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Homework Help: Working out signs of work and heat for entropy change question

  1. Jan 12, 2013 #1
    N.B. My set question is actually almost identical to another posted here https://www.physicsforums.com/showthread.php?t=368672 but my question is not answered there.

    1. The problem statement, all variables and given/known data

    The temperature inside a refrigerator is 275K. The room containing it has a
    temperature of 295K. In one hour, 95 kJ of heat are transferred
    from the interior to the room.

    Calculate the entropy change of (i) the contents
    of the refrigerator, and (ii) the room? (Assume that the heat transfer is

    2. Relevant equations

    ΔU = Q + W

    ΔS = [itex]\frac{Q}{T}[/itex]

    3. The attempt at a solution

    Tc = 275 K
    Th = 295 K

    Qc = -95 kJ (as heat is leaving the system)
    Qh = Qc + W

    So we need to find W





    W = -6.909 kJ

    Problem 1: Why is this value of W negative? Logically we know that this must be work being done on the system by the compressor in the refrigerator so surely work should be positive? I can't see a flaw in the logic of my maths though so I'll accept it for now and move on.

    Problem 2: Qh is heat flowing INTO the environment. As this is the opposite of what is happening inside the refrigerator it makes sense to me to consider

    Qh = |Qc + W|

    so we end up with a positive quantity for Qh, is this logic correct? is there a more rigorous mathematical argument I can follow instead of just using logic?


    Qh = 101.9 kJ

    which gives entropy changes

    ΔSc = [itex]\frac{Q_c}{T_c}[/itex] = -0.345 kJ /K

    ΔSh = [itex]\frac{Q_h}{T_h}[/itex] = 0.345 kJ /K

    Which equal 0 when added which is what I'd expect for this idealised reversible system. Assuming my answers are indeed right can someone please explain the two points of contention I have in the working above?

    Thanks in advance.
    Last edited: Jan 12, 2013
  2. jcsd
  3. Jan 12, 2013 #2

    Andrew Mason

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    You have to start with the correct definition of entropy:

    [tex]\Delta S = \int dS = \int \frac{dQ_{rev}}{T}[/tex]

    Ok so far. Your k is normally referred to as the coefficient of performance, COP.
    The definition of COP is:

    COP = output/input = Heat removed/work input = |Qc|/W where W is the work done ON the system.

    By using COP = Qc/W you are implicitly defining W as the work done BY the system.

  4. Jan 12, 2013 #3
    Dear Shakeybear,

    Welcome to Physics Forums.

    Shakey, your analysis looks Shakey.

    If 95 kJ of heat are removed from the refrigerator contents, then a greater amount of heat must be rejected to the room (because of the work done by the compressor). Since the refrigeration package is running in a cycle, the heat rejected to the room minus the heat removed from the contents must equal the work done by the compressor. Since the system is stated to operating reversibly, you must have that:

    -95/275 + Qroom/295 = 0

    As you correctly assumed, the overall change in entropy for the system plus surroundings is zero.
  5. Jan 12, 2013 #4

    Interesting. I can't use that formulation of COP though as I only know Qc. Do you think my conclusions, that for this ideal case the entropy changes should cancel is reasonable? If so then there's nothing very dramatic I can change in the working.

    If I understand you correctly what I have calculated is not Win at all but Wout. If correct then simply relabelling will solve that problem for me.

  6. Jan 12, 2013 #5
    Thanks for your reply.

    Which part looks shakey exactly? is your formulation

    -95/275 + Qroom/295 = 0

    Not equivalent to mine with Qh = Qroom?

    -95/275 = -0.345
    101.9/295 = 0.345
  7. Jan 12, 2013 #6

    rude man

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    As someone else said, your formula for S is wrong. Invoke the fact that total entropy of the universe is zero:

    Qh/Th = Qc/Tc. You know 3 of the 4 parameters already, solve for the 4th!
  8. Jan 12, 2013 #7
    Thanks, but to be clear the problem is not really how to solve this, as far as I'm aware I've already solved it, unless someone corrects me. The problem I have now is explaining why it works. Andrew provided some insight into this but I'm still not convinced as to why the value of work is negative.

    Yes, I made a typo in the entropy equation, thanks for pointing that out, I'll edit.
  9. Jan 12, 2013 #8

    Andrew Mason

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    Sure you can. W = Qc/COP where COP = Tc/(Th-Tc)

    Certainly. Since it is reversible, Qc/Tc = -Qh/Th.

    Just treat Qc = |Qc| when using COP. This is implicit in your statement that W = Qh-Qc. This is only true if you are taking the absolute values of the heat flows. You can see this from the first law:

    If you are using the first law form:

    ΔU = Q - W then W is the work done BY the system. Since in any complete cycle ΔU = 0 so W = Q = Qh+Qc. For an engine Qh is negative and Qc is positive. For a refrigerator, it is the opposite.

    So the only way you can say that W = Qh-Qc is if you are using the absolute values of the heat flows and assuming that W is the work done ON the system where it is a refrigerator.

    Last edited: Jan 12, 2013
  10. Jan 12, 2013 #9
    Shakeybear, Please forgive me. I didn't read your analysis carefully enough. You were , of course, correct. Sorry :redface:

  11. Jan 12, 2013 #10

    rude man

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    Depends on how you define work. Thermodynamicists almost always write U = Q - W.

    In thermodynamics it's traditionally W = ∫pdV. So for a refrigeration cycle, we have isothermal compression at a high p and isothermal expansion at a low p, so the net work is < 0 since ∫phighdV is negative and larger in magnitude than ∫plowdV which is > 0 (the area inside the loop on a p-V diagram is work; if the circulatin is ccw then w < 0 but if it's cw then W > 0.
  12. Jan 12, 2013 #11
    There is no work done on the air in the refrigerator, and there is no work done on the air in the room. The work gets done on the refrigerant fluid. Since the refrigerant operates in a cycle, the change in internal energy of the refrigerant is zero. That means that the work done on the refrigerant plus the net heat added to the refrigerant is zero (Q + W = 0). Since the refrigerant receives 95 kJ from the refrigerator contents, and rejects 101.9 kJ to the room, the net heat added to the refrigerant Q = 95 - 101.9 kJ = -6.9 kJ. Therefore, the work done by the compressor on the refrigerant is + 6.9 kJ.

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