Chapter 18 Heat and the First Law of Thermodynamics

Transcription

1 Chapter 8 Heat and the Frst Law o Thermodynamcs 9 [SSM] How much heat must be absorbed by 6. g o ce at.ºc to transorm t nto 6. g o at 4.ºC? Pcture the Problem e can nd the amount o heat that must be absorbed by addng the heat requred to warm the ce rom. C to C, the heat requred to melt the ce, and the heat requred to warm the ormed rom the ce to 4. C. Express the total heat requred: Substtute or each term to obtan: + + mc warm ce ce m c ΔT melt ce warm ( ΔT + L + c ΔT ce ce ce + ml + mc ΔT Substtute numercal values (See Tables 8- and 8- and evaluate : (.6 kg.5 ( C (..kj kg K kg K ( 4. C +.5 kj kg 5 Durng hs many appearances at the Tour de France, champ bcyclst Lance Armstrg typcally expended an average power o 4, 5. hours a day or days. hat quantty o, ntally at 4ºC, could be brought to a bol you could harness all o that energy? Pcture the Problem e can use mcδt to express the mass m o that can be heated through a temperature nterval ΔT by an amount o heat energy. e can then nd the amount o heat energy expended by Armstrg rom the dent o power. Express the amount o heat energy requred to rase the temperature o a mass m o by ΔT: Use the dent o power to relate the heat energy expended by Armstrg to the rate at whch he mcδt m cδt P PΔt Δ t 77

2 774 Chapter 8 expended the energy: Substtute or to obtan: evaluate m: PΔt m cδt J 6 s 4 5. s h m 4.84 kg K 4.5 kg h d d ( C 4 7 [SSM] A -g pece o ce at ºC s placed n 5 g o at ºC. Ths system s n a ctaner o neglgble heat capacty and s nsulated rom ts surroundngs. (a hat s the nal equlbrum temperature o the system? (b How much o the ce melts? Pcture the Problem Because we can not tell, wthout perormng a couple o calculats, whether there s enough heat avalable n the 5 g o to melt all o the ce, we ll need to resolve ths quest rst. See Tables 8- and 8- or specc heats and the latent heat o us o. (a Determne the energy requred to melt g o ce: melt ce mcel (. kg The energy avalable rom 5 g o at ºC s: 66.7 kj.5 kj kg avalable,max m c T 4.84 kj Δ kg K (.5 kg 4.84 ( C Because avalable, max < meltce : The nal temperature s C. (b Equate the energy avalable rom the to m ce L and avalable, max solve or m ce to obtan: m ce avalable,max L evaluate m ce : m 4.84 kj kj.5 kg ce 5g

3 Heat and the Frst Law o Thermodynamcs A well-nsulated bucket o neglgble heat capacty ctans 5 g o ce at ºC. (a I g o steam at ºC s njected nto the bucket, what s the nal equlbrum temperature o the system? (b Is any ce let ater the system reaches equlbrum? Pcture the Problem Frst you need to cvnce yoursel that there s enough energy n the steam to melt all the ce. Once you ve de that you can use cservat o energy to nd the nal equlbrum temperature. See Tables 8- and 8- or specc heats and the heats o us and vaporzat o. (a Frst, determne the energy melt ce m L requred to melt all the ce: (.5 kg(.5 kj/kg ce 5. kj Fnd the maxmum amount o energy avalable rom the steam s: steam, max m L + m c T steam v steam Δ max kg 45.4kJ 8.7 kj 5.5 kj kg K (. kg 57 + (. kg 4.84 ( C Because steam, max > melt ce, all the ce wll melt and the nal temperature wll be greater than C. Apply cservat o energy to obtan: or cdense + cool the + + melt warm the steam hot the ce all ce ( t + 5. kj + m c ( t 45.4 kj + m c ce hot Substtutng numercal values yelds: 45.4 kj + kg K (. kg 4.8 ( t kj kg K (.5 kg 4.84 ( t Solvng or t yelds: t 4.9 C (b Because the nal temperature s greater than C, no ce s let.

