apter - Heat Engines, Entropy, and te Seond Law o ermodynamis.1 (a).0 J e 0.069 4 or 6.94% 60 J (b) 60 J.0 J J. e eat to melt 1.0 g o Hg is 4 ml 1 10 kg 1.18 10 J kg 177 J e energy absorbed to reeze 1.00 g o aluminum is and te work output is 0 J e teoretial (arnot) eiieny is.6 O( rerigerator) ml 10 kg.97 10 J/ kg 97 J 0 J e 0.4, or.4% 97 J (a) I 10 J and O.00, ten 4.0 J 9 K 4.1 K 0.749 74.9% 9 K (b) Heat expelled Heat removed + ork done. + 10 J+ 4 J 144 J.7 O.00. ereore, e eat removed ea minute is t..00 ( 0.00 0 kg )( 4 186 J kg )(.0 ) ( 0.00 0 kg )(. 10 J kg ) + ( ) 4 0.00 0 kg 090 J kg 0.0 1.40 10 J min + or, J s t us, te work done per seond is J s 77.8..00.9 70 K 14 K (a) Δ 1 440 e 14 67.%
(b) 1.40 10 J, 0.40 4.88 10 J 8.8 k Δt 1 s.1 Isotermal expansion at Isotermal ompression at K K Gas absorbs 1 00 J during expansion. (a) 1 00 J 741 J (b) ( ) 1 00 741 J 49 J V γ γ V i i.1 (a) In an adiabati proess, V V i i. lso, i γ 1 γ Dividing te seond equation by te irst yields i i Sine γ or rgon, γ 1 0.400 and we ave γ γ γ ( ) 0.400 00 10 a ( 1 07 K ) 64 K 6 1.0 10 a (b) Δ Eint nv Δ 0, so nv Δ, and te power output is nv Δ or t t ( 80.0 kg )( 1 m ol/ 0.099 kg )( )( 8.14 J m ol K)( 64 1 07) K 60.0 s 1. 10 1 k () 64 K e 1 1 0.47 or 47.% 107 K.16 (a) e max 78 9 1 1.1 10.1%
6 (b) 7.0 10 J s Δt ereore, 6 11 7.0 10 J s 600 s.70 10 J From e we ind 11.70 10 J 1.7 10 J.7 J e.1 10 () s ossil-uel pries rise, tis way to use solar energy will beome a good buy..1 (a) For a omplete yle, Δ Eint 0 and ( ) 1 e text sows tat or a arnot yle (and only or a reversible yle) ereore, (b) e ave te deinition o te oeiient o perormane or a rerigerator, O Using te result rom part (a), tis beomes O O 0.100O.4 arnot yle or 1 0.100 0.100 arnot eiieny arnot yle 9 K 0.100 0.100 1.17 9 K 68 K FIG..4 us, 1.17 joules o energy enter te room by eat or ea joule o w ork done.
.8 (a), (b) e quantity o gas is 6 ( 100 10 a)( 00 10 m ) V n 0.00 mol R ( 8.14 J mol K )( 9 K ) 6 int, ( 100 10 a )( 00 10 E nr V m ) 1 J γ V B V B In proess, 1.40 int, B V nr 6 100 10 a 8.00 1.84 10 a 6 6 ( 1.84 10 a)( 00 10 m 8.00) ( 0.00 mol )( 8.14 J mol K ) B B B E 67 K nrb ( 0.00 mol )( 8.14 J mol K )( 67 K ) 87 J Δ E 87 J 1 J 16 J 0 16 J so int, out out roess B takes us to: int, int, B ( 0.00 mol )( 8.14 J mol K )( 1 0 K ) nr 6 V 6. 10 m E out 6.79 10 a nr ( 0.00 mol )( 8.14 J mol K )( 1 0 K ) 46 J E 46 J 87 J 149 J 0 B 149 J In proess D: 1.40 V 6 1 D (.79 10 a) 1. 10 a V D 8.00 int, D V nr int, D 6 ( 1. 10 a)( 00 10 m ) ( 0.00 mol )( 8.14 J mol K ) D D D E γ out 44 K nrd ( 0.00 mol )( 8.14 J mol K )( 44 K ) 190 J Δ E 190 J 46 J 46 J 0 out D 46 J Δ E E E 1 J 190 J 6.0 J 0 and int, D int, int, D out 6.0 J D
