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JEE Main202224 Jun 2022Morning ShiftPhysicsThermodynamicsActual

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 ?

Options

  1. AIncreases by 18 ° C
  2. BIncreases by 200 ° C
  3. CIncreases by 120 ° C
  4. DIncreases by 73 ° C

Correct answer

B. Increases by 200 ° C

Step-by-step solution

Efficiency of carnot engine is given by, η = 1 - T sink T source . For initial case: T sink = 27 + 273   ° C = 300   K So we can write, 1 4 = 1 - 300 T 1 ⇒ T 1 = 400   K Now for the second case: T sink = 300   K After increasing the previous efficiency by 100%, value of efficiency will get doubled. Therefore, 1 2 = 1 - 300 T 2 ⇒ T 2 = 600   K Increase in temperature required will be, 600   K - 400   K = 200   K

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