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Alright, I'm completely stuck on this and so is everyone else I've asked.

A freezer has a temperature $T_\mathrm{c} = -23\,\rm ^\circ C$. The air in the kitchen has a temperature $T_\mathrm{h} = 27\,\rm ^\circ C$. The freezer in not perfectly insulated and heat leaks through the walls of the freezer at a rate of $50\,\rm W$. Find the power of the motor that is needed to maintain the temperature in the freezer.

The answer to this question is $10\,\rm W$.

I don't understand how this is possible: $50\,\rm W$ is literally $\mathrm{J/s}$ in heat. To keep the temperature the same, the amount of heat in the freezer has to be the same. If the motor only removes heat at a rate of $10\,\rm J/s$, then $40\,\rm J$ is being added to the fridge every second, and the temperature changes in response.

Is there something I'm missing?

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  • $\begingroup$ Think of energy conservation, and think of the motor in the fridge adding heat to the inside of the refrigerator (cold reservoir) and flowing to the outside (hot reservoir). In other words, the motor takes heat away from the cold and brings it to the hot. Compare this to what you expect the "leaky walls" to do... $\endgroup$
    – levitopher
    Jul 4 '14 at 5:24
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You're mistaking the power to move the motor for the transferred heat.

It's possible that the refrigerant removes 50 W of heat from the freezer, then receives 10 W of mechanical power from the compressor (which you call motor). Afterwards, it dumps 60 W of heat to the environment. See, this cycle perfectly satisfies the First Law of Thermodynamics.

Note that, in refrigeration, this is the norm: removed heat is (desirably) much bigger than the input power. We call it the beta of the cycle. In your example, $\beta = \frac{50}{10}=5$. It's the ratio of "useful energy transfer" (in this case, removed heat) over "input power used" (in this case, electrical power to the compressor). It means that, for every watt spent, you get 5 watts of useful stuff.

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  • $\begingroup$ I think the confusion is coming from the fact that this was an extra credit question and we didn't go over thermodynamics at all. I thought this was just a general question about thermodynamics... never knew B was used that way. $\endgroup$
    – NmdMystery
    Jul 4 '14 at 5:13
  • $\begingroup$ In any case, I think the answer I gave should be understandable, right? $\endgroup$ Jul 4 '14 at 5:14
  • $\begingroup$ Yup! Makes perfect sense. $\endgroup$
    – NmdMystery
    Jul 4 '14 at 5:21

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