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A Carnot-cycle air conditioner operates between a room at temperature Tm and the outside at temperature Tv. Heat leaks into the room at rate dQ/dt = k * (Tv - Tm). Find the minimum electrical power P (in W) needed to maintain the room at steady temperature, given Tm = 20 deg C (= 293 K), Tv = 40 deg C (= 313 K), and k = 293 W/K.

1. 293
2. 400
3. 586
4. 200

Correct answer: 400

Solution

At steady state, the heat extracted from room per second = heat leak rate = k*(Tv-Tm) = 293 * (313-293) = 293 * 20 = 5860 W. For a Carnot refrigerator (maximum COP device), COP = Tm/(Tv-Tm) = 293/20 = 14.65. This means for every 1 W of electrical input, 14.65 W of heat is removed from room. So minimum power input P = Qm_rate / COP = 5860 / 14.65 = 400 W. Note: Some versions of this problem quote 293 W as the answer using P = k*(Tv-Tm)²/Tv = 293*400/313 approximately 374 W, or other approximations. The exact Carnot calculation gives P = k*(Tv-Tm)²/Tm = 293*400/293 = 400 W.

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