A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52K, its efficiency is doubled. The temperature in Kelvin of the source will be __________.

Solution

Physics sir · 3 steps

Write the Carnot efficiency twice, before and after the sink cools, and subtract the two.

1. Reversible means Carnot. A reversible engine working between source T₁ and sink T₂ has efficiency η = 1 - T₂/T₁. One-fourth of the heat becomes work, so η = 1/4, which gives T₂ = 3/4T₁.
2. After cooling the sink. The sink drops to T₂ - 52 and the efficiency doubles to 1/2. So 1 - (T₂ - 52)/T₁ = 1/2, which gives T₂ - 52 = 1/2T₁.

   3/4T₁ - 1/2T₁ = 52 ⇒ T₁/4 = 52

3. So. T₁ = 4 × 52 = 208 K, and the sink was at 156 K. The source temperature is 208.

The common slip: 156. Solving for the sink instead of the source gives 156, which is T₂ = 3/4 × 208. The question asks for the source.

The idea to keep. Cooling the sink by Δ T raises a Carnot engine's efficiency by exactly (Δ T)/T_(source). Use that and the source temperature comes out in one line.

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