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
JEE MAIN 2021, Diffcult
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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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