THERMAL PROPERTIES OF MATTER 12

HERMAL PROPERIES OF MAER Q.. Reason: he mass o a mole o a substance n grams equals the atomc or molecular mass o the substance. Snce neon has an atomc mass o 0, a mole o neon has a mass o 0 g. Snce N has a molecular mass o 8, a mole o N has a mass o 8 g. hus a mole o N has more mass than a mole o neon. Assess: Even though ntrogen atoms are lghter than neon atoms, ntrogen molecules are more massve, so a mole o ntrogen has more mass than a mole o neon. Q.. Reason: Hydrogen and helum can escape the atmosphere whle heaver elements such as oxygen and ntrogen cannot. hs s because hydrogen and helum, beng lghter, move aster. From Chapter, we know that the root-mean-square speed o a partcle s gven by the ollowng: v rms It ollows that the lghter a partcle s, the greater s ts rms speed. Assess: We can be thankul that our Earth s gravty holds on to the vtal nutrents oxygen and ntrogen. A smaller body than the earth mght not be able to hold on to these elements. For example, the moon s gravty s too weak or t to have an atmosphere o oxygen or ntrogen. kb m Q.. Reason: Snce there s almost pure helum n a helum balloon and almost no helum n the outsde ar, helum tends to duse out o the balloon. Smlarly, wth almost no oxygen or ntrogen n the balloon ntally and hgh concentratons o oxygen and ntrogen n the ar, these molecules tend to duse nto the balloon. However, snce helum atoms travel about three tmes aster than oxygen or ntrogen molecules and snce helum atoms are smaller, they duse much aster so gas leaves the balloon aster than t enters. An arlled balloon has the same partcles nsde as out and so the stated eect does not contrbute to the delaton o such balloons. Instead, there s a weaker eect whch s also at work or helum balloons: Hgher pressure nsde the balloon than outsde makes the nteror ar molecules duse aster. Assess: Because o the advantages helum atoms have n duson, helum balloons delate aster than ar-lled ones. Q.4. Reason: From the deal gas law, the lower the pressure o a contaner o gas, the hgher the volume. he pressure nsde the bag equals the pressure outsde. So when the atmospherc pressure goes down, the pressure n the bag goes down and the volume goes up. Q.5. Reason: From Equaton.7, we know that the temperature o a gas s drectly proportonal to the square o the rms speed o the molecules. hus, doublng the typcal speed o molecules n a gas ncreases the temperature by a actor o our. We also know, rom Equaton., that the pressure o a gas s drectly proportonal to ts temperature, so doublng the typcal speed o the molecules must also ncrease the pressure by -

- Chapter a actor o our. Increasng the speed o molecules n a gas ncreases the amount o orce a molecule exerts on the wall o the contaner and the rate o collsons wth the walls. Assess: It makes sense that the pressure depends on the square o the average velocty o molecules n a gas snce the orce and collson rate are proportonal to the velocty o a molecule. Q.6. Reason: Note that NVand / v rms are the same or both gases. (a) In the process o dervng the deal gas law we saw that N p mv V or that p m, so gven the condtons above, the gas wth the more massve molecules (gas ) wll have the hgher pressure. (b) he deal gas law can be rearranged as p > p p = N kb V whch shows that p, so gven the answer to part (a) the temperature o gas must be greater than the temperature o gas. > Assess: We conclude that, other thngs beng equal, the gas wth the more massve molecules wll have a greater pressure and a greater temperature. Q.7. Reason: (a) As dened n the chapter, a mole s 6.00 basc partcles, regardless o whch chemcal element we have. So there are equal numbers o partcles n a mole o helum gas and a mole o oxygen gas. (b) A mole o helum gas has a mass o 4 g, whle a mole o oxygen gas has a mass o g, so one gram o helum gas has 6.00 partcles, whle one gram o oxygen gas has 6.00 partcles. hereore, the 4 gram o helum gas has more partcles than the gram o oxygen gas. Assess: We note that the basc partcles or the helum gas are helum atoms whle the basc partcles or the oxygen gas are datomc oxygen molecules. hat s why the O molecule has a mass o u. 6 Q.8. Reason: From able., an atom o alumnum has a mass o 7 u 4.480 kg and an atom o 5 lead has a mass o 07 u.440 kg. he number o moles n 00 g o alumnum and lead s n 6 Al NAl / NA (0.00 kg)/(4.480 kg/atom)/(6.00 atom/mol).7 mol n N N 5 Pb Pb/ A (0.00 kg)/(.440 kg/atom)/(6.00 atom/mol) 0.48 mol he number o atoms n each s rms 6 4 N Al (0.00 kg)/(4.480 kg/atom). 0 atom 5 N Pb (0.00 kg)/(.440 kg/atom).9 0 atom Assess: hs makes sense, snce an atom o lead s much more massve than an atom o alumnum. Q.9. Reason: (a) he thermal energy o the gas does change. In Secton.4 we are gven Eth Nmv rms so the speed o every molecule s doubled then the thermal energy s ncreased by a actor o 4.

