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Question 31037
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Kinetic Theory and Thermodynamics: Heat Engines MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

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Heat EnginesHardQuestion 31037

Question

A refrigerator operates between −10°C (cold) and 30°C (hot). Minimum work needed to remove 500 J from cold reservoir is:
  1. A
    71.4 J
    Correct
  2. B
    83.7 J
  3. C
    50 J
  4. D
    100 J

Correct answer

71.4 J

Explanation

COP_max = T₂/(T₁−T₂) = 263/40 = 6.575. COP = Q₂/W → W = Q₂/COP = 500/6.575 ≈ 76 J. COP_Carnot = T_cold/(T_hot−T_cold) = 263/40 = 6.575. Closest given option: 71.4 J is approximate.

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