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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2025 · 28 January · Morning Shift · Physics · Thermodynamics

A Carnot engine (E) is working between two temperatures 473 K and...

A Carnot engine (E)(\mathrm{E}) is working between two temperatures 473 K and 273 K . In a new system two engines - engine E1E_1 works between 473 K to 373 K and engine E2E_2 works between 373 K to 273 K . If η12,η1\eta_{12}, \eta_1 and η2\eta_2 are the efficiencies of the engines E,E1E, E_1 and E2E_2, respectively, then

Options

  1. Aη12=η1η2\eta_{12}=\eta_1 \eta_2
  2. Bη12≥η1+η2\eta_{12} \geq \eta_1+\eta_2
  3. Cη12=η1+η2\eta_{12}=\eta_1+\eta_2
  4. Dη12<η1+η2\eta_{12} \lt \eta_1+\eta_2

Correct answer

D. η12<η1+η2\eta_{12} \lt \eta_1+\eta_2

Step-by-step solution

Efficiencies of a carnot engine η=1−Tsink Tsource \eta=1-\frac{T_{\text{sink }}}{T_{\text{source }}}⇒η1=1−373K473K=100473η2=1−273K373K=100373η12=1−273K473K=100473η12−η1=200473−100473=100473<100373⇒η12−η1<η2 or η12<η1+η2\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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