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

Solution

Physics sir · 4 steps

Efficiencies in series do not add, because E₂ only gets the heat that E₁ throws away.

1. Each efficiency. For a Carnot engine η = 1 - T_C/T_H. So η₁₂ = 1 - 273/473 = 200/473 ≈ 0.423, η₁ = 1 - 373/473 = 100/473 ≈ 0.211 and η₂ = 1 - 273/373 = 100/373 ≈ 0.268.
2. Compare the sum. η₁ + η₂ ≈ 0.211 + 0.268 = 0.480, which is larger than η₁₂ ≈ 0.423. The product η₁ η₂ ≈ 0.057 is far too small, so it cannot be the answer either.
3. Why it is less. The fraction of heat that is not converted multiplies through the two engines: 1 - η₁₂ = 373/473 × 273/373 = (1-η₁)(1-η₂). Expanding this gives the exact link below.

   η₁₂ = η₁ + η₂ - η₁ η₂

4. So. Since η₁ η₂ > 0, η₁₂ < η₁ + η₂. The answer is (d).

The idea to keep. For engines in series, multiply the (1-η) terms, not the efficiencies. The combined efficiency equals that of one engine between the extreme temperatures, and it is always less than the sum.

Answer from the NTA final answer key. Solution by PrepWiser.