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JEE Mains · Physics · STD 11 - 11. thermodynamics
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
- A \(\frac{2}{3} {T}_{1}+\frac{1}{3} T_{3}\)
- B \(\frac{3}{2} {T}_{1}+\frac{1}{3} {T}_{3}\)
- C \(\frac{1}{3} {T}_{1}+\frac{2}{3} {T}_{3}\)
- D \(\frac{2}{3} {T}_{1}+\frac{3}{2} {T}_{3}\)
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 \({W}_{{A}}=1-\frac{{Q}_{2}}{{Q}_{1}}=1-\frac{{T}}{{T}_{1}} \Rightarrow \frac{{Q}_{2}}{{Q}_{1}}=\frac{{T}}{{T}_{1}}\)…
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