A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine $\mathrm{E}_{1}$ works between 473 K to 373 K and engine $\mathrm{E}_{2}$ 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 $\mathrm{E}_{2}$, respectively, then

(A) $\eta_{12}<\eta_{1}+\eta_{2}$

(B) $\eta_{12}=\eta_{1}\eta_{2}$

(C) $\eta_{12}=\eta_{1}+\eta_{2}$

(D) $\eta_{12}\geq\eta_{1}+\eta_{2}$

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Heat & Thermodynamics

A Carnot engine (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

η_{12}

η_{1}

and

η_{2}

are the efficiencies of the engines

E

E_{1}

and

E_{2}

, respectively, then

Answer: (A)

The efficiency of a Carnot engine operating between temperatures

T_{H}

and

T_{C}

η = 1 - T_{C} / T_{H}

.

For engine E (473 K to 273 K):

η_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423

For engine

E_{1}

(473 K to 373 K):

η_{1} = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211

For engine

E_{2}

(373 K to 273 K):

η_{2} = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268

Adding,

η_{1} + η_{2} \approx 0.479

, which is greater than

η_{12} \approx 0.423

. So

η_{12} < η_{1} + η_{2}

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option