JEE Mains · Physics · STD 11 - 11. thermodynamics
A Carnot's engine working between \(400\, K\) and \(800\, K\) has a work output of \(1200\, J\) per cycle. The amount of heat energy supplied to the engine from the source in each cycle is ........... \(J\)
- A \(3200\)
- B \(1800\)
- C \(1600\)
- D \(2400\)
Answer & Solution
Correct Answer
(D) \(2400\)
Step-by-step Solution
Detailed explanation
\(\eta=\frac{ T _{2}}{ T _{1}}=\frac{ Q _{2}}{ Q _{1}}=\frac{ Q _{1}- W }{ Q _{1}} \quad\left(\because W = Q _{1}- Q _{2}\right)\) \(\frac{400}{800}=1-\frac{ W }{ Q _{1}}\) \(\frac{ W }{ Q _{1}}=1-\frac{1}{2}=\frac{1}{2}\) \(Q _{1}=2 W =2400 J\)
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