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JEE Main 2018 · Physics · Thermodynamics and Kinetic Theory of Gases · easy

A Carnot refrigerator works between a cold chamber at -3^circC and a room at 27^circC. In each cycle it extracts 4500 J of heat from the chamber. The…

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A Carnot refrigerator works between a cold chamber at −3∘C-3^\circ\mathrm{C} and a room at 27∘C27^\circ\mathrm{C}. In each cycle it extracts 4500 J4500\ \mathrm{J} of heat from the chamber. The heat it rejects to the room in each cycle is
  1. A. 500 J500\ \mathrm{J}
  2. B. 5500 J5500\ \mathrm{J}
  3. C. 4050 J4050\ \mathrm{J}
  4. D. 5000 J5000\ \mathrm{J}
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Answer: (D) 5000 J5000\ \mathrm{J}

Convert to kelvin: Tc=270 KT_c=270\ \mathrm{K}, Th=300 KT_h=300\ \mathrm{K}. For a reversible cycle QhQc=ThTc\dfrac{Q_h}{Q_c}=\dfrac{T_h}{T_c}, so Qh=4500×300270=5000 JQ_h=4500\times\dfrac{300}{270}=5000\ \mathrm{J}. The work supplied is W=Qh−Qc=500 JW=Q_h-Q_c=500\ \mathrm{J}, giving a coefficient of performance TcTh−Tc=9=4500500\dfrac{T_c}{T_h-T_c}=9=\dfrac{4500}{500}, consistent. Using Celsius values directly would be wrong.
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Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake