Q107 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A Carnot engine with efficiency 50% takes heat from a source at 600 K. To increase the efficiency by 20%, keeping temperature of the sink same, the new temperature of the source will be
- A$300\ \text{K}$
- B$900\ \text{K}$
- Correct answerC$1000\ \text{K}$
- D$360\ \text{K}$
Solution
The efficiency of a Carnot engine is given by $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given initial efficiency $\eta_1 = 50\% = 0.5$ and source temperature $T_1 = 600\ \text{K}$.
$0.5 = 1 - \dfrac{T_2}{600}$
$\dfrac{T_2}{600} = 0.5 \Rightarrow T_2 = 300\ \text{K}$
The efficiency is increased by $20\%$, so the new efficiency is $\eta_2 = 50\% + 20\% = 70\% = 0.7$.
Let the new source temperature be $T_1'$. The sink temperature $T_2$ remains the same.
$0.7 = 1 - \dfrac{300}{T_1'}$
$\dfrac{300}{T_1'} = 0.3$
$T_1' = \dfrac{300}{0.3} = 1000\ \text{K}$
Answer: $1000\ \text{K}$
Given initial efficiency $\eta_1 = 50\% = 0.5$ and source temperature $T_1 = 600\ \text{K}$.
$0.5 = 1 - \dfrac{T_2}{600}$
$\dfrac{T_2}{600} = 0.5 \Rightarrow T_2 = 300\ \text{K}$
The efficiency is increased by $20\%$, so the new efficiency is $\eta_2 = 50\% + 20\% = 70\% = 0.7$.
Let the new source temperature be $T_1'$. The sink temperature $T_2$ remains the same.
$0.7 = 1 - \dfrac{300}{T_1'}$
$\dfrac{300}{T_1'} = 0.3$
$T_1' = \dfrac{300}{0.3} = 1000\ \text{K}$
Answer: $1000\ \text{K}$
Hence, the correct answer is option: 'C'.