hermal Propertes o Matter - (b) No, the molar specc heat doesn t change. A careul readng o Secton.7 shows that the molar specc heat s only a uncton o the possble orms o thermal energy (translatonal, rotatonal, vbratonal; these are called degrees o reedom). Assess: In realty the molar specc heat does actually change a lttle bt as a uncton o temperature. hat s why the temperature s speced at the top o able.6. Q.0. Reason: Snce the helum atoms are ve tmes lghter than the neon atoms, they are travelng, on average, 5 tmes aster. For the sake o argument, we wll compare a sample neon atom to a sample helum atom whch s travelng aster by exactly 5 and n the same drecton. You mght thnk the helum atom would exert greater pressure snce t s aster. However, snce momentum depends on velocty and mass, ts momentum s actually less by a actor o 5. hs means that every tme the helum atom colldes wth the walls o the contaner, t mparts a orce whch s less than that exerted by the neon atom by atoms travel aster, they collde more oten, agan by 5. Fnally, snce the helum 5. he combned eect o a orce whch s weaker by 5 wth collsons whch occur 5 tmes more oten s that the tme-averaged orce s the same or both atoms. Consequently, the pressures exerted by the two derent gases are the same. Assess: It s nterestng to see how a number o eects have canceled so that the two gases have the same pressure. Smlarly, helum and neon atoms have the same average knetc energy even though the helum atoms are aster because they are correspondngly lghter. Q.. Reason: Equaton. apples. he number o molecules n the gas s constant snce the contaner s sealed. Equaton. can be wrtten as p NkB( / V). (a) I the volume s doubled and the temperature trpled, the pressure ncreases by a actor o /. (b) I the volume s halved and the temperature trpled, the pressure ncreases by a actor o sx. Assess: hs makes sense. Increasng the temperature ncreases the pressure n a gas as does decreasng the volume o the contaner. Q.. Reason: Assume the gas s an deal gas, and use the deal gas law nr. Snce the number o moles doesn t change and R s a constant, then Equaton.4 gves In each case we want to solve or (a) For V V and p p = = ( p)( V) = = 6 So the absolute temperature has ncreased by a actor o sx. (b) For V V and p =p ( p)( V) So the absolute temperature has ncreased by a actor o /. Assess: A shortcut s that and we double V and trple p, then s ncreased by a actor o 6.

