AP Physics 2 Exam Question:

A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. During each cycle, the engine absorbs 500 J of heat from the hot reservoir and discharges 300 J of heat to the cold reservoir. Determine the efficiency of this heat engine.

Answer:

Efficiency of a heat engine is given by:

$\text{Efficiency} = \frac{\text{Useful work output}}{\text{Heat input}}$

Given that the heat engine absorbs 500 J of heat from the hot reservoir (Qh) and discharges 300 J of heat to the cold reservoir (Qc).

Step 1: Calculate the net heat added to the engine.

$Q_{\text{net}} = Q_h - Q_c$

$Q_{\text{net}} = 500 \, \text{J} - 300 \, \text{J}$

$Q_{\text{net}} = 200 \, \text{J}$

Step 2: Calculate the efficiency.

$\text{Efficiency} = \frac{\text{Useful work output}}{\text{Heat input}}$

$\text{Efficiency} = \frac{Q_{\text{net}}}{Q_h}$

$\text{Efficiency} = \frac{200 \, \text{J}}{500 \, \text{J}}$

$\text{Efficiency} = 0.4$ or 40%

The efficiency of this heat engine is 40%.

Note: The efficiency of a heat engine is always less than 100% due to the second law of thermodynamics, which states that no heat engine can convert all heat into work without any energy loss.