AP Physics 2 Exam Question

A heat engine operates between two reservoirs at temperatures $T_{h}$ and $T_{c}$, where $T_{h} > T_{c}$. The engine absorbs energy Qh from the high-temperature reservoir and releases energy Qc to the low-temperature reservoir. The efficiency of the heat engine is given by the equation:

1. A heat engine operates between a high-temperature reservoir at 400°C and a low-temperature reservoir at 100°C. Find the efficiency of the heat engine.

Answer

We are given:

• The high-temperature reservoir temperature $T_{h} = 400°C$
• The low-temperature reservoir temperature $T_{c} = 100°C$

Using the formula for efficiency, we can substitute the given values:

Therefore, the efficiency of the heat engine is $\frac{3}{4}$ or 75%.

Explanation:

The efficiency of a heat engine is given by Carnot's efficiency formula, which relates the temperatures of the high and low-temperature reservoirs. The formula shows that the efficiency of a heat engine is always less than 1, and it decreases as the temperature difference between the reservoirs decreases.

In this question, we are given the high-temperature reservoir temperature $T_{h}$ and the low-temperature reservoir temperature $T_{c}$. We substitute these values into the efficiency formula and calculate to find the efficiency.

In the given case, the calculated efficiency is $\frac{3}{4}$ or 75%. This means that the heat engine is able to convert 75% of the absorbed energy from the high-temperature reservoir into useful work, while the remaining 25% is rejected to the low-temperature reservoir.