Q 11-11-108JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
In a carnot engine, the temperature of reservoir is $527^\circ$C and that of sink is $200$ K. If the workdone by the engine when it transfers heat from reservoir to sink is $12000$ kJ, the quantity of heat absorbed by the engine from reservoir is ______ $\times10^6$ J.
Numerical value type. Enter your answer.
Answer: 16
$T_1 = 527 + 273 = 800$ K, $T_2 = 200$ K.
$$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{200}{800} = \frac34$$
$$Q_1 = \frac{W}{\eta} = \frac{12000\ \text{kJ}}{3/4} = 16000\ \text{kJ} = 16\times10^6\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics