Giancoli's Physics: Principles with Applications, 7th Edition
15
The Laws of Thermodynamics
Change chapter15-1 and 15-2: First Law of Thermodynamics
15-3: Human Metabolism
15-5: Heat Engines
15-6: Refrigerators, Air Conditioners, Heat Pumps
15-7: Entropy
15-10: Statistical Interpretation
15-11: Energy Resources
Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 18
Q
A heat engine exhausts 8200 J of heat while performing 2600 J of useful work. What is the efficiency of this engine?
A
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VIDEO TRANSCRIPT
This is Giancoli Answers with Mister Dychko. Efficiency is the work done by the engine divided by the amount of heat added to it from the high temperature reservoir. Work done is also the difference between the heat added minus the heat lost of the low temperature reservoir. We can move this over to the left side and then switch the signs around and get Qh is W plus QL. And then we'll substitute that in for Qh up here. So, W plus QL, doing my substitution in red. So, that's 2600 joules of work done divided by 2600 plus 2800 joules. This is an efficiency of 24 percent.
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