JEE Main 2023 · 25 January · morning · Single correct · +4 / −1

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :

  1. (a) 300 K
  2. (b) 1000 K
  3. (c) 900 K
  4. (d) 360 K
Answer and solution

Answer (b) 1000 K

Solution

Physics sir · 3 steps

Use the first efficiency to find the sink temperature, then hold that sink fixed and solve for the new source.

  1. Carnot efficiency. For a Carnot engine the efficiency depends only on the two temperatures, in kelvin: η = 1 - T₂/T₁, where T₁ is the source and T₂ is the sink.
  2. Find the sink. With η = 0.5 and T₁ = 600 K, you get T₂/600 = 0.5, so T₂ = 300 K.

    0.7 = 1 - 300/T₁' ⇒ T₁' = 300/0.3 = 1000 K

  3. So. To reach 70% efficiency with the sink still at 300 K, the source must be raised to 1000 K, which is option (b).

The idea to keep. In Carnot problems, T₂/T₁ = 1 - η. Find whichever temperature stays fixed first, then use the new efficiency to get the one that changes.

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

Study Thermodynamics

Asked about 150 times in JEE Main since 2020. This question is from 2023.