-4 Chapter Q.. Reason: I the work done s equals the area under the graph, then there s no work done V s constant, because the graph would be a vertcal straght lne encompassng no area. Stated another way, Equaton.7 says W and V s constant then V 0 andw gas 0. gas Assess: As the chapter says, n order or a gas to do work, the volume must change. Q.4. Reason: hermal expanson wll make your sun-drenched tape longer than the shaded tape, so you and your coworker measure the same object yours wll read a smaller value. Assess: o the extent that ths s notceable, ths would be an error n the measurements made wth the hot tape, as they are calbrated to be correct at room temperature. Q.5. Reason: From able., we see that water has a sgncantly hgher coecent o thermal expanson than steel about sx tmes as much. As the water and steel get hotter, the water expands sx tmes more than the steel. hus the water wll overlow out o the radator. Assess: hs seems reasonable snce we expect gases to expand more than lquds and lquds to expand more than solds when the temperature ncreases. Q.6. Reason: Snce A and B are n a well-nsulated contaner, any heat leavng B goes entrely nto rasng the temperature o A so Solvng or the nal temperature, mc ( 0C) mc ( 00 C) A B 00C c / c Snce the specc heat o A s larger than the specc heat o B, the nal temperature s less than 00C. Assess: As expected, the specc heats o the two materals are equal, the nal temperature wll be 00C. Note that the temperature change depends only on the mass o materal, not the densty. Q.7. Reason: You are heatng both contaners (each wth n moles o ntrogen gas) and thereby ncrease the nternal energy o each by Q, but the temperatures do not rse by the same amount. See Equatons.5 and.6. Q ncva ncpb 0 J Consultng able.6 shows that C P > C V ; thereore A Band snce ( A) ( B), then ( A ) ( B ). Assess: Some o the energy n contaner B s used as work done n changng the volume. In contaner A the volume dd not change, so no work was done and all o the energy went nto changng the temperature. Q.8. Reason: From Equatons.5 and.6, the heat requred to rase the temperature o a gas depends on the number o moles o the gas, the molar specc heat o the gas, and whether the gas s heated at constant volume or constant pressure. Lookng at able.6, the molar specc heat o gasses s always smaller or gasses heated at constant volume than those heated at constant pressure. In order to use the least amount o heat, you should heat the gas at a constant volume. Assess: he constant pressure molar specc heat o a gas s generally hgher than the constant volume molar specc heat because gasses do work when ther volume ncreases and ths work must be pad or wth extra heat. Q.9. Reason: Snce ths s an sothermal process, the nternal energy o the gas does not change. Consequently, the rst law o thermodynamcs says: 0 Q W. So W and Q are negatves o one another. Q s the heat whch lows nto a system. Here, t s 600 J. So W 600 J. When W s negatve, t means that the system s dong work. In ths case, the gas does 600 J o work. Assess: I a gas does no work ts thermal energy ncreases when heat s added. But by dong work equal n amount to the heat added, the gas n ths problem spends the energy gven to t so that ts thermal energy and thereore ts temperature stay constant. Q.0. Reason: he chocolate starts at some temperature. As t s heated the temperature rses to a pont where the chocolate changes phase. Assumng the chocolate started as a sold, the chocolate melts durng the second porton o the graph. Durng meltng, the temperature s constant. Ater all the chocolate melts, the temperature rses as the lqud chocolate s heated. A B

hermal Propertes o Matter -5 Assess: Compare to Fgure.. Q.. Reason: At hgher elevatons the ar pressure s lower. he amount o leavenng agent should be reduced, snce bubbles wll orm more readly n the lower pressure. Water wll also vaporze at a lower temperature, so the bakng temperature should be decreased. Decreasng the temperature requres an ncrease n the bakng tmes. Assess: Most non-physcst cooks wll suggest ncreasng the amount o water n a recpe nstead o ncreasng bakng tme. Q.. Reason: Revew Example. to see how Q Mc s used n these calormetry problems. Qw Qm Mwcww Mmcmm 0 In our case we want to consder w n both the alumnum case and the ron case. Note that all the varables that can be the same or the two cases are. he alumnum has more thermal mass and so can gve up more energy to the water wthout lowerng ts own temperature as much, whereas the ron wth less thermal mass lowers ts temperature more as t gves up energy to the water. he alumnum rases the temperature o the water more than the ron rases the temperature o the water. he alumnum block ends up wth the warmer water. Assess: It should be noted n connecton wth ths queston that n rasng the two blocks o metal to the 00C temperature rom room temperature n the rst place, the alumnum block absorbed more energy or the same temperature change; thereore t had more thermal energy to gve up to the water. Q.. Reason: Part (a) o the