JEE Main 2020 · 7 January · evening · Single correct · +4 / −1

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperature, T₁ and T₂ . The temperature of the hot reservoir of the first engine is T₁ and the temperature of the cold reservoir of the second engine is T₂ . T is temperature of the sink of first engine which is also the source for the second which is also the source for the second engine. How is T related to T₁ and T₂ . If both engines perform equal amount of work?

  1. (a) T = 2T₁T₂/(T₁ + T₂)
  2. (b) T = √T₁T₂
  3. (c) T = (T₁ + T₂)/2
  4. (d) T = 0
Answer and solution

Answer (c) T = (T₁ + T₂)/2

Solution

Physics sir · 3 steps

Equal work means equal temperature drops, so T lands exactly halfway between T1 and T2.

  1. Heat ratios. Let Q₁ enter the first engine at T₁, Q pass from it to the second engine at T, and Q₂ leave at T₂. For a Carnot engine heat is proportional to reservoir temperature, so Q₁/T₁ = Q/T = Q₂/T₂.
  2. Work of each engine. The first engine does W₁ = Q₁ - Q and the second does W₂ = Q - Q₂. Writing every heat as Q/T times its temperature gives W₁ = Q/T(T₁ - T) and W₂ = Q/T(T - T₂).

    W₁ = W₂ ⇒ T₁ - T = T - T₂ ⇒ T = (T₁ + T₂)/2

  3. So. The intermediate temperature is the arithmetic mean, T = (T₁ + T₂)/2, which is option (c).

The idea to keep. In Carnot engines in series, work is proportional to the temperature drop. Equal work gives the arithmetic mean. Equal efficiency gives the geometric mean.

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

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Asked about 150 times in JEE Main since 2020. This question is from 2020.