4 776 Chapter 8 4 The specc heat o a -g block o a substance s to be determned. The block s placed n a 5-g copper calormeter holdng 6 g o ntally at ºC. Then, ml o at 8ºC are added to the calormeter. hen thermal equlbrum s reached, the temperature o the system s 54ºC. Determne the specc heat o the block. Pcture the Problem Let the subscrpt B denote the block, w the ntally n the calormeter, and w the ml o that s added to the calormeter vessel. e can use cservat o energy to nd the specc heat o the block. See Table 8- or specc heats. Apply cservat o energy to obtan: or m warm + warm the + warm w + cool w the block calormeter Δ T + mcuccuδt + mw c Δ Δ w T + m w c w T w B cb where ΔT s the comm temperature change o the calormeter, block, and ntally n the calormeter. Substtute numercal values to obtan: (. kg c ( 54 C + (.5kg.86 ( 54 C B + kg K (.6 kg 4.84 ( 54 C + kg K kg K (. kg 4.84 ( 54 C 8 Solvng or c B yelds: c B. kj/kg K 47 I a gas absorbs 84 J whle dg J o work, what s the change n the nternal energy o the gas? Pcture the Problem e can apply the rst law o thermodynamcs to nd the change n nternal energy o the gas durng ths process.

5 Heat and the Frst Law o Thermodynamcs 777 Apply the rst law o thermodynamcs to express the change n nternal energy o the gas n terms o the heat added to the system and the work de the gas: Δ E + nt n The work de by the gas s the negatve o the work de the gas. evaluate ΔE nt : ΔE nt 84 J J 54J 48 A lead bullet ntally at ºC just melts up strkng a target. Assumng that all o the ntal knetc energy o the bullet goes nto the nternal energy o the bullet, calculate the mpact speed o the bullet. Pcture the Problem e can use the dent o knetc energy to express the speed o the bullet up mpact n terms o ts knetc energy. The heat absorbed by the bullet s the sum o the heat requred to warm the bullet rom K to ts meltng temperature o 6 K and the heat requred to melt t. e can use the rst law o thermodynamcs to relate the mpact speed o the bullet to the change n ts nternal energy. See Table 8- or the specc heat and meltng temperature o lead. Usng the rst law o thermodynamcs, relate the change n the nternal energy o the bullet to the work de t by the target: Substtute or ΔE nt, K, and K to obtan: Δ E + nt n or, because n, ΔE ΔK K nt ( K ( mv mc PbΔ TPb + ml,pb mv or mc T T + ml Pb ( mv MP,Pb [ ] Solvng or v yelds: c ( T T v + L Pb MP,Pb evaluate v: kj v.8 kg K kg ork and the P Dagram or a Gas ( 6K K m/s

6 778 Chapter 8 5 The gas s allowed to expand at cstant pressure untl t reaches ts nal volume. It s then cooled at cstant volume untl t reaches ts nal pressure. (a Illustrate ths process a P dagram and calculate the work de by the gas. (b Fnd the heat absorbed by the gas durng ths process. Pcture the Problem e can nd the work de by the gas durng ths process rom the area under the curve. Because no work s de alg the cstant volume (vertcal part o the path, the work de by the gas s de durng ts sobarc expans. e can then use the rst law o thermodynamcs to nd the heat added to the system durng ths process. (a The path rom the ntal state ( to the nal state ( s shown the P dagram... P, atm...., L The work de by the gas equals the area under the shaded curve: PΔ 68J (. atm(.l.5 kpa m. atm.l atm L (b The work de by the gas s the negatve o the work de the gas. Apply the rst law o thermodynamcs to the system to obtan: evaluate n : n n ΔE nt ( Ent, Ent, ( ( Ent, Ent, + by gas ( 9J 456J.6 kj + 68J 5 [SSM] The gas s rst cooled at cstant volume untl t reaches ts nal pressure. It s then allowed to expand at cstant pressure untl t reaches ts nal volume. (a Illustrate ths process a P dagram and calculate the work de by the gas. (b Fnd the heat absorbed by the gas durng ths process. Pcture the Problem e can nd the work de by the gas durng ths process

