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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2022
  • Medium
  • Numerical

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 \(\times 10^{6} \; J\)

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Answer & Solution

Correct Answer

(A) \(16\)

Step-by-step Solution

Detailed explanation

\(\left(T_{2}\right) T_{\sin k}=200 \; K\)

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

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

\(Q_{1}=\) ?

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

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

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