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Question 31037
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Kinetic Theory and Thermodynamics: Heat Engines MCQ

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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 the cold reservoir is:
  1. A
    76.0 J
    Correct
  2. B
    71.4 J
  3. C
    83.7 J
  4. D
    100 J

Correct answer

76.0 J

Explanation

For a reversible refrigerator, COP=Tc/(Th−Tc)=263/40=6.575COP=T_c/(T_h-T_c)=263/40=6.575. Therefore W=Qc/COP=500/6.575≈<strong>76.0 J</strong>W=Q_c/COP=500/6.575≈<strong>76.0\text{ J}</strong>.

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