7 Heat and the Frst Law o Thermodynamcs 779 rom the area under the curve. Because no work s de alg the cstant volume (vertcal part o the path, the work de by the gas s de durng ts sobarc expans. e can then use the rst law o thermodynamcs to nd the heat absorbed by the gas durng ths process (a The path rom the ntal state ( to the nal state ( s shown the P dagram... P, atm...., L The work de by the gas equals the area under the curve: PΔ 45J (. atm(.l.5 kpa m. atm.l atm L (b The work de by the gas s the negatve o the work de the gas. Apply the rst law o thermodynamcs to the system to obtan: n ΔE nt ( Ent, Ent, ( ( Ent, Ent, + by gas evaluate n : ( 9 J 456 J + 45 J 86J n 5 The gas s allowed to expand sothermally untl t reaches ts nal volume and ts pressure s. atm. It s then heated at cstant volume untl t reaches ts nal pressure. (a Illustrate ths process a P dagram and calculate the work de by the gas. (b Fnd the heat absorbed by the gas durng ths process. Pcture the Problem e can nd the work de by the gas durng ths process rom the area under the curve. Because no work s de alg the cstant volume (vertcal part o the path, the work de by the gas s de durng ts sothermal expans. e can then use the rst law o thermodynamcs to nd the heat absorbed by the gas durng ths process.

8 78 Chapter 8 (a The path rom the ntal state ( to the nal state ( s shown the P dagram. P, atm......, L The work de by the gas equals the area under the curve: P P Pd nrt.l.l ln d L L d P [ ln ]. L. L evaluate :.5 kpa m.atm.l ln atm L 4J (b The work de by the gas s the negatve o the work de the gas. Apply the rst law o thermodynamcs to the system to obtan: n ΔE nt ( Ent, Ent, ( ( Ent, Ent, + by gas evaluate n : ( 9 J 456 J + 4 J 79 J n 5 The gas s heated and s allowed to expand such that t ollows a sngle straght-lne path a P dagram rom ts ntal state to ts nal state. (a Illustrate ths process a P dagram and calculate the work de by the gas. (b Fnd the heat absorbed by the gas durng ths process. Pcture the Problem e can nd the work de by the gas durng ths process rom the area under the curve. e can then use the rst law o thermodynamcs to nd the heat absorbed by the gas durng ths process.

9 Heat and the Frst Law o Thermodynamcs 78 (a The path rom the ntal state ( to the nal state ( s shown the P dagram: P, atm......,l The work de by the gas equals the area under the curve: (.atm +.atm (.L A trapezod.5j 5.atm L atm L 57J (b The work de by the gas s the negatve o the work de the gas. Apply the rst law o thermodynamcs to the system to obtan: n ΔE nt ( Ent, Ent, ( ( Ent, Ent, + by gas evaluate n : ( 9J 456J + 57 J 96J n Remarks: You could use the lnearty o the path cnectng the ntal and nal states and the coordnates o the endpts to express P as a unct o. You could then ntegrate ths unct between. and. L to nd the work de by the gas as t goes rom ts ntal to ts nal state. uas-statc Adabatc Expans o a Gas 69 [SSM] A.5-mol sample o an deal matomc gas at 4 kpa and K, expands quas-statcally untl the pressure decreases to 6 kpa. Fnd the nal temperature and volume o the gas, the work de by the gas, and the heat absorbed by the gas the expans s (a sothermal and (b adabatc. Pcture the Problem e can use the deal-gas law to nd the ntal volume o the gas. In Part (a we can apply the deal-gas law or a xed amount o gas to nd the nal volume and the express or the work de n an sothermal process. Applcat o the rst law o thermodynamcs wll allow us to nd the heat absorbed by the gas durng ths process. In Part (b we can use the

