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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.......\(J\)

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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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