gure represents a constant pressure, or sobarc, expanson o the gas. Part (b) represents a constant volume reducton o pressure o the gas. Durng part (b), the temperature also decreases, rom Equaton.. Part (c) represents a decrease n volume along wth an ncrease n pressure. However, part (c) s not sothermal snce the graph s a straght lne. Isothermal processes are hyperbolae on dagrams. Assess: A correct dagram would look lke the ollowng gure. Q.4. Reason: Accordng to able.7 slver has a much greater thermal conductvty than stanless steel, so the slver spoon wll eel hot rst. Assess: he same atomc crystallne structure and electrons ree to move n the metal make slver a good conductor o both thermal energy and electrcty. Q.5. Reason: he trees help prevent the energy rom beng radated out nto space on a cold clear nght; the trees relect back down some o the nrared radaton and keep the ground under them warmer. In contrast, the open ground radates ts thermal energy nto space wthout the blanket o the trees or clouds to keep the energy n. Assess: Gardeners n northern clmes know to cover ther plants on clear all nghts to keep the radaton n and keep the plants rom reezng. hese early rst rosts n the all are even called radaton rosts. hey take place on clear nghts wth calm wnds. Another type, called advectve reeze, occurs when very cold ar moves n (by convecton); advectve reezes can take place wth wnds and clouds present, and are much harder to protect plants aganst. Q.6. Reason: From the equaton denng gauge pressure, p p g atm. he absolute pressure s hgher nsde the tre. he correct choce s C. Assess: Gauge pressure s the pressure above atmospherc pressure.

-6 Chapter Q.7. Reason: Apply the deal gas law n Equaton.: Nk. B We are told that s kept constant and assume that V s also. So p N; the number o atoms s doubled then the pressure s also. he correct answer s D. Assess: hs makes sense because wth twce as many atoms there wll be twce as many atomc collsons n a gven amount o tme, so the pressure s doubled. Q.8. Reason: I a gas s compressed sothermally, then rom the deal gas law, we know that PV s constant. hus, the volume decreases by a actor o two, the pressure must ncrease by a actor o two. he answer s C. Assess: We see that or an sothermal process, pressure and volume are nversely proportonal. You may have heard ths result called Boyle s law. Q.9. Reason: In an adabatc compresson, the temperature rses. From the deal gas law, we can say: P nr / V. I the temperature were constant, halvng the volume would double the pressure. But snce the temperature rses and temperature s n the numerator o the pressure ormula, the pressure wll more than double. he answer s D. Assess: It makes sense that an adabatc compresson would rase the temperature more than an sothermal compresson such as we saw n the prevous queston. hs s because adabatc processes are generally processes whch occur rapdly so that heat does not have tme to escape. Ater the heat starts to escape, the gas partcles have less energy and exert less pressure on the walls o the contaner. Q.0. Reason: o measure the specc heat o a materal by puttng t n water, you need to know the mass o the sample, the mass o the water, the ntal temperatures and the nal temperature. he mass o the pennes was measured, but one was lost so that the mass o the sample was overestmated. When you solve the calormetry equaton, you obtan the ollowng: Snce you are usng a value o m c m c 0 J c p p p p w w w m c m w w w m p whch s too hgh, and snce p p m p s n the denomnator, your value o c p wll be too low. hus you wll underestmate the specc heat. he answer s A. Assess: Another way to look at ths s that snce you are usng too ew pennes, the temperature o the water wll rse less than t would have you had used all o them. In a calormetry problem, the greater the specc heat o a sample, the more nluence t has on the nal temperature. Snce the sample wll change the temperature o the water less than t would have wth all the pennes, you wll underestmate ts specc heat. Q.. Reason: We ll use Equaton. Q Mc and solve or M. We are gven 0C 0 K and able.4 provdes c 490 J/kg K. 60 s Q energy power tme (00 W)( mn) 6000 J mn Q 6000 J M 0.07 kg 7 g c (490 J/kg K)(0 K) he correct answer s A. Assess: For the water (wth a large specc heat) to rse n temperature by that much n one mnute the mass must be pretty small (7 g o water has a volume o about / o a cup). o bol a lter (just over a quart, and almost 4 tmes the answer to ths queston) o water on your stove n just a ew mnutes you need to delver much more than 00 W to t. Q.. Reason: Assume that the beakers are well-nsulated. he specc heat o water s much hgher than the specc heat o alumnum, so more heat wll be taken rom the beaker wth the water added than the beaker wth the alumnum added. In the beaker wth ce added, the ce rst melts and then the meltwater wll have ts temperature rased. Snce the meltng takes extra heat, the second beaker wll end up wth the lowest temperature. he correct choce s B. Assess: hs s as expected rom experence.

