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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2023
  • Medium
  • Numerical

Work done by a Carnot engine operating between temperatures \(127^{\circ}\,C\) and \(27^{\circ}\,C\) is \(2\,kJ\). The amount of heat transferred to the engine by the reservoir is \(........\,kJ\)

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Answer & Solution

Correct Answer

(C) \(8\)

Step-by-step Solution

Detailed explanation

\(\eta=1-\frac{ T _2}{ T _1}=\frac{ W }{ Q _1}\)

\(\Rightarrow \frac{ W }{ Q _1}=1-\frac{300}{400}=\frac{1}{4}\)

\(\Rightarrow \frac{2 kJ }{ Q _1}=\frac{1}{4}\)

\(\Rightarrow Q _1=8\,kJ\)

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