AP Physics 2 Exam Question

Consider a heat engine that operates between two reservoirs at temperatures $T_H = 600 \, \text{K}$ and $T_C = 300 \, \text{K}$, respectively. The engine absorbs $100 \, \text{J}$ of heat from the hot reservoir and discharges $60 \, \text{J}$ of heat into the cold reservoir per cycle.

1. Calculate the efficiency of this heat engine.
2. Determine the entropy change of the hot and cold reservoirs during one cycle.
3. If the engine operates at a rate of $4 \, \text{cycles/s}$, calculate the power output of the engine.

Answer and Explanation

1. The efficiency of a heat engine is given by the equation:

   $$\text{Efficiency} = \frac{{\text{Work Done by Engine}}}{{\text{Heat Absorbed from Hot Reservoir}}}$$

   The work done by the engine is given by the difference between the heat absorbed and the heat discharged:

   $$\text{Work Done by Engine} = \text{Heat Absorbed} - \text{Heat Discharged}$$

   Plugging in the given values:

   $$\text{Work Done by Engine} = 100 \, \text{J} - 60 \, \text{J} = 40 \, \text{J}$$

   Now we can calculate the efficiency:

   $$\text{Efficiency} = \frac{{\text{Work Done by Engine}}}{{\text{Heat Absorbed from Hot Reservoir}}} = \frac{{40 \, \text{J}}}{{100 \, \text{J}}} = 0.4 \, \text{or} \, 40\%$$

   Therefore, the efficiency of this heat engine is 40%.

2. The entropy change of a reservoir is given by the equation:

   $$\Delta S = \frac{{\text{Heat Transferred}}}{T}$$

   For the hot reservoir, the heat transferred is the amount absorbed:

   $$\Delta S_{\text{hot}} = \frac{{\text{Heat Absorbed from Hot Reservoir}}}{{T_{\text{hot}}}} = \frac{{100 \, \text{J}}}{{600 \, \text{K}}}$$

   For the cold reservoir, the heat transferred is the amount discharged:

   $$\Delta S_{\text{cold}} = \frac{{\text{Heat Discharged into Cold Reservoir}}}{{T_{\text{cold}}}} = \frac{{60 \, \text{J}}}{{300 \, \text{K}}}$$

   Plugging in the given values:

   $$\Delta S_{\text{hot}} = \frac{{100 \, \text{J}}}{{600 \, \text{K}}} = \frac{1}{6} \, \text{J/K}$$
   $$\Delta S_{\text{cold}} = \frac{{60 \, \text{J}}}{{300 \, \text{K}}} = \frac{1}{5} \, \text{J/K}$$

   Therefore, the entropy change of the hot reservoir is $\frac{1}{6} \, \text{J/K}$ and the entropy change of the cold reservoir is $\frac{1}{5} \, \text{J/K}$.

3. The power output of the engine can be calculated using the equation:

   $$\text{Power} = \text{Work Done by Engine} \times \text{Number of Cycles per Second}$$

   Plugging in the given values:

   $$\text{Power} = 40 \, \text{J} \times 4 \, \text{cycles/s} = 160 \, \text{W}$$

   Therefore, the power output of the engine is 160 Watts.