JEE Main 2022 · 24 June · morning · Single correct · +4 / −1

A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

  1. (a) Increases by 18°C
  2. (b) Increases by 200°C
  3. (c) Increases by 120°C
  4. (d) Increases by 73°C
Answer and solution

Answer (b) Increases by 200°C

Solution

Physics sir · 4 steps

Put the sink in kelvin, find the source for 25%, then find it again for 50%.

  1. Carnot efficiency. For a Carnot engine η = 1 - T_C/T_H, so T_H = T_C/(1-η). Both temperatures must be in kelvin, so the sink is T_C = 27 + 273 = 300 K.
  2. Original source. With η = 0.25, T_H = 300/0.75 = 400 K.
  3. New efficiency. Raising the efficiency by 100% of its original value doubles it, from 25% to 50%. Then T_H' = 300/0.5 = 600 K.
  4. So. The source must go from 400 K to 600 K, an increase of 200 K. A change of 200 K is a change of 200°C, so the answer is (b).

The idea to keep. Carnot efficiency uses absolute temperatures only. Convert to kelvin first. A temperature difference is the same in kelvin and in degrees Celsius.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 150 times in JEE Main since 2020. This question is from 2022.