Q104 / 150 · MHT CET 2026 (11 April Shift 2) · Physics

Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $0.2$. When $T_2$ is lowered by $45$ K, its efficiency becomes $0.5$. Temperatures $T_1$ and $T_2$ are respectively

  • Correct answer
    A
    $150$ K, $120$ K
  • B
    $120$ K, $150$ K
  • C
    $60$ K, $80$ K
  • D
    $80$ K, $60$ K

Solution

Efficiency of a Carnot engine is given by $\eta = 1 - \dfrac{T_2}{T_1}$

Given $\eta = 0.2$

$1 - \dfrac{T_2}{T_1} = 0.2 \Rightarrow \dfrac{T_2}{T_1} = 0.8 \Rightarrow T_2 = 0.8 T_1$

When $T_2$ is lowered by $45$ K, efficiency becomes $0.5$

$1 - \dfrac{T_2 - 45}{T_1} = 0.5 \Rightarrow \dfrac{T_2 - 45}{T_1} = 0.5$

$T_2 - 45 = 0.5 T_1$

Substituting $T_2 = 0.8 T_1$

$0.8 T_1 - 45 = 0.5 T_1$

$0.3 T_1 = 45 \Rightarrow T_1 = 150$ K

$T_2 = 0.8 \times 150 = 120$ K

Answer: $150$ K, $120$ K

Hence, the correct answer is option: 'A'.

Relevant Questions