hermal Propertes o Matter -7 Q.. Reason: o condense the steam to lqud at 00C would requre extractng an amount o energy o 5 Qcond MLv (0. kg)(.60 J/kg) 6000 J. 5 he amount o energy requred to melt the ce s Qmelt ML (0. kg)(.0 J/kg) 00 J. he amount o energy requred to rase the temperature o the melted ce (now a lqud) rom 0C to 00C s Q Mc (0.kg)(490 J/kg K)(00 K) 4900 J. he steam has more than enough energy to melt the ce and rase ts temperature by 00C. Some o the steam would not even need to condense to a lqud; t would stay as steam. he correct answer s E. Assess: he key to ths queston s the large derence between L v and L or water. Had the derence been so large the other way the answer would have been A. Q.4. Reason: he amount o heat needed can be calculated wth Equaton.. he heat the mcrowave oven provdes s gven by the power absorbed multpled by the tme t has been on, Q Pt. Settng ths equal to the heat needed to rase the water s temperature and solvng or the tme, he correct choce s B. Assess: hs makes sense, rom common experence. Mc (0.5 kg)(490 J/kg K)(60 K) t 00 s P (600 W) Q.5. Reason: some o the ce. We ll rst brng the ce up to 0C rom 0C and then see how much energy s let to melt Q Mcce (.00 kg)(090 J/kg K)(0K) 0900 J Now we subtract: 40000J 0900J 900J; ths s the energy let to melt the ce. Q 900 J M 0.0574 kg L 5. 0 J/kg he correct answer s B. Assess: here was enough energy to rase the water to 0, but not enough to melt t all. Q.6. Reason: Choce A can t be rght because the queston explctly states both the steam and lqud water are at 00C. But as the steam condenses on the skn t transers a lot o heat to the skn; ths causes the more severe burn. he correct answer s B. Assess: he specc heat o steam s actually less than that o lqud water. Problems P.. Prepare: For each element, one mole o atoms has a mass n grams equal to the atomc mass number; or example, snce the atomc mass number o carbon s then there s one mole o carbon atoms n grams; lkewse, there s one mole o argon atoms n 40 grams o argon. See able.. he only catch s that hydrogen gas s datomc, so one mole o datomc hydrogen gas molecules has a mass o g. For hydrogen: 0 g mol/ g 5 mol For carbon: 00 g mol/ g 8. mol For lead: 50 g mol/07 g 0.4 mol he answer s that 00 g o carbon has the most moles. Assess: Notce that we count atoms or the solds, but molecules or the datomc gas. P.. Prepare: oxygen atoms. A water molecule has a sngle oxygen atom, whle an oxygen molecule s made up o two

-8 Chapter A sngle mole o oxygen gas wll contan two moles o oxygen atoms, snce there are two oxygen atoms per molecule. here s one oxygen atom per molecule o water. In order to have the same number o oxygen atoms n the water as n one mole o gas, there must be two moles o water. A molecule o water conssts o two hydrogen atoms and one oxygen atom and has a mass ou u 6 u 8 u. wo moles o water has a mass o m 7 (8 u)(.66 0 kg/u)( mol)(6.0 0 mol ).60 0 kg he mass o water requred s 6.0 g. Assess: Note that snce an oxygen molecule has a mass o 6 u 6 u u, the mass o one mole o oxygen s a lttle less than the mass o water requred. P.. Prepare: We ll rst compute how many moles o hydrogen peroxde molecules there are n 00 g and then use Equaton. to nd how many partcles that s. he molecular mass number or HO s 6 4. (00 g)( mol/4 g).94 mol o hydrogen peroxde molecules. However, there are two hydrogen atoms n each molecule o hydrogen peroxde, so there are.94 mol 5.88 mol o hydrogen atoms. N nn A (5.88 mol)(6.00 mol ).54 0 4 Assess: hree trllon trllon hydrogen atoms s a lot, but one gets used to huge numbers n these types o problems. P.4. Prepare: One lter s 000 cubc centmeters. 