Q90 / 90 · JEE Main 2021 (16 Mar Shift 2) · Physics

For an ideal heat engine, the temperature of the source is $127 ° C$ . In order to have $60 %$ efficiency the temperature of the sink should be __________ ${}^{\circ}C$ . (Round off to the nearest integer)

Correct answer: -113

Solution

$n = 0.60 = 1 - \frac{T_{L}}{T_{H}}$

$\frac{T_{L}}{T_{H}} = 0.4 \Rightarrow T_{L} = 0.4 \times 400$

$= 160 K$

$= - 113 ° C$

Hence, the correct answer is -113.

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