JEE Main 2022 · 25 July · evening · Single correct · +4 / −1

Let η₁ is the efficiency of an engine at T₁=447° C and T₂=147° C while η₂ is the efficiency at T₁=947° C and T₂=47° C The ratio η₁/η₂ will be :

  1. (a) 0.41
  2. (b) 0.56
  3. (c) 0.73
  4. (d) 0.70
Answer and solution

Answer (b) 0.56

Solution

Physics sir · 3 steps

Convert both temperatures to kelvin first, then the Carnot efficiency formula does the rest.

  1. Kelvin first. Carnot efficiency η = 1 - T₂/T₁ works only with absolute temperatures. Add 273 to every Celsius value: 447°C = 720 K, 147°C = 420 K, 947°C = 1220 K, 47°C = 320 K.
  2. Each efficiency. η₁ = 1 - 420/720 = 300/720 = 5/12 ≈ 0.417. η₂ = 1 - 320/1220 = 900/1220 = 45/61 ≈ 0.738.

    η₁/η₂ = (5/12)/(45/61) = (5 × 61)/(12 × 45) = 61/108 ≈ 0.565

  3. So. The ratio is about 0.56, which is option (b).

The idea to keep. Any formula with a temperature ratio, like Carnot efficiency, needs kelvin. Celsius values give a wrong ratio because the zero is in the wrong place.

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 2022.