JEE Mains · Physics · STD 11 - 11. thermodynamics
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 \(\times 10^{6} \; J\)
- A \(16\)
- B \(26\)
- C \(36\)
- D \(46\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\(\left(T_{2}\right) T_{\sin k}=200 \; K\) \(\left(T_{1}\right) T_{\text {Reservoir }}=527+273=800 \; K\) \(W=12000 K J=12 \times 10^{6} \; J\) \(Q_{1}=\) ? \(\eta=1-\frac{T_{2}}{T_{1}}=\frac{W}{Q_{1}}=1-\frac{200}{800}=\frac{12 \times 10^{6}}{Q_{1}}\)…
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