# chap 19 - THE SECOND LAW OF THERMODYNAMICS 19 EXERCISES...

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19.1 THE SECOND LAW OF THERMODYNAMICS EXERCISES 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, TT hc > , is given by Equation 19.3. (a) eT T ch =− = 1 1 273 373 26 8 // . % . (b) eTTT T T h h = = = () / . % . 21298 7 05 (c) With room temperature at Te c =° = = 20 980 1273 77 0 C, / . %. 15. INTERPRET This problem is about the thermal efficiency of a heat engine. DEVELOP If it were a reversible engine, its efficiency would be that of a Carnot engine, given by Equation 19.3: e T T Carnot c h 1 EVALUATE Substituting the values given in the problem, we obtain e T T Carnot c h K K =− =− × 11 27 5600 1 4 82 10 99 9 4 . .. 5 % ASSESS 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 T = (. ) ( . 1 1 0 777 4 25 K) 0.948 K. 17. INTERPRET This problem is about a Carnot engine that operates via the Carnot cycle. DEVELOP The efficiency of an engine, by definition is, e W Q h = where WQ h and are the work done and heat absorbed per cycle. EVALUATE (a) From the equation above, the efficiency of the engine is e W Q h == = 350 900 38 9 J J .% (b) h and are related to the heat rejected per cycle by the first law of thermodynamics (since U per cycle is zero). The relation is QQW = =−= 900 350 550 JJJ (c) For a Carnot engine operating between two temperatures, QQ h c , = so Q Q h c = = = 283 900 550 463 K J J K = ° 190 C ASSESS The maximum temperature T h is greater than T c , as our calculation confirms. Note that Carnot’s theorem applies to the ratio of absolute temperatures. 18. From Equation 19.4, COP rev = = TT T ch c /( ) / . . 273 30 9 10 19. INTERPRET This problem is about the work done by a refrigerator to freeze water. Heat of transformation is involved in the phase change. 19

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19.2 Chapter 19 DEVELOP The amount of heat that must be extracted in order to freeze the water is Qm L cf == = (. ) ( ) 0 67 334 224 kg kJ/kg kJ The work consumed by the refrigerator while extracting this heat is given by Equation 19.4, WQ c = /COP. EVALUATE Substituting the values given, we obtain W Q c = COP kJ kJ 224 42 53 3 . . ASSESS 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 ∆∆ SQ T m L T f = // () ( ) / 1 334 273 kg kJ/kg K 1.22 kJ/K. ×= 21. INTERPRET This problem asks for the entropy increase after heating up water. DEVELOP For a substance with constant specific heat (in this case at constant pressure), dQ mcdT = , and the change in entropy is S dQ T mc dT T mc T T T T = ∫∫ 1 2 2 1 1 2 ln EVALUATE Substituting the values given, we have Sm c T T = =⋅ ( . )( . )ln 2 1 0 25 4 184 3 kg kJ/kg K 68 275 K 283 K J/K = ASSESS The final entropy of the system has increased.
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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap 19 - THE SECOND LAW OF THERMODYNAMICS 19 EXERCISES...

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