A Carnot engine has efficiency of 50 %. If the temperature of sink is reduced by 40° C, its efficiency increases by 30 %. The temperature of the source will be:

Solution

Physics sir · 4 steps

Read "increases by 30%" as 30% of 50%, so the new efficiency is 65%, and a 40 °C drop is a 40 K drop.

1. New efficiency. The efficiency goes up by 30% of its own value: 50% × 1.3 = 65%. So η₁ = 0.5 and η₂ = 0.65. A change of 40°C is the same as a change of 40 K.
2. Write both cases. With source temperature T₁ and sink temperature T₂, η = 1 - T₂/T₁. The source stays the same and the sink becomes T₂ - 40.

   1 - T₂/T₁ = 0.5, 1 - (T₂ - 40)/T₁ = 0.65

3. Subtract. Subtracting the first equation from the second leaves 40/T₁ = 0.15. So T₁ = 40/0.15 = 266.7 K, and the sink was at T₂ = 133.3 K.
4. So. The source is at 266.7 K, which is option (c).

The idea to keep. In Carnot problems, changing only the sink by Δ T changes the efficiency by Δ T / T_(source). Temperature differences in °C and in K are the same.

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