PHYS153lectureNote - Q2: What is the efficiency of this...

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1 Q2: What is the efficiency of this cycle? A. 33.3% B. 50% C. 66.7% D. 100% E. Not enough information given. *** Write down your reasoning on the activity sheet ***
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Answer to question 2 is C The reason: e = W / Q H is the efficiency of any heat engine. Q H is defined to be when Q > 0. From the graph in Question 2, Q is only positive for b to c. So Q H is the area under the line b to c. The total heat for the cycle is the area enclosed in the shape abcd. The first law of Thermodynamics says that dQ = dU +dW. For a complete cycle, the change internal energy is zero. Thus, Q cycle = W cycle . Finally, e = W / Q H = Q cycle / Q H = (Area of region abcd ) / (Area under line ab) Using the diagram, e = 66.7% 2
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3 Q3: Is this cycle more or less efficient than a Carnot engine with a hot reservoir at 1200K and cold reservoir at 300K. A. More B. Less *** Write down your reasoning on the activity sheet ***
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The answer to Question 3 is B. The reason: A Carnot engine’s efficiency run between temperatures T H and T C is e = 1-T C /T H. A Carnot run between a hot reservoir at 1200K and cold reservoir at 300K is e = 1-T C /T H = 1- 300K / 1200K =1 - 3/12 = 1 – ¼ = ¾ e = 75% > 66.7% 4
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5 Q1: What is the highest efficiency of the Carnot cycle? A. 100%, if T C is absolute zero. B. No limit, because one can take T C <0. C. 100%, if you make T H large enough. D. Not enough information given.
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The answer to Question 1 is A. A Carnot engine’s efficiency run between temperatures T H and T C is 1-T C /T H. So when T C =0, e = 1 =100%. The definition of the temperature does not allow T C <0. e = 1 =100% is the best possible efficiency. 6
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7 Q3: Can any heat engine ever be more efficient than a Carnot engine operating between the same two temperatures? A. Yes. Heat engines can be made that are more efficient than a Carnot operating between the same two temperatures. B. No. No heat engine can be more efficient than a Carnot operating between the same two temperatures.
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Reason: The second law of thermodynamics says that for an isolated system, Δ S 0. This is equivalent to the fact that no engine operating with a hot reservoir of temperature T H and cold reservoir of temperature T C can be more efficient than the Carnot engine operating between the same two temperatures. Please re-read this in your book readings.
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This note was uploaded on 11/16/2010 for the course PHYS 153 PHYS 153 taught by Professor Sarahburke,donaldwitt,michaeldhasinoff,andrzejkotlicki during the Spring '10 term at The University of British Columbia.

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PHYS153lectureNote - Q2: What is the efficiency of this...

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