Please type out and or diagram Your solution in a way that is easy to read I have bad eyesight **Question 18: (I)**A heat engine exhausts 8200 J of heat while performing 2600 J of useful work. What is the efficiency of this engine?**Solution:**To find the efficiency of a heat engine, you can use the formula:\[ \text{Efficiency} (\eta) = \frac{\text{Useful Work Output}}{\text{Total Heat Input}} \times 100\% \]In this case:- The useful work output \( W = 2600 \text{ J} \)- The total heat input \( Q = W + \text{Heat Exhausted} \) So, \[ Q = 2600 \text{ J} + 8200 \text{ J} = 10800 \text{ J} \]Then the efficiency (\(\eta\)) is:\[ \eta = \frac{2600 \text{ J}}{10800 \text{ J}} \times 100\% \]\[ \eta = \frac{2600}{10800} \times 100\% \]\[ \eta \approx 24.07\% \]Therefore, the efficiency of this engine is approximately 24.07%.

Given:- Amount of heat exhausted by the engine is Q=8200 J Work done by the engine is W=2600 J…

Answered: (I) A heat engine exhausts 8200 J of…

(I) A heat engine exhausts 8200 J of heat while performing 2600 J of useful work. What is the efficiency of this engine?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter18: Heat Engines, Entropy, And The Second Law Of Thermodynamics

Section: Chapter Questions

Problem 8P

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