Question

Part 1: An engine absorbs 2182 J from a hot reservoir and expels 1192 J to a cold reservoir in each cycle. What is the engine’s efficiency? Answer with an efficiency in decimal form.

Part 2: How much work is done in each cycle? Answer in units of J.

Part 3: What is the mechanical power output of the engine if each cycle lasts for 0.15 s? Answer in units of kW.

Answer 1

To solve the problems in parts 1, 2, and 3, we will use the definitions of efficiency, work done, and power output.

### Part 1: Engine Efficiency
The efficiency (\( \eta \)) of an engine is defined as the ratio of the work output to the heat input. It can also be expressed using the heat absorbed from the hot reservoir (\( Q_H \)) and the heat expelled to the cold reservoir (\( Q_C \)) with the formula:

\[
\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H}
\]

Where:
- \( Q_H = 2182 \, \text{J} \)
- \( Q_C = 1192 \, \text{J} \)

First, we calculate the work done (\( W \)):

\[
W = Q_H - Q_C = 2182 \, \text{J} - 1192 \, \text{J} = 990 \, \text{J}
\]

Now we substitute \( W \) into the efficiency formula:

\[
\eta = \frac{990 \, \text{J}}{2182 \, \text{J}} \approx 0.454
\]

So, the engine’s efficiency in decimal form is:

\[
\eta \approx 0.454
\]

### Part 2: Work Done in Each Cycle
From the calculations above, we have already found the work done in each cycle:

\[
W = 990 \, \text{J}
\]

### Part 3: Mechanical Power Output
Power (\( P \)) is defined as the work done per unit time. The formula is:

\[
P = \frac{W}{t}
\]

Where:
- \( W = 990 \, \text{J} \)
- \( t = 0.15 \, \text{s} \)

Substituting these values in gives:

\[
P = \frac{990 \, \text{J}}{0.15 \, \text{s}} = 6600 \, \text{W}
\]

To convert this to kilowatts (kW):

\[
P = \frac{6600 \, \text{W}}{1000} = 6.6 \, \text{kW}
\]

### Summary of Answers:
1. Efficiency: \( \eta \approx 0.454 \)
2. Work Done: \( 990 \, \text{J} \)
3. Power Output: \( 6.6 \, \text{kW} \)