Chapter 20 - Heat and the First Law of Thermodynamics

Transcription

1 hapter 0 - Heat and the Frst Law o hermodynamcs 0.1 akng m 1.00 kg, we have Ug mgh But Δ 1.00 kg 9.80 m s 0.0 m 90 Δ U Q mcδ 1.00 kg 186 kg Δ 90 so Δ g +Δ he contaner s thermally nsulated, so no energy lows by heat: Q 0 and ΔE nt Q+W nput 0+W nput mgh he work on the allng weghts s equal to the work done on the water n the contaner by the rotatng blades. hs work results n an ncrease n nternal energy o the water: mgh ΔE nt m water cδ mgh 1.0 kg 9.80 m s.00 m 88. Δ m c 0.00 kg 186 kg 87 water 0.10 FIG Q cold Q hot ( mcδ ) ( mcδ ) water ( )( ) ( )( ) 9.6 ron 0.0 kg 186 kg kg 8 kg (a) ( )( mgh) mcδ ( 0.600)( kg )( 9.80 m s )( 0.0 m ).186 cal Δ ;.8 (.00 g )( 0.09 cal g )( ) Δ he nal temperature does not depend on the mass. Both the change n potental energy, and the heat that would be requred rom a stove to produce the temperature change, are proportonal to the mass; hence, the mass dvdes out n the energy relaton We do not know whether the alumnum wll rse or drop n temperature. he energy the water can absorb n rsng to 6 s mcδ 0. kg he kg

2 energy the copper can put out n droppng to 6 s mcδ 0.1 kg Snce 679 >86, the nal temperature s kg less than 6. We can wrte Q Q as h c Qwater + Q + Qu 0 0. kg 186 ( 0 ) + 0. kg 900 ( 6 ) kg kg kg 87 ( 100 ) 0 kg Vessel one contans oxygen descrbed by V nr : V 1.7( a ) m nc 1.19 mol R 8.1 Nm mol K 00 K Vessel two contans ths much oxygen: n h mol 1.6 mol (a) he gas comes to an equlbrum temperature accordng to ( mc Δ ) ( mc Δ ) cold ( nmc 00 K + nmc 0 K 0 hot c h he molar mass M and specc heat dvde out: K K K 80 K.9 ) he pressure o the whole sample n ts nal state s nr.9 mol K a.0 atm V m ol K m 0.1 he bullet wll not melt all the ce, so ts nal temperature s 0. hen where 1 mv +mc Δ m w bullet m w L s the melt water mass

3 m m w w ( ) + ( )( ) kg 0 m s kg 18 kg 0.0. kg g 000 kg 0.16 (a) heat to melt all the ce kg. kg 1.67 Q 1 Q ( heat to rase temp o ce to 100 ) ( kg )( 186 kg )( 100 ).09 hus, the total heat to melt ce and rase temp to heat avalable 6 Q ( kg )(.6 kg ).6 as steam condenses hus, we see that Q >Q 1, but Q <Q +Q. 1 hereore, all the ce melts but <100. Let us now nd Q cold Q hot ( kg )(. kg ) + ( kg )( 186 kg )( 0 ) 6 ( kg )(.6 kg ) ( kg )( 186 kg )( 100 ) From whch, 0. Q 1 heat to melt all ce [See part (a)] heat gven up 6 Q ( 10 kg )(.6 kg ).6 as steam condenses heat gven up as condensed Q ( 10 kg )( 186 kg )( 100 ) 19 steam cools to 0 Note that Q +Q <Q 1. hereore, the nal temperature wll be 0 wth some ce remanng. Let us nd the mass o ce whch must melt to condense the steam and cool the condensate to 0. ml Q +Q hus,.68 m kg 8.0 g o ce melts. kg hereore, there s.0 g o ce let over, also at 0.

