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Chapter 20 Entropy and the Second Law of Thermodynamics
In this chapter we will introduce the second law of thermodynamics. The following topics will be covered: Reversible processes Entropy The Carnot engine Refrigerators Real engines (20–1)
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Example 1: An isothermal process is one in which Ti = Tf which implies ln (Tf /Ti) = 0. Therefore, Eq leads to
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Example 2: (a) The energy absorbed as heat is given by Eq Using Table 19-3, we find We use the following relation derived in Sample Problem 20-2: (b) With Tf = K and Ti = K, we obtain
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Example 3:
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(b) Now temperatures must be given in Kelvins: Tai = 393 K, Twi = 293 K, and Tf = 330 K. For the aluminum, dQ = macadT and the change in entropy is (c) The entropy change for the water is (d) The change in the total entropy of the aluminum-water system is DS = DSa + DSw = J/K J/K = +2.8 J/K.
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Example 4: (a) This may be considered a reversible process (as well as isothermal), so we use DS = Q/T where Q = Lm with L = 333 J/g from Table Consequently, (b) The situation is similar to that described in part (a), except with L = 2256 J/g, m = 5.00 g, and T = 373 K. We therefore find DS = 30.2 J/K.
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Example 5:
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Example 7:
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Example A particular engine has a power output of 5000 W and an efficiency of 25%. If the engine expels 8000 J of heat in each cycle, find (a) the heat absorbed in each cycle and (b) the time for each cycle 10,667 J 2667 J 0.53 s
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Example The efficiency of a Carnot engine is 30%. The engine absorbs 800 J of heat per cycle from a hot temperature reservoir at 500 K. Determine (a) the heat expelled per cycle and (b) the temperature of the cold reservoir 240 J 560 J 350 K
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Example An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?
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(20–11)
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