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Physics - Heat Pump - 2nd Law of Thermodynamics

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    A heat pump has a coefficient of performance of 3.70 and operates with a power consumption of 6.91 103 W.
    (a) How much energy does it deliver into a home during 1 h of continuous operation?
    9.2x10^7 J

    (b) How much energy does it extract from the outside air?
    ? J

    2. Relevant equations
    COP(coefficient of performance) = |Q_h|/W
    P = W/t
    W = P*t

    e = W_eng/|Q_h| = (|Q_h|-|Q_c|)/|Q_h| = 1 - |Q_c|/|Q_h|
    3. The attempt at a solution

    (a) I got this part correct.
    COP = |Q_h|/W
    P = W/t
    W = P*t

    |Q_h| = COP*W = COP * P*t = 3.70(6.91x10^3 J/s)(3600s) = 9.2x10^7 J

    (b) I need help on part (b). I can't get the correct answer.

    I tried using e = 1 - |Q_c|/|Q_h|

    |Q_c| = |Q_h|(e-1) = 9.2x10^7 J (3.70-1)

    but that answer doesn't work.
  2. jcsd
  3. Sep 17, 2010 #2
    Use W=H1-H2
    arranging equations

    U have got H2 and W on part a.
    Now u need to find H1. Just plus them.
    Last edited: Sep 17, 2010
  4. Sep 17, 2010 #3
    Thanks for replying. I actually figured it out by looking at a previous problem I did where it was |Q_c| = |Q_h| + W_eng.
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