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Homework Help: How to convert power to energy in a heat pump problem?

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A heat pump fitted with a 120W electric motor supplies 600W of heat per second to a room from outdoors. Calculate: a) the coefficient of performance of the heat pump
    b) the heat gained from outdoors.

    2. Relevant equations
    Coeff of performance for heat pump = Qin/ W = Qin/(Qin-Qout)
    Qin = Qout + W

    3. The attempt at a solution
    P = 600J/s
    W = 120

    P = Wt --> 600 = 120t --> t = 0.2s
    Qout = Pt = 600 x 0.2 = 120J

    Qin = 120 + 120 = 240

    Qin/Qin - Qout = 240/240-120 = 2

    The answer is 5.


    The answer is 480W

    Please help....
    Last edited: Apr 15, 2015
  2. jcsd
  3. Apr 15, 2015 #2


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    Dear T,

    Unfortunately, the errors start already with the rendering of the problem statement: you can't "supply 600W of heat per second" :
    either you supply 600 Watt or you supply 600 J/s. No wonder, because ##1W \equiv 1J/s##.

    My guess: you misread or mistyped.

    It gets a little worse when you write "P = Wt --> 600 = 120t --> t = 0.2s" . This is the wrong calculation, but on top of that it's executed wrong. You should be able to check the results from your calculator, e.g. 600 divided by approximately 100 should be approximately 6. (Or you know that 5 x 12 = 60 so 5 x 120 = 600).

    And -- considering the above -- including the dimensions: 600 W / 120 W = 5 (without a dimension: it's a ratio).

    Funny that this 5 looks a lot like the answer to (a). Any idea why ?

    (b) is easy now, isn't it ?
  4. Apr 15, 2015 #3

    So 600 is the Qin, and work done 120?
    I want to use the equation somehow but I just literally cannot understand what part of the equation those numbers are...
  5. Apr 15, 2015 #4
    You know what, its funny because I mistakenly said its not 600W but 900W in my first tries, which is why I failed to get the right answer...
  6. Apr 15, 2015 #5


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    I take it this exercise was given to you in the context of a lecture or a chapter on heat pumps or so ?

    For a heat pump, in ##CoP = {Q\over W}## Q is the heat delivered and W is the work consumed by the heat pump.

    It's kind of the opposite of a Carnot engine, where Qin is picked up, Qout is given off and W is delivered.

    Carnot efficiency can never be > 1 , but a CoP can.

    Work a bit slower, check all and every dimensions; before using a calculator, estimate the outcome of calculations. Try not to work too fast.
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