AP Physics 2 Exam Question

A heat engine operates between two heat reservoirs at temperatures $T_1$ and $T_2$, where $T_1 > T_2$. The engine can convert a portion of the input heat $Q_1$ into useful work $W$, while the remaining heat $Q_2$ is expelled into the cold reservoir. In this process, the total entropy change of the system should be zero.

a) Derive an expression for the efficiency $η$ of the heat engine in terms of the temperatures $T_1$ and $T_2$.

b) Consider a heat engine that takes in heat at $T_1 = 400 \, \mathrm{K}$ and rejects heat at $T_2 = 300 \, \mathrm{K}$. If the heat engine expels $Q_2 = 600 \, \mathrm{J}$ into the cold reservoir, calculate the amount of work $W$ done by the engine and its efficiency $η$.

Answer:

a) The efficiency of a heat engine is defined as the ratio of the useful work output to the input heat: [ η = \frac{W}{Q_1}]

According to the second law of thermodynamics, the total entropy change of a system undergoing a reversible process is given by: [ \Delta S = \frac{Q_1}{T_1} - \frac{Q_2}{T_2}]

Since the total entropy change of the system should be zero (assuming a reversible process), [ \Delta S = 0 = \frac{Q_1}{T_1} - \frac{Q_2}{T_2}]

Solving for $Q_1$, we get: [ Q_1 = \frac{T_1}{T_2} \cdot Q_2]

Substituting this into the expression for efficiency $η$, we find: [ η = \frac{W}{Q_1} = \frac{W}{\frac{T_1}{T_2} \cdot Q_2} = \frac{T_2}{T_1}]

Therefore, the efficiency of the heat engine is given by $η = \frac{T_2}{T_1}$.

b) Given $T_1 = 400 \, \mathrm{K}$, $T_2 = 300 \, \mathrm{K}$, and $Q_2 = 600 \, \mathrm{J}$, we can use the efficiency formula derived in part a) to find the amount of work done by the engine.

Using $η = \frac{T_2}{T_1}$, we have: [ η = \frac{300 , \mathrm{K}}{400 , \mathrm{K}} = 0.75]

The efficiency $η$ of the heat engine is 0.75, which means it converts 75% of the input heat into useful work.

To find the amount of work $W$, we can use the formula $η = \frac{W}{Q_1}$ and rearrange it to solve for $W$: [ W = η \cdot Q_1]

Substituting the given values for $η$ and $Q_1$, we get: [ W = 0.75 \cdot \left(\frac{T_1}{T_2} \cdot Q_2\right) = 0.75 \cdot \left(\frac{400 , \mathrm{K}}{300 , \mathrm{K}} \cdot 600 , \mathrm{J}\right) = 800 , \mathrm{J}]

Therefore, the amount of work done by the heat engine is $W = 800 \, \mathrm{J}$.