A heat engine operates with the cold reservoir at temperature 324 K. The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is ____________ K.

Solution

Physics sir · 3 steps

No engine beats Carnot, so its actual efficiency sets the lowest possible hot-reservoir temperature.

1. Actual efficiency. The engine takes Q_H = 300 J and rejects Q_C = 180 J, so the work per cycle is W = 300 - 180 = 120 J. Its efficiency is η = 120/300 = 0.4.
2. Carnot limit. No engine working between T_H and T_C can be more efficient than a Carnot engine. So η must be at most 1 - T_C/T_H.

   1 - T_C/T_H ≥ 0.4 ⇒ T_H ≥ T_C/0.6 = 324/0.6

3. So. T_H ≥ 540 K. The minimum temperature of the hot reservoir is 540 K, where the engine works exactly as a Carnot engine.

The common slip: 810. Taking the 180 J as the work instead of the heat rejected gives an efficiency of 0.6 and T_H = 324/0.4 = 810 K, not 540.

The idea to keep. For a minimum or maximum reservoir temperature, set the engine's real efficiency equal to the Carnot efficiency 1 - T_C/T_H. The Carnot value is the ceiling no engine can cross.

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