Fridge door open makes the room hotter, COP of 9 seems odd, and fridge cooling lowers entropy. Does it break the second law?

7 answers & replies

• [+][–]Nisha Verma12thSample· 3d ago

  Bookkeeping for the open door: the fridge pulls $Q_2$ out of the air inside (which is the room) and rejects $Q_1 = Q_2 + W$ into the room through the coils at the back. With the door open, the "inside" is the room itself, so the net effect on the room is that it receives $W$ extra:

  $Q_1 - Q_2 = W > 0$

  The electrical work done by the compressor ends up as heat in the room. So the room warms. A fridge cools one place only by dumping more heat elsewhere.

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• [+][–]Arjun Nair12thSample· 3d ago

  COP isn't an efficiency, it's the ratio of what you want (heat removed from inside, $Q_2$) to what you pay (work $W$). For an engine, the useful output (work) can't exceed the input heat, so efficiency < 1. For a fridge, the fridge isn't creating the heat it moves, it is just moving it from cold to hot using a little work. So $Q_2$ can be much larger than $W$, and $\alpha = 9$ means for every 1 J of work, 9 J of heat is pumped out of the inside.

  Note that the smaller the temperature difference between inside and outside, the larger the COP. As $T_1 \to T_2$, $\alpha \to \infty$, and that's why a fridge set to a less cold temperature is cheaper to run.

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• [+][–]Kabir Joshi11thSample· 3d ago

  Inside a real fridge: the compressor does the work W, and the refrigerant circulates. In the evaporator (inside, cold) it evaporates and absorbs $Q_2$. The compressor raises its pressure and temperature, and in the condenser coils at the back it gives out $Q_1$ to the room and liquefies. Then it passes through a narrow valve and the pressure falls again. So the warm coils at the back are literally $Q_1 = Q_2 + W$.

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• [+][–]Zoya Khan12thSample· 3d ago

  On energy: COP 9 doesn't break conservation, because the 9 J isn't created. It's taken from the cold space, and the total delivered to the room is $Q_1 = Q_2 + W = 9 + 1 = 10$ J. Energy in = 1 J (work) + 9 J (from inside) = 10 J out. Balanced.

  Rahul, for 3, entropy. The inside loses $Q_2$ at $T_2$, so its entropy changes by $-Q_2/T_2$. The room gains $Q_1$ at $T_1$ ($T_1 > T_2$), so its entropy changes by $+Q_1/T_1$. Total:

  $\Delta S = \frac{Q_1}{T_1} - \frac{Q_2}{T_2} \geq 0$

  For an ideal (Carnot) fridge it's exactly zero, for a real one it's positive. The second law applies to the whole (fridge plus surroundings), not to the interior alone. Entropy can decrease locally as long as it increases more somewhere else.

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  • [+][–]Rahul DubeyDropperSample· 3d agoAuthor

    That bookkeeping with 1 J and 9 J and 10 J is the cleanest I've seen. And "second law applies to the whole system" is the line I'd been missing. Thanks, everyone.

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  • [+][–]Kabir Joshi11thSample· 3d ago

    ok so the 9 J isn't made, just moved. nice way to see it. COP is not efficiency

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• [+][–]Vivek Mishra12thSample· 3d ago

  In simple words: the second law (Clausius statement) says heat won't flow from cold to hot by itself. A refrigerator makes it happen, but only because you give work, using a compressor. So a refrigerator is a heat engine running backwards. If it could work with no external work, that would break the law; with work, no problem. And that is why the back of the fridge is warm.

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