Answer to Question #87605 in Molecular Physics | Thermodynamics for Lesly

Question

An engine absorbs 2364 J from a hot reservoir and expels 881 J to a cold reservoir in each cycle.
1. What is the engine’s efficiency?
2. How much work is done in each cycle? Answer in units of J.
3. What is the mechanical power output of the engine if each cycle lasts for 0.32s? Answer in units of kW.

Answer 1

1. The engine's efficiency is calculated by using the formula

$$\eta = \frac{Q_2 - Q_1}{Q_2} = 1 - \frac{Q_1}{Q_2} \, ,$$

where $Q_1 = 881\, \text{J}$ and $Q_2 = 2364\, \text{J}$ are the amounts of heat, respectively, expelled and absorbed by the system in one cycle. We obtain $\eta = 1 - 881 / 2364 \approx 0.627$.

Answer: 0.627

2. The work done in each cycle is $W = Q_2 - Q_1 = 1483\, \text{J}$.

Answer: 1483 J.

3. The mechanical power output is $W / t$, where $t = 0.32\, \text{s}$ is the duration of one cycle. We have $W / t = 1483\, \text{J} / 0.32\, \text{s} \approx 4634\, \text{W} = 4.634\, \text{kW}$.

Answer: 4.634 kW.