Q73 / 90 · JEE Main 2021 (31 Aug Shift 1) · Physics

A reversible engine has an efficiency of $\frac{1}{4} .$ If the temperature of the sink is reduced by $58 ° C,$ its efficiency becomes double. Calculate the temperature of the sink:

A
$180.4 ° C$
B
$382 ° C$
Correct answer
C
$174 K$
D
$280 ° C$

Solution

$η = 1 - \frac{T_{L}}{T_{H}}$
$\frac{1}{4} = 1 - \frac{T_{L}}{T_{H}}$
$\frac{T_{L}}{T_{H}} = \frac{3}{4}$
$2 η = \frac{1}{2} = 1 - \frac{T_{L} - 58}{T_{H}}$
$\frac{T_{L} - 58}{T_{H}} = \frac{1}{2}$
$\frac{T_{L}}{T_{L} - 58} = \frac{3 / 4}{1 / 2}$
$2 T_{L} = 3 T_{L} - 174$
$T_{L} = 174 K$

This question is bonus as the value given by NTA in option was in degree celsius.

Hence, the correct answer is option: 'C'.

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A reversible engine has an efficiency of 14. If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink:

Solution

Relevant Questions

A reversible engine has an efficiency of $\frac{1}{4} .$ If the temperature of the sink is reduced by $58 ° C,$ its efficiency becomes double. Calculate the temperature of the sink: