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NEET 2016 (Phase 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from NEET 2016 (Phase 1)

Q67NEET 2016 · PHASE 1Medium

Heat Engines, Refrigerators & Second Law

A refrigerator works between 4oC and 30oC. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 Joules)

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Answer: (a)

  1. T2=4∘C=277T_2 = 4^\circ\text{C} = 277 K, T1=30∘C=303T_1 = 30^\circ\text{C} = 303 K, so β=T2T1−T2=27726≈10.65\beta = \frac{T_2}{T_1 - T_2} = \frac{277}{26} \approx 10.65.
  2. Heat removed per second: Q2=600×4.2=2520Q_2 = 600 \times 4.2 = 2520 W
  3. Power P=Q2β=2520×26277≈236.5P = \frac{Q_2}{\beta} = \frac{2520 \times 26}{277} \approx 236.5 W
  4. So the answer is (a) 236.5 W.
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