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JEE Main 2012 (AIEEE) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2012 (AIEEE)

Q74JEE Main 2012 · AIEEEMedium

Heat Engines, Refrigerators & Second Law

A Carnot engine, whose efficiency is 40%40\%, takes in heat from a source maintained at a temperature of 500 K.500\ K. It is desired to have an engine of efficiency 60%.60\%. Then, the intake temperature for the same exhaust (sink) temperature must be:

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Answer: (c)

  1. T2=(1−0.4)(500)=300 KT_2 = (1 - 0.4)(500) = 300\text{ K}
  2. For η=0.6\eta = 0.6: T1′=3000.4=750 KT_1' = \frac{300}{0.4} = 750\text{ K}
  3. So the answer is (c) 750 K750\ K.
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