Heat Engine; Carnot Engine and refrigerator

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Question

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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Solution

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

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Heat Engine; Carnot Engine and refrigerator

Remember concepts with our Masterclasses.

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 ——

η=1-T2T1Case 1 : If heat engine converts one – fourth of the heat input into workthen its efficiency =1414=1-T2T1⇒34=T2T1⇒T2=34T1Case 2 : If the temperature of the sink is reduced by 52K, its efficiency is doubled24=1-T2-52T1⇒T2-52T1=12⇒T2-52=T12⇒34T1-52=T12 ⇒T1=208 K

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