Q 11-11-132JEE MainJEE Main 2020 (8 Jan, Shift 2)Easy
A Carnot engine having an efficiency of $\dfrac{1}{10}$ is being used as a refrigerator. If the work done on the refrigerator is $10$ J, the amount of heat absorbed from the reservoir at a lower temperature is
Answer: (D) $90$ J
$\eta = 1 - \dfrac{T_2}{T_1} = \dfrac{1}{10}$, so $\dfrac{T_2}{T_1} = \dfrac{9}{10}$.
Coefficient of performance of the reversed engine:
$$\beta = \frac{T_2}{T_1 - T_2} = \frac{9}{1} = 9$$
Heat taken from the cold reservoir: $Q_2 = \beta W = 9\times10 = 90$ J.
Solution by Sreeraj P, M.Sc Physics