JEE Main 2022 · 27 July · evening · Single correct · +4 / −1
In 1ˢᵗ case, Carnot engine operates between temperatures 300 K and 100 K. In 2ⁿᵈ case, as shown in the figure, a combination of two engines is used. The efficiency of this combination (in 2ⁿᵈ case) will be :
- (a) same as the 1ˢᵗ case.
- (b) always greater than the 1ˢᵗ case.
- (c) always less than the 1ˢᵗ case.
- (d) may increase or decrease with respect to the 1ˢᵗ case.
Answer and solution
Answer (a) same as the 1ˢᵗ case.
Solution
Physics sir · 4 steps
Follow the heat through both engines: the 200 K reservoir only passes it on, so the 2/3 stays.
- First case. A single Carnot engine between 300 K and 100 K has η = 1 - 100/300 = 2/3.
- Each engine separately. The upper engine works between 300 K and 200 K, so η₁ = 1 - 200/300 = 1/3. The lower engine works between 200 K and 100 K, so η₂ = 1 - 100/200 = 1/2.
- Track the heat. Let the upper engine take Q from 300 K. It does work Q/3 and rejects 2Q/3 to 200 K. The lower engine takes that 2Q/3 and does work 1/2 × 2Q/3 = Q/3.
η = (W₁ + W₂)/Q = (Q/3 + Q/3)/Q = 2/3
- So. The combination has efficiency 2/3, exactly the same as the first case. The answer is (a).
The idea to keep. Carnot engines in series, where the second uses all the heat the first rejects, act like one engine between the hottest and coldest reservoirs: η = 1 - T_(cold)/Tₕₒₜ.
Answer from the NTA final answer key. Solution by PrepWiser.