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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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