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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 since I have bad eyesight and it is very difficult for me to read handwriting 

**Problem Statement:** A heat engine receives 7000 J of heat and loses 3000 J in each cycle. What is the efficiency of this engine? **Options:** A) 43% B) 30% C) 57% D) 70% **Explanation:** To determine the efficiency of a heat engine, you can use the formula: \[ \text{Efficiency} (\eta) = \frac{\text{Work Output}}{\text{Heat Input}} \times 100 \] From the problem statement: - Heat received by the engine (Heat Input) \( Q_{in} = 7000 \) J - Heat lost by the engine (Heat Output or Heat Rejected \( Q_{out}) = 3000 \) J First, calculate the Work Output: \[ \text{Work Output} = Q_{in} - Q_{out} = 7000 \text{ J} - 3000 \text{ J} = 4000 \text{ J} \] Then, substitute the values into the efficiency formula: \[ \eta = \frac{4000 \text{ J}}{7000 \text{ J}} \times 100 \] \[ \eta = \frac{4000}{7000} \times 100 \] \[ \eta = 0.5714 \times 100 \] \[ \eta = 57.14\% \] Thus, considering the given options, the closest value is: **Correct Option:** C) 57%
Transcribed Image Text:**Problem Statement:** A heat engine receives 7000 J of heat and loses 3000 J in each cycle. What is the efficiency of this engine? **Options:** A) 43% B) 30% C) 57% D) 70% **Explanation:** To determine the efficiency of a heat engine, you can use the formula: \[ \text{Efficiency} (\eta) = \frac{\text{Work Output}}{\text{Heat Input}} \times 100 \] From the problem statement: - Heat received by the engine (Heat Input) \( Q_{in} = 7000 \) J - Heat lost by the engine (Heat Output or Heat Rejected \( Q_{out}) = 3000 \) J First, calculate the Work Output: \[ \text{Work Output} = Q_{in} - Q_{out} = 7000 \text{ J} - 3000 \text{ J} = 4000 \text{ J} \] Then, substitute the values into the efficiency formula: \[ \eta = \frac{4000 \text{ J}}{7000 \text{ J}} \times 100 \] \[ \eta = \frac{4000}{7000} \times 100 \] \[ \eta = 0.5714 \times 100 \] \[ \eta = 57.14\% \] Thus, considering the given options, the closest value is: **Correct Option:** C) 57%
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