Chapter 22 - Heat Engines, Entropy, and the Second Law of Thermodynamics

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "Chapter 22 - Heat Engines, Entropy, and the Second Law of Thermodynamics"

Αἰνέας Αγγελόπουλος
5 χρόνια πριν
Προβολές:

1 apter - Heat Engines, Entropy, and te Seond Law o ermodynamis.1 (a).0 J e 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 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 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 % 9 K (b) Heat expelled Heat removed + ork done J+ 4 J 144 J.7 O.00. ereore, e eat removed ea minute is t..00 ( kg )( J kg )(.0 ) ( kg )(. 10 J kg ) + ( ) kg 090 J kg J min + or, J s t us, te work done per seond is J s K 14 K (a) Δ e %

2 (b) J, 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) ( ) 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 and we ave γ γ γ ( ) a ( 1 07 K ) 64 K 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/ kg )( )( 8.14 J m ol K)( ) K 60.0 s k () 64 K e or 47.% 107 K.16 (a) e max %

3 6 (b) J s Δt ereore, J s 600 s J From e we ind J 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 arnot eiieny arnot yle 9 K K 68 K FIG..4 us, 1.17 joules o energy enter te room by eat or ea joule o w ork done.

4 .8 (a), (b) e quantity o gas is 6 ( a)( m ) V n 0.00 mol R ( 8.14 J mol K )( 9 K ) 6 int, ( a )( E nr V m ) 1 J γ V B V B In proess, 1.40 int, B V nr a a 6 6 ( a)( 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 m E out 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 ( a) a V D 8.00 int, D V nr int, D 6 ( a)( 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

5 For te entire yle, Δ Eint, net 16 J e net work is net 16 J J J J J 84. J e tables look like: State (K) (ka) V ( m ) E int (J) B D roess (J) output (J) Δ Eint (J) B D D D () 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 rev min γ e 1 1.% 6.00 (b) I atual eiieny e 1.0% losses in system are e e 6.%

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.

76 J K Vi ere is no ange in temperature or an ideal gas. FIG.

.40.49 (a) For an isotermal proess, nrln V V1 ereore, ( ) 1 nr i ln

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 S Δ J K 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 Rln Vi.40 Δ ( ) Δ S ln 0.07 J K FIG (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 ( ) or nr i ln FIG..49

7 (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 ( + ) F K 74.8 K 9.0 (a) ( ) ( ) 98.6 F ( ) ( ) 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 ( ) Δ Sbody al K 10.1 Δ S al K (b) ( 4.6)( 1)( 74.8) ( )( 1)( 10.1 ) us, F ( ) and F 09.9 K F F F 09.9 Δ S ie water 4.6ln 4. al K Δ S body ( ) ln 1.9 al K 09.9 Δ S sys al K is is signiiantly less tan te estimate in part (a).

8 .1 / Δt e 1 / t Δ : ( 1 ) Δ t + : Δt Δt Δt Δt mδ : Δm Δ Δt Δt Δ m Δt Δ ( ) 9 ( )( 00 K ) Δ m Δt 00 K J kg kg s *.4 (a) For te isotermal proess, te work on te gas is V B V ln V 0.0 ( a)( m ) ln J were we ave used and 1.00 atm a 1.00 L m FIG..4 ( ) ( ) B BΔ V a m J 0 and B J 4.10 kj (b) Sine is an isotermal proess, Δ Eint, 0 and For an ideal monatomi gas, R V and J R B V B B nr R R

9 lso, V nr R R n V Δ 1.00 R kj R so te total energy absorbed by eat is kj kj 14. kj n Δ nrδ Δ V () ( ) B B B B J 10.1 kj ( ) ( ) (d) e J J 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

10 *.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 () Δ S J K (d) Yes, te mixing is irreversible. Entropy as inreased.

Σχετικά έγγραφα

2. Chemical Thermodynamics and Energetics - I

2. Chemical Thermodynamics and Energetics - I . Chemical Thermodynamics and Energetics - I 1. Given : Initial Volume ( = 5L dm 3 Final Volume (V = 10L dm 3 ext = 304 cm of Hg Work done W = ext V ext = 304 cm of Hg = 304 atm [... 76cm of Hg = 1 atm]