JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2020
- Medium
- Numerical
A heat engine is involved with exchange of heat of \(1915\, J,\) \(-40\, J ,+125\, J\) and \(-Q\,J\), during one cycle achieving an efficiency of \(50.0 \%\). The value of \(Q\) is.......\(J\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(980\)
Step-by-step Solution
Detailed explanation
\(\eta=\frac{\text { Work done }}{\text { Heat supplied }}\)
\(\frac{1}{2}=\eta=\frac{1915-40+125- Q }{1915+125}\)
\(\frac{1}{2}=\frac{2000- Q }{2040}\)
\(2040=4000-2 Q\)
\(2 Q =1960\)
\(Q =980 J\)
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