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Unformatted text preview: THE SECOND LAW OF THERMODYNAMICS E XERCISES Sections 19.2 and 19.3 The Second Law of Thermodynamics and Its Applications 14. The efficiency of a reversible engine, operating between two absolute temperatures, , h c T T is given by Equation 19.3. (a) 1 / 1 273/373 26.8%. c h e T T = = = (b) ( )/ / 21/298 7.05%. h c h h e T T T T T = = = = (c) With room temperature at 20 C, 980/1273 77.0%. c T e = = = 15. I NTERPRET This problem is about the thermal efficiency of a heat engine. D EVELOP If it were a reversible engine, its efficiency would be that of a Carnot engine, given by Equation 19.3: c Carnot h 1 T e T = E VALUATE Substituting the values given in the problem, we obtain 4 c Carnot h 2.7 K 1 1 1 4.82 10 99.95% 5600 K T e T = = = A SSESS The engine efficiency is almost 100%. This is too good to be true. 16. We can solve Equation 19.3 for the low temperature to find (1 ) (1 0.777)(4.25 K) 0.948 K. c h T e T = = = 17. I NTERPRET This problem is about a Carnot engine that operates via the Carnot cycle. D EVELOP The efficiency of an engine, by definition is, h W e Q = where and h W Q are the work done and heat absorbed per cycle. E VALUATE (a) From the equation above, the efficiency of the engine is 350 J 38.9% 900 J h W e Q = = = (b) and h W Q are related to the heat rejected per cycle by the first law of thermodynamics (since U per cycle is zero). The relation is 900 J 350 J 550 J c h Q Q W = = = (c) For a Carnot engine operating between two temperatures, / / , h c h c T T Q Q = so 900 J (283 K) 463 K 190 C 550 J h h c c Q T T Q = = = = A SSESS The maximum temperature h T is greater than , c T as our calculation confirms. Note that Carnots theorem applies to the ratio of absolute temperatures. 18. From Equation 19.4, rev COP /( ) 273/30 9.10. c h c T T T = = = 19. I NTERPRET This problem is about the work done by a refrigerator to freeze water. Heat of transformation is involved in the phase change. D EVELOP The amount of heat that must be extracted in order to freeze the water is 19.1 19 19.2 Chapter 19 (0.67 kg)(334 kJ/kg) 224 kJ c f Q mL = = = The work consumed by the refrigerator while extracting this heat is given by Equation 19.4, /COP. c W Q = E VALUATE Substituting the values given, we obtain 224 kJ 53.3 kJ COP 4.2 c Q W = = = A SSESS A COP of 4.2 means that each unit of work can transfer 4.2 units of heat from inside the refrigerator. A smaller COP would mean that more work is required to freeze the water. Section 19.4 Entropy and Energy Quality 20. Since the temperature is constant during a change of phase, Equation 19.6 gives / / f S Q T mL T = = = (1 kg) (334 kJ/kg)/273 K 1.22 kJ/K. = 21. I NTERPRET This problem asks for the entropy increase after heating up water....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics, Second Law Of Thermodynamics

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