AP Physics 2 Exam Question

Consider a heat engine that operates between two heat reservoirs. The engine absorbs 3000 J of energy from the high-temperature reservoir at 600 K, and it exhausts 2000 J of energy to the low-temperature reservoir at 300 K. Assume the engine is operating in a Carnot cycle.

(a) Calculate the efficiency of the heat engine. (b) Calculate the work done by the engine during one complete cycle. (c) Based on the laws of thermodynamics, is it possible for this heat engine to have an efficiency greater than 50%? Justify your answer.

Answer with Step-by-Step Explanation

(a) To calculate the efficiency of the heat engine, we can use the formula:

.

where is the energy absorbed and is the energy absorbed from the high-temperature reservoir.

Given that and , we can substitute these values into the formula:

.

Therefore, the efficiency of the heat engine is 0.667 or 66.7%.

(b) To calculate the work done by the engine during one complete cycle, we can use the formula:

.

Therefore, the work done by the engine during one complete cycle is 1000 J.

(c) According to the second law of thermodynamics, the Carnot efficiency is given by:

,

where is the temperature of the cold reservoir and is the temperature of the hot reservoir.

In this case, TC = 300 K and TH = 600 K. Substituting these values into the formula:

.

The Carnot efficiency is 0.5 or 50%.

Since the efficiency of the heat engine in question (66.7%) is greater than the Carnot efficiency (50%), it is possible for this heat engine to have an efficiency greater than 50%. This suggests that the heat engine is not reversible and may have additional mechanisms or inefficiencies beyond the ideal Carnot engine.