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NEET 2016 (Phase 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from NEET 2016 (Phase 2)

Q66NEET 2016 · PHASE 2Easy

Heat Engines, Refrigerators & Second Law

The temperature inside a refrigerator is t2 oC and the room temperature is t1 oC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

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Answer: (b)

Ideal refrigerator: $\frac{Q_1}{T_1} = \frac{Q_2}{T_2}$ and $W = Q_1 - Q_2$
Heat to room per joule: $\frac{Q_1}{W} = \frac{T_1}{T_1 - T_2}$
$T_1 = t_1 + 273$ and $T_1 - T_2 = t_1 - t_2$ , so $\frac{Q_1}{W} = \frac{t_1 + 273}{t_1 - t_2}$
So the answer is (b).

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