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586_PartUniversity Physics Solution

586_PartUniversity Physics Solution - THE SECOND LAW OF...

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20-1 T HE S ECOND L AW OF T HERMODYNAMICS 20.1. I DENTIFY : For a heat engine, H C . W Q Q = H . W e Q = H 0, Q > C 0. Q < S ET U P : 2200 J. W = C 4300 J. Q = E XECUTE : (a) H C 6500 J. Q W Q = + = (b) 2200 J 0.34 34%. 6500 J e = = = E VALUATE : Since the engine operates on a cycle, the net Q equal the net W . But to calculate the efficiency we use the heat energy input, H . Q 20.2. I DENTIFY : For a heat engine, H C . W Q Q = H . W e Q = H 0, Q > C 0. Q < S ET U P : H 9000 J. Q = C 6400 J. Q = E XECUTE : (a) 9000 J 6400 J 2600 J. W = = (b) H 2600 J 0.29 29%. 9000 J W e Q = = = = E VALUATE : Since the engine operates on a cycle, the net Q equal the net W . But to calculate the efficiency we use the heat energy input, H . Q 20.3. I DENTIFY and S ET U P : The problem deals with a heat engine. 3700 W W = + and H 16,100 J. Q = + Use Eq.(20.4) to calculate the efficiency e and Eq.(20.2) to calculate C . Q Power / . W t = E XECUTE : (a) H work output 3700 J 0.23 23%. heat energy input 16,100 J W e Q = = = = = (b) H C W Q Q Q = = Heat discarded is C H 16,100 J 3700 J 12,400 J. Q Q W = = = (c) H Q is supplied by burning fuel; H c Q mL = where c L is the heat of combustion. H 4 c 16,100 J 0.350 g. 4.60 10 J/g Q m L = = = × (d) 3700 J W = per cycle In 1.00 s t = the engine goes through 60.0 cycles. / 60.0(3700 J)/1.00 s 222 kW P W t = = = 5 (2.22 10 W)(1 hp/746 W) 298 hp P = × = E VALUATE : C 12,400 J. Q = − In one cycle tot C H 3700 J. Q Q Q = + = This equals tot W for one cycle. 20.4. I DENTIFY : H C . W Q Q = H . W e Q = H 0, Q > C 0. Q < S ET U P : For 1.00 s, 3 180 10 J. W = × E XECUTE : (a) 3 5 H 180 10 J 6.43 10 J. 0.280 W Q e × = = = × (b) 5 5 5 C H 6.43 10 J 1.80 10 J 4.63 10 J. Q Q W = = × × = × E VALUATE : Of the 5 6.43 10 J × of heat energy supplied to the engine each second, 5 1.80 10 J × is converted to mechanical work and the remaining 5 4.63 10 J × is discarded into the low temperature reservoir. 20

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20-2 Chapter 20 20.5. I DENTIFY : H C . W Q Q = H . W e Q = H 0, Q > C 0. Q < Dividing by t gives equivalent equations for the rate of heat flows and power output. S ET U P : / 330 MW. W t = H / 1300 MW. Q t = E XECUTE : (a) H H / 330 MW 0.25 25%. / 1300 MW W W t e Q Q t = = = = = (b) C H Q Q W = so C H / / / 1300 MW 330 MW 970 MW. Q t Q t W t = = = E VALUATE : The equations for e and W have the same form when written in terms of power output and rate of heat flow. 20.6. I DENTIFY : Apply 1 1 1 . e r γ = C H 1 . Q e Q = S ET U P : In part (b), H 10,000 J. Q = The heat discarded is C . Q E XECUTE : (a) 0.40 1 1 0.594 59.4%. 9.50 e = = = (b) C H (1 ) (10,000 J)(1 0.594) 4060 J. Q Q e = = = E VALUATE : The work output of the engine is H C 10,000 J 4060 J 5940 J W Q Q = = = 20.7. I DENTIFY : 1 1 1 . e r γ = S ET U P : 1.40 γ = and 0.650. e = E XECUTE : 1 1 1 0.350. e r γ = = 0.40 1 0.350 r = and 13.8. r = E VALUATE : e
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