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

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answer is 208.

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Detailed Solution

$\begin{array}{l} η = 1 - \frac{T_{2}}{T_{1}} \\ C a s e 1 : I f h e a t e n g i n e c o n v e r t s 1 / 4^{t h} \\ o f t h e h e a t i n p u t i n t o w o r k t h e n, \\ e f f i c i e n c y = \frac{1}{4} \\ \frac{1}{4} = 1 - \frac{T_{2}}{T_{1}} \\ \Rightarrow \frac{3}{4} = \frac{T_{2}}{T_{1}} \\ \Rightarrow T_{2} = \frac{3}{4} T_{1} \\ C a s e 2 : I f t h e t e m p e r a t u r e o f t h e \sin k i s r e d u c e d b y 52 K, \\ i t s e f f i c i e n c y i s d o u b l e d . \\ \frac{2}{4} = 1 - \frac{(T_{2} - 52)}{T_{1}} \\ \Rightarrow \frac{(T_{2} - 52)}{T_{1}} = \frac{1}{2} \\ \Rightarrow T_{2} - 52 = \frac{T_{1}}{2} \\ \Rightarrow \frac{3}{4} T_{1} - 52 = \frac{T_{1}}{2} \\ {\Rightarrow T}_{1} = 208 K \end{array}$