Q170 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
Carnot engine has an efficiency of $\dfrac{1}{7}$. When the temperature of the sink is reduced by $60$ K, its efficiency is doubled. The temperature of the source is
- A$47^\circ\text{C}$
- Correct answerB$147^\circ\text{C}$
- C$43^\circ\text{C}$
- D$143^\circ\text{C}$
Solution
The efficiency of a Carnot engine is given by $\eta = 1 - \dfrac{T_2}{T_1}$
Given $\eta = \dfrac{1}{7}$
$\dfrac{1}{7} = 1 - \dfrac{T_2}{T_1} \Rightarrow \dfrac{T_2}{T_1} = \dfrac{6}{7}$
When the sink temperature is reduced by $60$ K, the efficiency is doubled:
$\dfrac{2}{7} = 1 - \dfrac{T_2 - 60}{T_1} \Rightarrow \dfrac{T_2 - 60}{T_1} = \dfrac{5}{7}$
Subtracting the second equation from the first:
$\dfrac{T_2}{T_1} - \dfrac{T_2 - 60}{T_1} = \dfrac{6}{7} - \dfrac{5}{7}$
$\dfrac{60}{T_1} = \dfrac{1}{7}$
$T_1 = 420$ K
Converting to Celsius:
$T_1 = 420 - 273 = 147^\circ\text{C}$
Answer: $147^\circ\text{C}$
Given $\eta = \dfrac{1}{7}$
$\dfrac{1}{7} = 1 - \dfrac{T_2}{T_1} \Rightarrow \dfrac{T_2}{T_1} = \dfrac{6}{7}$
When the sink temperature is reduced by $60$ K, the efficiency is doubled:
$\dfrac{2}{7} = 1 - \dfrac{T_2 - 60}{T_1} \Rightarrow \dfrac{T_2 - 60}{T_1} = \dfrac{5}{7}$
Subtracting the second equation from the first:
$\dfrac{T_2}{T_1} - \dfrac{T_2 - 60}{T_1} = \dfrac{6}{7} - \dfrac{5}{7}$
$\dfrac{60}{T_1} = \dfrac{1}{7}$
$T_1 = 420$ K
Converting to Celsius:
$T_1 = 420 - 273 = 147^\circ\text{C}$
Answer: $147^\circ\text{C}$
Hence, the correct answer is option: 'B'.