer of the heat, it is undergoing a reversible isothermal process You need to consider the surroundings as just another system.
  11. Oct 3, 2012 #10
    What you just said is true for both reversible and irreversible processes assuming that the surroundings are a "reservoir". A reservoir is a system that won't change its temperature no matter what happens. If the system is a reservoir, then that is the formula.

    The entropy change of the surroundings is given by that formula regardless of whether the entropy was created or transferred into the surroundings. I have seen that same formula used in the analysis of a Carnot cycle, which is by definition has only reversible processes.
    Your quotation from the textbook doesn't sound quite right. I suspect that you are misinterpreting what the textbook said. However, it is quite possible that the textbook made a wrong statement. They do that, sometimes.

    My suspicion is that either you or the textbook are using the word "heat" incorrectly. My experience is that both teachers and students are often confused by the colloquial word heat. The word heat is used ambiguously because there is more than one physical meaning of the word. For instance, heat can be either the internal energy or the entropy carried energy.

    I will conjecture as to what went wrong on the hypothesis that either you or the textbook misused the word "heat".

    You used two suspicious phrases. One suspicious phrase in your statement is "heat transferred to the surroundings". A second suspicious phrase is "entropy change in the surroundings." The two quantities are not always related.

    Let me speculate on what was meant by the what you said. I won't make any speculate as to whether it was you or the textbook that messed it up. My money is on the textbook.

    The "heat transferred to the surroundings" may refer to the internal energy of the surroundings. If this is what was meant, then I can give you an answer that depends on a contingency. It won't be simple. However, the difficulty is in language not science.
    The internal energy in the surroundings can be increased either due to the either heat conduction into the surroundings or due to work done on the surroundings. This is the contingency.

    If the internal energy of the surroundings are increased by work, then entropy is created in the surroundings. The creation of entropy is irreversible. To put it another way, the only way work can increase the entropy in a reservoir is if the work was done by frictional forces. So if the "heat energy" was created by friction, then what you are describing is an irreversible process.

    If the internal energy of the surroundings are increased by heat conduction, the entropy has been transferred to the surroundings. The words "heat conduction" always refers to the energy that is transferred along with the entropy. Whatever entropy has been transferred by heat conduction can be transferred back by work.

    If the internal energy of the surrounding has been increased by heat conduction, then the process may be reversible. You don't know whether or not the process is reversible unless you know how much entropy was created.

    Also note that in an irreversible process, entropy can be created anywhere in the system. There could be friction in the piston. The gas in the container may have entropy created by friction.

    That is my conjecture. My advice is to read that part of the textbook and write down what you think the word "heat" means at each point. I particular, ask yourself this. At each point where the word "heat" is used, does it mean heat conduction or internal energy?

    Heat conduction is completely different from internal energy, although the same word is often used for both. This has created a lot of confusion (entropy?-) and difficulty (work?-).
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