Certainly! Here's an AP Physics 2 exam question on the Laws of Thermodynamics:

Question:

A heat engine operates on a Carnot cycle, and the hot reservoir has a temperature of 500 K while the cold reservoir is maintained at 300 K. The engine does 1500 J of work per cycle. Calculate the amount of heat transfer between the hot reservoir and the cold reservoir during each cycle.

Answer:

To solve this problem, we can apply the laws of thermodynamics that govern the efficiency of a Carnot engine.

First, let's calculate the efficiency of the Carnot engine using the formula:

Where: $\eta$ = Efficiency of the engine $T_C$ = Temperature of the cold reservoir $T_H$ = Temperature of the hot reservoir

Plugging in the given values: [ \eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 ]

Next, we can use the efficiency to find the heat transfer between the reservoirs using the formula:

Where: $Q_H$ = Heat transfer from the hot reservoir $W$ = Work done per cycle $\eta$ = Efficiency

Plugging in the given values: [ Q_H = \frac{1500}{0.4} = 3750 , \text{J} ]

Finally, to find the heat transfer to the cold reservoir, we use the relationship: [ Q_H - Q_C = W \implies Q_C = Q_H - W ]

Plugging in the values: [ Q_C = 3750 - 1500 = 2250 , \text{J} ]

Therefore, the amount of heat transfer from the hot reservoir to the cold reservoir during each cycle is 2250 J.

This question assesses the student's understanding of the Carnot cycle, efficiency of a heat engine, and heat transfer between reservoirs based on temperature differences.