A heat engine operates between a cold reservoir at temperature T₂ = 400 K and a hot reservoir at temperature T₁. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be ______________ K.

Solution

Physics sir · 4 steps

No real engine can beat a Carnot engine, so the Carnot limit tells you how hot the source must be.

1. Actual efficiency. The engine takes Q₁ = 300 J and rejects Q₂ = 240 J, so the work done is W = 60 J. Its efficiency is η = W/Q₁ = 60/300 = 0.2.
2. Carnot limit. Between the same two reservoirs, the Carnot efficiency 1 - T₂/T₁ is the highest possible. So the actual efficiency cannot be more than this value.

   1 - T₂/T₁ ≥ 1 - Q₂/Q₁ ⇒ T₂/T₁ ≤ Q₂/Q₁ = 0.8

3. Solve for T1. 400/T₁ ≤ 0.8 gives T₁ ≥ 400/0.8 = 500 K.
4. So. The smallest allowed hot-reservoir temperature is the one at which the engine just reaches Carnot efficiency, which is 500 K.

The common slip: 2000. Treating 240 J as the work makes the efficiency 0.8 instead of 0.2, which gives 2000 K. Here 240 J is the heat rejected, so the work is only 60 J.

The idea to keep. For a minimum or maximum temperature question, set the real engine's efficiency equal to the Carnot efficiency. Efficiency is always work over heat taken in, 1 - Q₂/Q₁.

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