JEE Main 2021 · 25 July · evening · Single correct · +4 / −1

A heat engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency get doubled. The temperature of the source is :

  1. (a) 124°C
  2. (b) 37°C
  3. (c) 62°C
  4. (d) 99°C
Answer and solution

Answer (d) 99°C

Solution

Physics sir · 4 steps

Write the efficiency twice, once before and once after cooling the sink, then subtract to find T1.

  1. Before the change. With absolute temperatures, η = 1 - T₂/T₁. So 1 - T₂/T₁ = 1/6, which gives T₂/T₁ = 5/6.
  2. After the change. A drop of 62°C is the same as a drop of 62 K. The efficiency doubles to 1/3, so (T₂ - 62)/T₁ = 2/3.

    T₂/T₁ - (T₂ - 62)/T₁ = 5/6 - 2/3 ⇒ 62/T₁ = 1/6

  3. Solve. T₁ = 6 × 62 = 372 K. The sink is at T₂ = 5/6 × 372 = 310 K.
  4. So. The source is at 372 - 273 = 99°C, which is option (d).

The idea to keep. Efficiency formulas need kelvin, but a temperature change is the same in kelvin and Celsius. Subtract the two efficiency equations so that only the known change is left.

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 2021.