JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2025
- Medium
- MCQ
A Carnot engine \((\mathrm{E})\) is working between two temperatures 473 K and 273 K . In a new system two engines - engine \(E_1\) works between 473 K to 373 K and engine \(E_2\) works between 373 K to 273 K .If \(\eta_{12}, \eta_1\) and \(\eta_2\) are the efficiencies of the engines \(E, E_1\) and \(E_2\), respectively, then
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\eta_{12} \lt \eta_1+\eta_2\)
Step-by-step Solution
Detailed explanation
Efficiencies of a carnot engine \(\eta=1-\frac{T_{\text {sink }}}{T_{\text {source }}}\)
\(\begin{aligned}
\Rightarrow & \eta_1=1-\frac{373 \mathrm{~K}}{473 \mathrm{~K}}=\frac{100}{473} \\
& \eta_2=1-\frac{273 \mathrm{~K}}{373 \mathrm{~K}}=\frac{100}{373} \\
& \eta_{12}=1-\frac{273 \mathrm{~K}}{473 \mathrm{~K}}=\frac{100}{473} \\
& \eta_{12}-\eta_1=\frac{200}{473}-\frac{100}{473}=\frac{100}{473} < \frac{100}{373} \\
\Rightarrow & \eta_{12}-\eta_1 < \eta_2 \\
\text { or } & \eta_{12} < \eta_1+\eta_2
\end{aligned}\)
\(\begin{aligned}
\Rightarrow & \eta_1=1-\frac{373 \mathrm{~K}}{473 \mathrm{~K}}=\frac{100}{473} \\
& \eta_2=1-\frac{273 \mathrm{~K}}{373 \mathrm{~K}}=\frac{100}{373} \\
& \eta_{12}=1-\frac{273 \mathrm{~K}}{473 \mathrm{~K}}=\frac{100}{473} \\
& \eta_{12}-\eta_1=\frac{200}{473}-\frac{100}{473}=\frac{100}{473} < \frac{100}{373} \\
\Rightarrow & \eta_{12}-\eta_1 < \eta_2 \\
\text { or } & \eta_{12} < \eta_1+\eta_2
\end{aligned}\)
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