Q81 / 90 · JEE Main 2021 (25 Feb Shift 2) · Physics

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by $52 K,$ its efficiency is doubled. The temperature in Kelvin of the source will be ___ .

Correct answer: 208

Solution

Efficiency of reversible engine is $η = \frac{1}{4} = \frac{W}{Q_{in}} = 1 - \frac{T_{2}}{T_{1}}$

From above, $\frac{T_{2}}{T_{1}} = \frac{3}{4}$

When the temperature of the sink is reduced by $52 K$ then its efficiency is doubled.

Then, $\frac{T_{2} - 52}{T_{1}} = \frac{1}{2}$

$\Rightarrow \frac{T_{2}}{T_{1}} - \frac{52}{T_{1}} = \frac{1}{2} \Rightarrow \frac{3}{4} - \frac{52}{T_{1}} = \frac{1}{2} \Rightarrow \frac{52}{T_{1}} = \frac{3}{4} - \frac{1}{2} = \frac{1}{4} \Rightarrow T_{1} = 52 \times 4 = 208 K$

Hence, the correct answer is 208.

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A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ___ .

Solution

Relevant Questions

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by $52 K,$ its efficiency is doubled. The temperature in Kelvin of the source will be ___ .