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A Carnot engine E is working between two temperatures 473 K and 273 K. In a new system, two engines are used: engine E₁ works between 473 K and 373 K, and engine E₂ works between 373 K and 273 K. If η₁₂, η₁ and η₂ are the efficiencies of the engines E, E₁ and E₂ respectively, then

Asked in JEE Main 28th Jan 1st Shift 2025 · Entropy and the Carnot Engine

Answer: (1) η₁₂<η₁+η₂

Step-by-step solution

η₁₂=1-(273)/(473)=0.423.

η₁=1-(373)/(473)=0.211 and η₂=1-(273)/(373)=0.268.

η₁+η₂=0.479>0.423.

The two efficiencies do not simply add because E₂ works on the heat E₁

rejects, not on the original input.

Why the other options are wrong

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