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

A Carnot engine operating between two reservoirs has efficiency \(\frac{1}{3}\). When the temperature of cold reservoir raised by \(x\), its efficiency decreases to \(\frac{1}{6}\). The value of \(x\), if the temperature of hot reservoir is \(99^{\circ}\,C\), will be \(........\,K\)

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Answer & Solution

Correct Answer

(D) \(62\)

Step-by-step Solution

Detailed explanation

\(T _{ H }=99^{\circ} C =99+273\)

\(\qquad=372\,K\)

\(1-\frac{ T _{ C }}{ T _{ H }}=\frac{1}{3}\)

\(\frac{ T _{ C }}{ T _{ H }}=\frac{2}{3} \quad(1) \Rightarrow T _{ C }=\frac{2}{3} \times 372\)

\(1-\frac{ T _{ C }+ X }{ T _{ H }}=\frac{1}{6}\)

\(\frac{5}{6}=\frac{ T _{ C }+ X }{ T _{ H }}\)

\(\frac{5}{6}=\frac{248+ X }{372}\)

\(248+ X =5 \times 62\)

\(X =310-248=62\,K\)

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