### Example Problem: Understanding Heat Engine PerformanceIn one cycle, a heat engine absorbs **1.71 kJ** from a hot reservoir at **277°C** and expels **1.18 kJ** to a cold reservoir at **27°C**.#### Questions:**(a) What is the engine's thermal efficiency (in percent)?**[ Input box for answer ] %**(b) How much work is done by the engine (in Joules) in each cycle?**[ Input box for answer ] J**(c) What is the engine's power output (in kW) if each cycle lasts **0.302 s**?**[ Input box for answer ] kWPlease attempt the questions above with the given data. If you need help understanding any of the concepts, click on the "Read It" button below for additional resources and explanations.[ **Need Help?** Read It ]---### Explanation of Concepts:When dealing with heat engines, several key terms and formulas are essential:1. **Thermal Efficiency**: This is the measure of how well a heat engine converts the heat absorbed from the hot reservoir into work. It is given by: \[ \text{Thermal Efficiency} (\eta) = \left( \frac{W}{Q_h} \right) \times 100 \] where $ W $ is the work done by the engine (in Joules), and $ Q_h $ is the heat absorbed from the hot reservoir (in Joules).2. **Work Done**: The work done by the engine in one cycle can be determined using the difference between the heat absorbed from the hot reservoir $ Q_h $ and the heat expelled to the cold reservoir $ Q_c $: \[ W = Q_h - Q_c \]3. **Power Output**: This is the work done by the engine per unit of time and can be calculated by dividing the work done by the cycle time: \[ \text{Power Output} (P) = \frac{W}{\text{Cycle Time}} \]Make sure to convert units where necessary (e.g., 1 kJ = 1000 J).#### Additional Resources:For a more in-depth explanation and step-by-step guidance on how to solve these types of problems, please refer to our educational articles and video tutorials by clicking the "

Given Data Qin=1.71 kW Qout=1.18 kW

In one cycle, a heat engine absorbs 1.71 kJ from a hot reservoir at 277°C and expels 1.18 kJ to a cold reservoir at 27°C. (a) What is the engine's thermal efficiency (in percent)? (b) How much work is done by the engine (in J) in each cycle? (c) What is the engine's power output (in kW) if each cycle lasts 0.302 s? kW

In one cycle, a heat engine absorbs 1.71 kJ from a hot reservoir at 277°C and expels 1.18 kJ to a cold reservoir at 27°C. (a) What is the engine's thermal efficiency (in percent)? (b) How much work is done by the engine (in J) in each cycle? (c) What is the engine's power output (in kW) if each cycle lasts 0.302 s? kW

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

### Example Problem: Understanding Heat Engine Performance

In one cycle, a heat engine absorbs **1.71 kJ** from a hot reservoir at **277°C** and expels **1.18 kJ** to a cold reservoir at **27°C**.

#### Questions:

**(a) What is the engine's thermal efficiency (in percent)?**

[ Input box for answer ] %

**(b) How much work is done by the engine (in Joules) in each cycle?**

[ Input box for answer ] J

**(c) What is the engine's power output (in kW) if each cycle lasts **0.302 s**?**

[ Input box for answer ] kW

Please attempt the questions above with the given data. If you need help understanding any of the concepts, click on the "Read It" button below for additional resources and explanations.

[ **Need Help?** Read It ]

---

### Explanation of Concepts:

When dealing with heat engines, several key terms and formulas are essential:

1. **Thermal Efficiency**: This is the measure of how well a heat engine converts the heat absorbed from the hot reservoir into work. It is given by:

\[
\text{Thermal Efficiency} (\eta) = \left( \frac{W}{Q_h} \right) \times 100
\]

where $ W $ is the work done by the engine (in Joules), and $ Q_h $ is the heat absorbed from the hot reservoir (in Joules).

2. **Work Done**: The work done by the engine in one cycle can be determined using the difference between the heat absorbed from the hot reservoir $ Q_h $ and the heat expelled to the cold reservoir $ Q_c $:

\[
W = Q_h - Q_c
\]

3. **Power Output**: This is the work done by the engine per unit of time and can be calculated by dividing the work done by the cycle time:

\[
\text{Power Output} (P) = \frac{W}{\text{Cycle Time}}
\]

Make sure to convert units where necessary (e.g., 1 kJ = 1000 J).

#### Additional Resources:

For a more in-depth explanation and step-by-step guidance on how to solve these types of problems, please refer to our educational articles and video tutorials by clicking the "

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