In a carnot engine, the temperature of reservoir is $527^{\circ} C$ and that of $\operatorname{sink}$ 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 $\times 10^{6} \; J$
JEE MAIN 2022, Medium
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$\left(T_{2}\right) T_{\sin k}=200 \; K$

$\left(T_{1}\right) T_{\text {Reservoir }}=527+273=800 \; K$

$W=12000 K J=12 \times 10^{6} \; J$

$Q_{1}=$ ?

$\eta=1-\frac{T_{2}}{T_{1}}=\frac{W}{Q_{1}}=1-\frac{200}{800}=\frac{12 \times 10^{6}}{Q_{1}}$

$\frac{3}{4}=\frac{12 \times 10^{6}}{Q_{1}}=Q_{1}=16 \times 10^{6} J$

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