4 0.1 W dv he work done on the gas s the negatve o the area under the curve between and. αv V V 1 W αv dv α V V V V 1.00 m.00 m W O α V 1 6 ( ).00 atm m a atm.00 m 1.00 m 1.18 M 1.00 m.00 m FIG. 0.1 V nr Δ Δ W V ( ) nr 0.7 ΔE nt Q +W Q ΔE nt W he negatve sgn ndcates that postve energy s transerred rom the system by heat. 0.8 W ( V V ) ( ) B B B.00 atm m 9. k Δ E Q+ W E E nt nt, nt, E k nt, B E.79 k nt, B (atm) B A D Snce s constant, E nt, D E nt, 0 WDA D VA VD 1.00 atm m k Ent, A E nt, D 10 k k 8.7 k Now, E nt, B Ent, A ( Ent, Ent, B ) + ( Ent, D Ent, ) + ( Ent, A E nt, D ) [ ] Ent, B E nt, A.79 k k.9 k FIG. 0.8 V(m ) 0.0 (a) V V W nrln V ln V V so V V W 000 exp exp m V 0.0 0( ) ( ) V a m nr 1.00 mol 8.1 K mol 0 K

5 0.1 (a) E Q V Δ nt Δ 1. k.0 ka m 7.0 k V 1 1 V V K 900 K V (a) W Δ V [ αvδ] 1.00 kg kg m 6 1 ( N m ) (.0 10 ) ( 18.0 ) W 8.6 m Q cm Δ 900 kg 1.00 kg k (c) ΔE nt Q +W 16. k 8.6 m 16. k 0. (a) he work done durng each step o the cycle equals the negatve o the area under that segment o the V curve. W W + W + W + W DA AB B D W V V + 0 V V + 0 V he ntal and nal values o or the system are equal. hereore, ΔE nt 0 and Q W V. (c) W V nr k FIG ( W m )(.00 m )(.0 ) kaδ W 10.0 kw L m 0.9 In the steady state conon, so that Δ Δ k A k A Δx Δx In ths case A A Δ ( 80.0 ) and Δ ( 0.0) where s the temperature o the juncton. hereore, k ( 80.0 ) k ( 0.0) Δ x Δx And 1. FIG. 0.9

6 *0.7 Intensty s dened as power per area perpendcular to the drecton o energy low. he drecton o sunlght s along the lne rom the sun to the object. he perpendcular area s the projected lat crcular area enclosed by the termnator the lne that separates day rom nght on the object. he object radates nrared lght outward n all drectons. he area perpendcular to ths energy low s ts sphercal surace area. he sphere o radus R absorbs sunlght over area π R. It radates over area π R. hen, n steady state, n out e R e ( ) 170 W m π σ π R he emssvty e, the radus R, and π all cancel. hereore, 1 70 W m 79 K 6 8 ( W m K ) It s chlly, well below room temperatures we nd comortable LρAdx Δ ka x 8.00 Lρ xdx kδ x Lρ Δt kδδt ( m ) ( m ). kg 917 kg m (.00 W m )( 10.0 ) Δt Δ t s 10. h A A end walls + A ends o attc + A sde walls + A roo 1 A ( 8.00 m.00 m ) +.00 m (.00 m ) tan m + ( 10.0 m.00 m ) + ( 10.0 m ) cos7.0 A 0 m ( kw m )( 0 m )(.0 ) kaδ 17. kw.1 kcal s L 0.10 m hus, the energy lost per day by heat s.1 kcal s s.9 10 kcal day. he gas needed to replace ths loss s.9 10 kcal day 9 00 kcal m 8.6 m day.

7 0.6 See the dagram appearng wth the next problem. For a cylndrcal shell o radus r, heght L, and thckness dr, the equaton or thermal conducton, thereore, dq d ka dx dq becomes k( π rl) Under equlbrum conons, dq dq 1 dr d π kl r b a dq 1 bdr d π kl a r and dq 1 b b a ln π kl a But, so a > b s constant; dq π kl ln d dr ( a b) ( b a) *0.6 Q cold Q hot or Q ( Q + Q ) water calo ( ) ( w w + c c)( ) m c m c m c ( 0.00 kg ) c ( + 9. ) 0.00 kg ( 186 kg ) kg ( 60 kg ) (.70 ) c kg 800 kg w hs ders rom the tabulated value by ( )/900 11%, so the values agree wthn 1%.