0 mm 0 mm 6 L 000 cm 000 cm 0 mm cm cm Assess: Note that the entre converson actor must be cubed. P.5. Prepare: he volume s clearly the product o the three length measurements; the ssue s convertng the unts. Frst multply LW H to get the number o cm, then convert to m. V (00 cm)(40 cm)(.0 cm) 4,000 cm Now remember that whlem 00 cm, m 00 cm. Instead, m,000,000 cm. m 4,000 cm (4,000 cm ) 0.04 m,000,000 cm Assess: he answer s small not a very bg racton o one cubc meter; however, ths s reasonable gven the small heght. he converson actor comes rom (m/00 cm). P.6. Prepare: We can solve ths problem usng the deal gas law. But rst we need to nd the number o moles n.0 kg o dry ce. Snce carbon has an atomc mass o and oxygen has an atomc mass o 6, the molecular mass o CO s 44. hus mole o CO has a mass o 44 g. Fnally,.0 kg o dry ce s (000 g)( mol/44 g).7 mol. We are also gven the temperature: 0 9 K and we wll use 5 P atm.0 0 Pa. Solvng Equaton. or V, we have: nr (.7 mol)(8.5 J/mol K)(9 K) V 0.55 m 5 P.00 Pa Assess: From the chapter, we know that mole o gas at SP has a volume o about 0 L. he condtons o ths problem are not SP but t s a good approxmaton. We would expect 0 moles to have a volume o about 400 L. hs s close to the value we obtaned, whch converts to 550 L. P.7. Prepare: he absolute pressure s the gauge pressure plus one atmosphere at sea level. atm 4.7ps.

hermal Propertes o Matter -9 p p g atm 5.0 ps 4.7 ps 49.7 ps Assess: he derence between p and p g s due to the act that your tre gauge measures pressure derences. P.8. Prepare: We ll assume that ar s an deal gas so we can use the deal gas law, nr. We are gvenv 5.0 L 0.0050 m, patm 0. kpa, and 7C 0 K. Also recall that R 8.J/(molK) and oxygen makes up 0% o the ar. Solve Equaton. or n, the number o moles o ar. (0. kpa)(0.0050 m ) n 0.0 mol R (8.J/(mol K))(0 K) Multply the number o moles o ar by 0% to get the number o moles o oxygen: (0.0 mol)(0.0) 0.040 mol o oxygen. Assess: he answer s a small number o moles o oxygen, but a large number o molecules o oxygen. P.9. Prepare: Equaton.9 gves the orce due to a pressure appled over an area. he prelmnary calculaton s to compute the cross secton area o the tube. We are gven p 6.0 kpa. 0.05 m 4 A R.770 m F pa 4 (6.0 kpa)(.77 0 m ). N Assess:. N s not a large orce, but t s pushng a lght dart, so the dart acheves a respectable acceleraton. P.0. Prepare: Equaton.9 gves the orce due to a pressure appled over an area. he prelmnary calculaton s to compute the cross-secton area o the tube. We are gven p 45 kpa. 0.0084 m 5 A R 5.540 m F pa 5 (45 kpa)(5.54 0 m ).5 N Assess:.5 N s not a large orce, but t s pushng on delcate tssue, so t pays to be careul. F PA (45,000 Pa) (.004 m).5 N P.. Prepare: In order to use the deal gas law (Equaton.) we need to know the number o helum atoms n the gas. N nn A (7.5 mol)(6.00 mol ) 4.5 0 m V 5L 0.05m 000L 4 As a urther prelmnary calculaton add atm to the gauge pressure to gve the absolute pressure and convert the pressure to SI unts. atm 0. kpa p pg atm 65 ps 4.7 ps 79.7 ps 549 kpa 4.7 ps atm (a) Solve Equaton. or.