10 78 Chapter 8 relatshp between the pressures and volumes or a quas-statc adabatc process to nd the nal volume o the gas. e can apply the deal-gas law to nd the nal temperature and, as n (a, apply the rst law o thermodynamcs, ths tme to nd the work de by the gas. Use the deal-gas law to express the ntal volume o the gas: nrt P evaluate : J mol K 4kPa (.5mol 8.4 ( K.8 m (a Because the process s sothermal: T T K Use the deal-gas law or a xed amount o gas to express : P T P T or, because T cstant, P P evaluate : (.8 L 7.8L 4 kpa L 6 kpa Express the work de by the gas durng the sothermal expans: nrt ln evaluate : (.5 mol ( K.4 kj J 8.4 mol K 7.795L ln.8l Notng that the work de by the gas durng the process equals the negatve o the work de the gas, apply the rst law o n ΔE nt.4 kj (.4 kj

11 thermodynamcs to nd the heat absorbed by the gas: Heat and the Frst Law o Thermodynamcs 78 (b Usng γ 5/ and the relatshp between the pressures and volumes or a quas-statc adabatc process, express : P P γ γ P P γ evaluate : (.8L 5.4L 4 kpa 6 kpa L Apply the deal-gas law to nd the nal temperature o the gas: evaluate T : T T P nr ( 6 kpa( 5.4 m (.5 mol 8K J 8.4 mol K For an adabatc process: n Apply the rst law o thermodynamcs to express the work de the gas durng the adabatc process: ΔEnt n CΔT nrδt evaluate : (.5 mol( 8.4 J/mol K ( 8K K 574J Because the work de by the gas equals the negatve o the work de the gas: ( 574J 574J Cyclc Processes 7 A.-mol sample o N gas at.ºc and 5. atm s allowed to expand adabatcally and quas-statcally untl ts pressure equals. atm. It s then heated at cstant pressure untl ts temperature s agan.ºc. Ater t

12 784 Chapter 8 reaches a temperature o.ºc, t s heated at cstant volume untl ts pressure s agan 5. atm. It s then compressed at cstant pressure untl t s back to ts orgnal state. (a Cstruct a P dagram showng each process n the cycle. (b From your graph, determne the work de by the gas durng the complete cycle. (c How much heat s absorbed (or released by the gas durng the complete cycle? Pcture the Problem To cstruct the P dagram we ll need to determne the volume occuped by the gas at the begnnng and endng pts or each process. Let these pts be A, B, C, and D. e can apply the deal-gas law to the startng pt (A to nd A. To nd the volume at pt B, we can use the relatshp between pressure and volume or a quas-statc adabatc process. e can use the deal-gas law to nd the volume at pt C and, because they are equal, the volume at pt D. e can apply the rst law o thermodynamcs to nd the amount o heat added to or subtracted rom the gas durng the complete cycle. (a Usng the deal-gas law, express the volume o the gas at the startng pt A o the cycle: A nrt P A A evaluate A : A J mol K.5kPa 5.atm atm 4.88 m 4.88L (. mol 8.4 ( 9K Use the relatshp between pressure and volume or a quasstatc adabatc process to express the volume o the gas at pt B; the end pt o the adabatc expans: B A P P A B γ 5. atm.atm evaluate B : B ( 4.88 L 5.8L.4 Usng the deal-gas law or a xed amount o gas, express the volume occuped by the gas at pts C and D: C D nrt P C C

13 Heat and the Frst Law o Thermodynamcs 785 evaluate C : C J mol K.5kPa.atm atm 4.4 m 4.4 L (. mol 8.4 ( 9K The complete cycle s shown n the dagram. P, atm A... B C 5 5 5, L (b Note that or the paths A B and B C,, the work de by the gas, s postve. For the path D A, s negatve, and greater n magntude than A C. Thereore the total work de by the gas s negatve. Fnd the area enclosed by the cycle by notng that each rectangle o dotted lnes equals 5 atm L and countng the rectangles: ( rectangles 5. ( 65atm L 6.6kJ atm L rectangle.5j J atm L (c The work de the gas equals the negatve o the work de by the gas. Apply the rst law o thermodynamcs to nd the amount o heat added to or subtracted rom the gas durng the complete cycle: n ΔE nt 6.6kJ ( 6.59kJ because ΔE nt or the complete cycle.