JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2021
- Medium
- Numerical
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 ........... \(J\)
Pick an option to check your answer.
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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