-0 Chapter (b) Now use Equaton.5 or K (549 kpa)(0.05 m ) K 05 C 4 Nk (4.50 )(.80 J/K) B ave. K ave (.8 0 J/K)( K).7 0 k J B Assess: he answer to part (a) s a cold temperature, but t needs to be to get that much gas n that volume. P.. Prepare: We need the mass o a mole o CO. Snce carbon has an atomc mass o and oxygen has an atomc mass o 6, the molecular mass o CO s 44. Hence a mole o CO has a mass o 44 g or 0.044 kg. We use Equaton.0 or rms speed but mody t by multplyng numerator and denomnator by Avogadro s number. v k m R m rms / / ( mol) (8.5 J/(mol K))(0 K)/(44 0 kg/mol) 50 m/s Assess: hs s a typcal speed or a gas molecule wth a temperature n the hundreds o Kelvns. For example, the rms speed o an O molecule at room temperature s about 480 m/s. P.. Prepare: Equaton. apples. We must convert all quanttes to SI unts. Convertng unts, Usng Equaton., 0 7 5 K 0 m V (.0 L).00 m L nr (.0 mol)(8. J/(mol K))(5 K) p V.00 m Assess: Note that ths s about twenty tmes atmospherc pressure. 6.9 0 Pa P.4. Prepare: he atomc mass o a ntrogen molecule s 8.0 so mole o ntrogen gas has a mass o 8.0 g or 0.080 kg. We are gven the rms speed o the ntrogen gas: 65 m/s. We solve Equaton.0 or to obtan the ollowng: hs converts to (78.8 7.5) C 94 C. mvrms mmolvrms (0.080 kg)(65 m/s) 78.8 K k R (8.5 J/mol K) B Assess: It s remarkable that at such a low temperature, the molecules are stll movng as ast as a jet. One way to put ths n perspectve s to see that, n Kelvns, ths temperature s about one-ourth o room temperature. Consequently, rom Equaton.0, the rms speed o the molecules wll be about hal that at room temperature. Indeed, the rms speed o N molecules at 0 C s about 50 m/s or around twce the speed o the jet. P.5. Prepare: We assume that the steam s an deal gas and solve the deal gas law or V. We are gven 00C 7 K and p 0. kpa, but we don t yet know n. In the sealed bag the number o water molecules won t change as the lqud s boled nto steam, so we ll compute the number o water molecules (wth molecular mass number 8) n 0 g. n (0 g)(mol/8 g) 0.556 mol

hermal Propertes o Matter - nr (0.556 mol)(8. J/(mol K))(7 K) V p 0. kpa 0.07 m 7 L Assess: he answer (over hal a cubc oot) s 7,000 tmes the orgnal volume o the lqud water. It s a good thng the bag was very lexble. he lesson s that a gas occupes a whole lot more volume than the same mass o lqud. P.6. Prepare: he gas s assumed to be deal. As a general rule, we must convert all quanttes nto SI unts. In the present case, however, we wll be dealng wth the rato o the nal and the ntal value o V, so we do not have to convert L nto m. he beore-and-ater relatonshp o an deal gas s V.0 L 600 K p p (.4 atm).6 atm V 9.0 L 00 K P.7. Prepare: he gas s assumed to be deal and t expands sothermally. (a) Isothermal expanson means the temperature stays unchanged. hat s. (b) he beore-and-ater relatonshp o an deal gas under sothermal condtons s p V p V p V p p V V Assess: he gas has a lower pressure at the larger volume, as we would expect. P.8. Prepare: In an sochorc process, the volume o the contaner stays unchanged. Argon gas n the contaner s assumed to be an deal gas. We must rst convert the volumes and temperatures to