College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please show all calculations and type them out So that I will be able to read them and understand what you're saying I have bad eyesight and it is very difficult for me to read handwriting
![**Question 39: Calculating the Efficiency of a Heat Engine**
A heat engine operates by receiving 7000 J (joules) of heat and loses 3000 J of heat in each cycle. To determine the efficiency of this engine, we can use the efficiency formula for a heat engine:
\[ \text{Efficiency} (\eta) = \left( \frac{\text{Work output}}{\text{Heat input}} \right) \times 100 \]
First, we find the work output. The work output (W) is calculated by the difference between the heat input (Q_in) and the heat lost (Q_out):
\[ W = Q_{\text{in}} - Q_{\text{out}} \]
Substituting the given values:
\[ W = 7000\ \text{J} - 3000\ \text{J} = 4000\ \text{J} \]
Now, we can calculate the efficiency:
\[ \eta = \left( \frac{W}{Q_{\text{in}}} \right) \times 100 = \left( \frac{4000\ \text{J}}{7000\ \text{J}} \right) \times 100 \]
\[ \eta \approx 57\% \]
Therefore, the efficiency of this engine is approximately 57%.
**Answer Choices:**
A) 43%
B) 30%
C) 57%
D) 70%
The correct answer is **C) 57%**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8f72c006-9b48-426b-aa7e-60007b6c336c%2Ff8ca4177-3ba4-4bde-8d1a-5e662c43a946%2Fqrw7b2_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 39: Calculating the Efficiency of a Heat Engine**
A heat engine operates by receiving 7000 J (joules) of heat and loses 3000 J of heat in each cycle. To determine the efficiency of this engine, we can use the efficiency formula for a heat engine:
\[ \text{Efficiency} (\eta) = \left( \frac{\text{Work output}}{\text{Heat input}} \right) \times 100 \]
First, we find the work output. The work output (W) is calculated by the difference between the heat input (Q_in) and the heat lost (Q_out):
\[ W = Q_{\text{in}} - Q_{\text{out}} \]
Substituting the given values:
\[ W = 7000\ \text{J} - 3000\ \text{J} = 4000\ \text{J} \]
Now, we can calculate the efficiency:
\[ \eta = \left( \frac{W}{Q_{\text{in}}} \right) \times 100 = \left( \frac{4000\ \text{J}}{7000\ \text{J}} \right) \times 100 \]
\[ \eta \approx 57\% \]
Therefore, the efficiency of this engine is approximately 57%.
**Answer Choices:**
A) 43%
B) 30%
C) 57%
D) 70%
The correct answer is **C) 57%**.
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