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Thermodynamics - Carnot efficiency

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    On a hot day a house is kept cool by an air conditioner. The outside temperature is 32°C and the inside temperature is 21°C. Heat leaks into the house at the rate of 9000 kcal/h. If the air conditioner has the efficiency of a Carnot engine, what is the mechanical power that it requires to hold the inside temperature constant?

    3. The attempt at a solution

    e= 1-294/305 = 0.036066


    Power = 9000 kCal/hour --> 37656000 J/hour --> 10460 J/s

    .036066=W/10460 ---> W (really power)=377.426 J/s

    This problem seems pretty straightforward...but the back of the book has an answer of 390 W, which is ~13 more J/s than my answer, so I must have made a mistake....yet there are only two equations, and one conversion! Is this a rounding error from the book?
  2. jcsd
  3. Feb 14, 2010 #2


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    I think you're right.
  4. Feb 14, 2010 #3

    Andrew Mason

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    The Coefficient of Performance is defined as:

    [tex]COP = \frac{Q_c}{W} = \frac{Q_c}{Q_h-Q_c}[/tex]

    which, for a Carnot refrigerator is:

    [tex]COP = \frac{T_c}{T_h-T_c}[/tex]

    In this case COP = 294/11 = 26.7

    Since Qc = 9000 Kcal/hr (ie. heat removed from the cold reservoir) or 37680 kJ/hr or 10470 watts, W = Qc/COP = 10470/26.7 = 392 watts

  5. Feb 16, 2010 #4
    I looked through the thermodynamics whole chapter, and I couldn't find a single reference to this equation. Is it derived from the Carnot engine efficiency -> e=W/Q? Your solution makes sense, I'm just puzzled why the equation is not even in the book.

    Thanks for responding. I'm glad I didn't submit the assignments yet...
  6. Feb 16, 2010 #5

    Andrew Mason

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    Performance measurement depends on the function of the device.

    Efficiency (output/input or W/Qh) is used for heat engines since it produces work from heat input. For heat pumps, which deliver heat by moving it from colder to warmer reservoirs, a coefficient of performance (COP) is used of Qh/W (output/input being heat output to work input). For a refrigerator, which inputs work to remove heat from a cold reservoir, the performance is measured by Qc/W (output/input or heat removed to work input).

    But you don't really need to use COP explicity. In all cases: W = Qh-Qc and for Carnot cycles, [itex]\Delta S = Q_c/T_c - Q_h/T_h = 0[/itex] so Qc/Qh = Tc/Th. Note W = Qh-Qc = Qc(Qh/Qc - 1) = Qc(Th/Tc-1)

    In this case you are given the amount of heat that must be removed from the cold reservoir (Qc). You can just use the above line to find W from Qc.

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