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JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2021
- Hard
- MCQ
Two carnot engines \(A\) and \(B\) operate in series such that engine \(A\) absorbs heat at \(T_{1}\) and rejects heat to a sink at temperature \(T\). Engine \(B\) absorbs half of the heat rejected by engine \(A\) and rejects heat to the sink at \({T}_{3}\). When workdone in both the cases is equal, the value of \({T}\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{2}{3} {T}_{1}+\frac{1}{3} T_{3}\)
Step-by-step Solution
Detailed explanation
carnot engine is as shown
\({W}_{{A}}=1-\frac{{Q}_{2}}{{Q}_{1}}=1-\frac{{T}}{{T}_{1}} \Rightarrow \frac{{Q}_{2}}{{Q}_{1}}=\frac{{T}}{{T}_{1}}\)
\({W}_{{B}}=1-\frac{{Q}_{3}}{\left({Q}_{2} / 2\right)}=1-\frac{{T}_{3}}{{T}} \Rightarrow \frac{2 {Q}_{3}}{{Q}_{2}}=\frac{{T}_{3}}{{T}}\)
Now, \({W}_{{A}}={W}_{{B}}\)
\({Q}_{1}-{Q}_{2}=\frac{{Q}_{2}}{2}-{Q}_{3}\)
\(\Rightarrow \frac{2 {Q}_{1}}{{Q}_{2}}+\frac{2 {Q}_{3}}{{Q}_{2}}=3\)
\(\Rightarrow \frac{2 {T}_{1}}{{T}}+\frac{{T}_{3}}{{T}}=3\)
\(\frac{2 {T}_{1}}{3}+\frac{{T}_{3}}{3}={T}\)
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