AP Physics 2 Exam Question:

Consider a Carnot engine operating between two temperatures, $T_1 = 600 \, \text{K}$ and $T_2 = 300 \, \text{K}$.

a) Calculate the efficiency of the engine.

b) If the Carnot engine absorbs $500 \, \text{J}$ of heat while operating between $T_1$ and $T_2$, determine the work done by the engine.

Useful information:

• The efficiency of a heat engine is given by the expression: \begin{align*} \text{Efficiency} = 1 - \frac{T_2}{T_1} \end{align*} where $T_1$ is the high-temperature reservoir and $T_2$ is the low-temperature reservoir.

• The work done by the engine can be found using the equation: \begin{align*} \text{Work} = \text{Change in Heat} = Q_1 - Q_2 \end{align*} where $Q_1$ is the heat absorbed and $Q_2$ is the heat released.

Answer with Step-by-Step Explanation:

a) To determine the efficiency of the Carnot engine, we can use the equation for efficiency:

\begin{align*} \text{Efficiency} = 1 - \frac{T_2}{T_1} \end{align*}

Let's substitute the known values into the equation and calculate the efficiency:

\begin{align*} \text{Efficiency} = 1 - \frac{300 , \text{K}}{600 , \text{K}} = 1 - 0.5 = 0.5 \end{align*}

Therefore, the efficiency of the engine is 0.5 or 50%.

b) To determine the work done by the Carnot engine, we need to calculate the change in heat:

\begin{align*} \text{Change in Heat} = Q_1 - Q_2 \end{align*}

Given that the engine absorbs $500 \, \text{J}$ of heat, we can substitute this value into the equation:

\begin{align*} \text{Change in Heat} = 500 , \text{J} \end{align*}

Since the Carnot engine is operating between $T_1$ and $T_2$, we can use the efficiency equation to find the ratio of $Q_2$ to $Q_1$:

\begin{align*} \frac{Q_2}{Q_1} = \frac{T_2}{T_1} = \frac{300 , \text{K}}{600 , \text{K}} = 0.5 \end{align*}

Substituting the ratio back into the equation for change in heat, we have:

\begin{align*} 500 , \text{J} = Q_1 - 0.5Q_1 \end{align*}

Simplifying, we find:

\begin{align*} 0.5Q_1 = 500 , \text{J} \implies Q_1 = \frac{500 , \text{J}}{0.5} = 1000 , \text{J} \end{align*}

Finally, we can substitute the value of $Q_1$ back into the equation for work done:

\begin{align*} \text{Work} = Q_1 - Q_2 = 1000 , \text{J} - 0.5 \times 1000 , \text{J} = 500 , \text{J} \end{align*}

Therefore, the work done by the Carnot engine is $500 \, \text{J}$ of energy.