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Carnot efficiency: is heat always positive?

  1. Mar 22, 2010 #1
    This is a very quick question but in the equation


    do we treat both Q's as positive even though heat goes in in one case and out in the other? Or is there some detail I'm missing that would otherwise make the result less than 1.
  2. jcsd
  3. Mar 22, 2010 #2
    The answer is yes. The minus sign takes into account the difference you stated above.
  4. Mar 23, 2010 #3

    Andrew Mason

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    Giordanus is correct. Heat flow is measured relative to the engine (gas): Qh is into the gas - positive - and Qc is out of the gas - negative. The equation derives from the first law:

    [tex]\Delta Q = \Delta U + W[/tex]

    After one complete cycle, there is no change in internal energy of the gas so:

    [tex]\Delta Q = W[/tex]

    This means that the heat flow into the gas minus the heat flow out of the gas equals the work done by the gas:

    [tex]\Delta Q = Q_h - Q_c = W[/tex]

    Since efficiency is Output/Input

    [tex]\eta = W/Q_h = (Q_h - Q_c)/Q_h = 1 - \frac{Q_c}{Q_h}[/tex]

  5. Mar 23, 2010 #4
    Thanks Andrew, I get the overall idea that you want to measure the ratio of heat lost to heat gained. I was running into confusion because, as you stated QH is positive and QC is negative. If I used this convention then the efficiency equation would give a value greater than 1...

    So would it not be more accurate to say that QH is the heat put into the system, and QC is the heat expelled by the system, and so both quantities are treated as positive?
  6. Mar 23, 2010 #5

    Andrew Mason

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    Efficiency is not a measure of heat lost to heat gained. It is a measure of the rate of work done to heat energy supplied. The heat lost is the heat delivered to the cold reservoir. So efficiency is a measure of the rate at which the heat is NOT lost to the cold reservoir to the heat supplied.

    Qc and Qh refer to the magnitude of the heat flows to/from the reservoirs to the engine. If that is all you are saying, you are correct.

    Last edited: Mar 24, 2010
  7. Apr 14, 2010 #6
    giordanus and mason are both right, since w=qh-qc, the efficiency(e)=Weng/Qh,

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