Question:

A heat engine operates on a Carnot cycle between two reservoirs at temperatures 600 K and 300 K respectively. The engine absorbs heat Q1 from the high-temperature reservoir and transfers heat Q2 to the low-temperature reservoir. The net work output of the engine is 200 J.

a) Calculate the efficiency of the engine.

b) Calculate the heat Q1 absorbed by the engine.

c) Calculate the heat Q2 rejected by the engine.

Answer:

a) The efficiency (η) of the Carnot engine is given by the formula:

η = 1 - (T2 / T1)

Where T1 is the temperature of the high-temperature reservoir and T2 is the temperature of the low-temperature reservoir.

Given that the high-temperature reservoir temperature (T1) is 600 K and the low-temperature reservoir temperature (T2) is 300 K, we can substitute these values into the equation to find the efficiency:

η = 1 - (300K / 600K) η = 1 - 0.5 η = 0.5

Therefore, the efficiency of the engine is 0.5 or 50%.

b) The heat Q1 absorbed by the engine can be calculated using the formula:

Q1 = W / η

Where W is the net work output of the engine and η is the efficiency of the engine.

Given that the net work output of the engine (W) is 200 J and the efficiency (η) is 0.5, we can substitute these values into the equation to find the heat Q1 absorbed by the engine:

Q1 = 200 J / 0.5 Q1 = 400 J

Therefore, the heat Q1 absorbed by the engine is 400 J.

c) The heat Q2 rejected by the engine can be calculated using the formula:

Q2 = Q1 - W

Where Q1 is the heat absorbed by the engine and W is the net work output.

Given that Q1 is 400 J and W is 200 J, we can substitute these values into the equation to find the heat Q2 rejected by the engine:

Q2 = 400 J - 200 J Q2 = 200 J

Therefore, the heat Q2 rejected by the engine is 200 J.