A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine E₁ works between 473 K to 373 K and engine E₂ works between 373 K to 273 K. If η₁₂, η₁ and η₂ are the efficiencies of the engines E, E₁ and E₂, respectively, then (a) η₁₂ = η₁ η₂ (b) η₁₂ = η₁ + η₂ (c) η₁₂ ≥ η₁ + η₂ (d) η₁₂ < η₁ + η₂

Answer: (d). η₁₂ = 1 - (273)/(473) = (200)/(473) ≈ 0.423 η₁ = (100)/(473) ≈ 0.211, η₂ = (100)/(373) ≈ 0.268 η₁ + η₂ ≈ 0.479 > η₁₂ So the answer is (d) η₁₂ < η₁ + η₂.

A Carnot engine (E) is working between two temperatures 473 K and 273… | JEE Main 2025 (28 Jan, Shift 1)

Q11JEE Main 2025 · 28 JAN · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

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

Show answer and solution

Answer: (d)

$\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$
$\eta_1 = \frac{100}{473} \approx 0.211$ , $\eta_2 = \frac{100}{373} \approx 0.268$
$\eta_1 + \eta_2 \approx 0.479 > \eta_{12}$
So the answer is (d) $\eta_{12} < \eta_1 + \eta_2$ .

Report a mistake

Heat Engines, Refrigerators & Second Law question from JEE Main 2025 (28 Jan, Shift 1)

More questions on Heat Engines, Refrigerators & Second Law