Q121 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
A Carnot engine, whose efficiency is 40% takes heat from a source maintained at temperature 600K. To have an efficiency 60%, the intake temperature for the same exhaust temperature should be
- A$1800$ K
- Correct answerB$900$ K
- C$720$ K
- D$360$ K
Solution
The efficiency of a Carnot engine is given by $\eta = 1 - \dfrac{T_2}{T_1}$
Given $\eta_1 = 40\% = 0.4$ and $T_1 = 600$ K
$0.4 = 1 - \dfrac{T_2}{600}$
$\dfrac{T_2}{600} = 0.6 \Rightarrow T_2 = 360$ K
For the new efficiency $\eta_2 = 60\% = 0.6$ with the same exhaust temperature $T_2 = 360$ K, let the new intake temperature be $T_1'$
$0.6 = 1 - \dfrac{360}{T_1'}$
$\dfrac{360}{T_1'} = 0.4$
$T_1' = \dfrac{360}{0.4} = 900$ K
Answer: $900$ K
Given $\eta_1 = 40\% = 0.4$ and $T_1 = 600$ K
$0.4 = 1 - \dfrac{T_2}{600}$
$\dfrac{T_2}{600} = 0.6 \Rightarrow T_2 = 360$ K
For the new efficiency $\eta_2 = 60\% = 0.6$ with the same exhaust temperature $T_2 = 360$ K, let the new intake temperature be $T_1'$
$0.6 = 1 - \dfrac{360}{T_1'}$
$\dfrac{360}{T_1'} = 0.4$
$T_1' = \dfrac{360}{0.4} = 900$ K
Answer: $900$ K
Hence, the correct answer is option: 'B'.