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39) A heat engine receives 7000 J of heat and loses 3000 J in each cycle. What is the efficiency of this engine? A) 43% B) 30% C) 57% D) 70%

39) A heat engine receives 7000 J of heat and loses 3000 J in each cycle. What is the efficiency of this engine? A) 43% B) 30% C) 57% D) 70%

College Physics
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%**.
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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