JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2020
- Medium
- MCQ
Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2} .\) The temperature of the hot reservoir of the first engine is \(\mathrm{T}_{1}\) and the temperature of the cold reservoir of the second engine is \(\mathrm{T}_{2} . T\) is temperature of the sink of first engine which is also the source for the second engine. How is \(T\) related to \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\), if both the engines perform equal amount of work?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\mathrm{T}=\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{L}}}=\frac{\mathrm{T}_{1}}{\mathrm{T}}\) and \(\mathrm{W}=\mathrm{Q}_{\mathrm{H}}-\mathrm{Q}_{\mathrm{L}}\)
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{L}}^{\prime}}=\frac{\mathrm{T}}{\mathrm{T}_{2}}\) and \(\mathrm{W}=\mathrm{Q}_{\mathrm{L}}-\mathrm{Q}_{\mathrm{L}}\)
From \((1)\) and \((2)\)
we get \(\mathrm{T}=\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\)
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