For te entire yle, Δ Eint, net 16 J+ 149 46 6.0 0. e net work is net 16 J+ 0 + 46 J+ 0 84. J 0 + 149 J+ 0 6.0 J 84. J e tables look like: State (K) (ka) V ( m ) E int (J) 9 100 00 1 B 67 1 840 6. 87 1 0 790 6. 46 D 44 1 00 190 9 100 00 1 roess (J) output (J) Δ Eint (J) 0 16 16 B 149 0 149 D 0 46 46 D 6.0 0 6.0 D 84. 84. 0 () e input energy is 149 J, te waste is 6.0 J, and 84. J. (d) e eiieny is: 84. J e 149 J 0.6 (e) Let represent te angular speed o te ranksat. en wi we obtain work in te amount o 84. J/yle: is te requeny at.9 ompression ratio 6. 00, γ 1.40 (a) 1000 Js ( 84. J yle) 000 J s 84. J yle V Eiieny o an Otto-ine e 1 V1.7 rev s 1.4 10 rev min γ 1 0.400 1 e 1 1.% 6.00 (b) I atual eiieny e 1.0% losses in system are e e 6.%
.0 For a reezing proess, ( 0.00 kg)(. 10 J kg ) Δ Δ S 610 J K 7 K d md. Δ S m ln i i i Δ S 0 g ( 1.00 al g ) ln 46.6 al K 19 J K 9.4 1 1 000 1 000 S Δ J K 1 90 700.7 J K V.9 Δ S nrln Rln.76 J K Vi ere is no ange in temperature or an ideal gas. FIG..9 V S nrln 0.044 0 Rln Vi.40 Δ ( ) Δ S 0.088 0 8.14 ln 0.07 J K FIG..40.49 (a) For an isotermal proess, nrln V V1 ereore, ( ) 1 nr i ln 1 nr i ln and ( ) int, i i 4 Δ Eint, 4 nr i i e net energy by eat transerred is ten For te onstant volume proesses, Δ E nr( ) and ( ) 1+ + + 4 or nr i ln FIG..49
(b) positive value or eat represents energy transerred into te system. ereore, + nr ( + ) 1 4 i 1 ln Sine te ange in temperature or te omplete yle is zero, Δ E int 0 and ereore, te eiieny is ln e 1 ln ( + ) 0.7.0 F.0.0 1.67 + 7.1 K 74.8 K 9.0 (a) ( ) ( ) 98.6 F ( 98.6.0) ( 7.0 + 7.1 ) K 10.1 K 9 d d 10.1 Δ Sie water ( 4.6 g )( 1.00 al g K ) 4.6ln 4.86 al K 74.8 system body 10.1 74.8 ( 10.1 74.8 ) Δ Sbody 4.6 1.00 1.67 al K 10.1 Δ S 4. 86 1.67.19 al K (b) ( 4.6)( 1)( 74.8) ( 70.0 10 )( 1)( 10.1 ) us, F ( ) 70.0 + 0.4 6 10 70.0 10.1 + 0.4 6 74.8 10 and F 09.9 K 6.77 98.19 F F F 09.9 Δ S ie water 4.6ln 4. al K 74.8 10.1 Δ S body ( 70.0 10 ) ln 1.9 al K 09.9 Δ S sys 4. 1.9.9 al K is is signiiantly less tan te estimate in part (a).
.1 / Δt e 1 / t Δ : ( 1 ) Δ t + : Δt Δt Δt Δt mδ : Δm Δ Δt Δt Δ m Δt Δ ( ) 9 ( 1.00 10 )( 00 K ) Δ m Δt 00 K 4 186 J kg 6.00 4.97 10 kg s *.4 (a) For te isotermal proess, te work on te gas is V B V ln V 0.0 ( 1.01 10 a)( 10.0 10 m ) ln 10.0 8.1 10 J were we ave used and 1.00 atm 1.01 10 a 1.00 L 1.00 10 m FIG..4 ( ) ( ) B BΔ V + 1.01 10 a 10.0 0.0 10 m 4.0 10 J 0 and B 4.10 10 J 4.10 kj (b) Sine is an isotermal proess, Δ Eint, 0 and For an ideal monatomi gas, R V and 8.1 10 J R B V 1.01 10 0.0 10 B B.06 10 nr R R
lso, V 1.01 10 10.0 10 1.01 10 nr R R n V Δ 1.00 R.06 10 1.01 10 6.08 kj R so te total energy absorbed by eat is + 8.1 kj+ 6.08 kj 14. kj n Δ nrδ Δ V () ( ) B B B B 1.01 10 10.0 0.0 10 1.01 10 4 J 10.1 kj ( ) ( ) (d) e + 1.4 10 J 4.10 10 J 4 0.88 or 8.9% (e) arnot ine operating between ot 060/R and old 1010/R as eiieny 1 / 1 1/ 80.0%. e tree-proess ine onsidered in tis problem as mu lower eiieny..9 e eat transer over te pats D and B is zero sine tey are adiabati. Over pat B: n ( ) > 0 B B B diabati roesses Over pat D: n ( ) < 0 ereore, D D V D and B D e eiieny is ten V i V i V ( ) ( ) e 1 1 D V B FIG..9 1 D e 1 γ B
*.61 (a) 0.0 (b) 9 9 Δ S mln + mln 1.00 kg( 4.19 kj kg K) ln + ln ( 4.19 kj K) ln 1 1 8 0 () Δ S + 4.88 J K (d) Yes, te mixing is irreversible. Entropy as inreased.