SI unts wth V 50 cm 50 0 6 m, 0C (7 + 0)9 K, and 00C (00 7)K 57 K. (a) he contaner has only argon nsde wth n 0. mol. he pressure beore heatng s nr (0.0 mol)(8. J/(mol K))(9 K) p 6 V 500 m 6 4.87 0 Pa 4870 kpa An deal gas process has p V / p V /. Isochorc heatng to a nal temperature has V V, so the nal pressure s (b) V 57 K p p 4870 kpa 9500 kpa 9 K V Assess: Note that t s essental to express temperatures n Kelvns. Increase n temperature at a constant volume leads to ncreased pressure, as would be expected. P.9. Prepare: he sobarc heatng means that the pressure o the argon gas stays unchanged. Argon gas n the contaner s assumed to be an deal gas. We must rst convert the volumes and temperatures to SI unts wth V 50 cm 50 0 6 m, 0C = (7 + 0)K9 K, and 00C (00 7)K 57 K. (a) he contaner has only argon nsde wth n 0.0 mol. hs produces a pressure nr (0.0 mol)(8. J/(mol K))(9 K) p 6 V 500 m 6 4.87 0 Pa 4870 kpa

- Chapter An deal gas process has p V / p V /. Isobarc heatng to a nal temperature 00C 57 K has p p, so the nal volume s (b) p 57 V V 50 cm 97.8 cm p 9 P.0. Prepare: In an sothermal expanson, the temperature stays the same. he argon gas n the contaner s assumed to be an deal gas. We must rst convert the volumes and temperature to SI unts: V 50 cm 50 0 6 m, V 00 cm 00 0 6 m, 0C = (7 + 0)K9 K. (a) he contaner has only argon nsde wth n 0.0 mol, V 50 cm 50 0 6 m, and 0 C 9 K. hs produces a pressure nr (0.0 mol)(8. J/(mol K))(9 K) 6 p 4.870 Pa 48.07 atm 6 V 500 Pa An deal gas process obeys p V / p V /. Isothermal expanson to V 00 cm gves a nal pressure (b) V 50 p p 48.07 atm atm 00 V P.. Prepare: Assume the gas to be an deal gas. Please reer to Fgure P.. We wll make use o the ollowng conversons: atm.0 0 5 Pa and cm 0 6 m. (a) Because the volume stays unchanged, the process s sochorc. (b) he deal-gas law nr gves 5 6 (.00 Pa)(00 0 m ) nr (0.0040 mol)(8. J/(mol K)) he nal temperature s calculated as ollows or an sochorc process: p p 94 K p atm (94 K) 00 K p atm P.. Prepare: Assume that the gas s an deal gas. Please reer to Fgure P.. We wll make use o the ollowng conversons: atm.0 0 5 Pa and cm 0 6 m. (a) he graph shows that the pressure s nversely proportonal to the volume. he process s sothermal.

hermal Propertes o Matter - (b) From the deal-gas law, 5 6 (.00 Pa)(00 0 m ) nr (0.0040 mol)(8. J/(mol K)) 94 K s also 94 K, because the process s sothermal. (c) he beore-and-ater relatonshp o an deal gas under sothermal condtons s p atm V V (00 cm ) 00 cm p atm P.. Prepare: Assume that the gas s deal. Please reer to Fgure P.. We wll make use o the ollowng conversons: atm.0 0 5 Pa and cm 0 6 m. (a) Because the process s at a constant pressure, t s sobarc. (b) For an deal gas at constant pressure, V V (c) Usng the deal-gas law p V nr, n V 00 cm 9 K V [(7 900) K] 00 cm 5 6 (.00 Pa)(00 0 m ) 9.40 mol R (8.J/(mol K))(9K) P.4. Prepare: Please reer to Fgure P.4. Snce the process s occurrng at constant pressure, t s sobarc and the work done by a gas s the area under the p-versus-v curve or gven by Equaton.7. he gas s compressng, so we expect the work by the gas to be negatve or the work on the gas to be postve. he work done by the gas s W 6 5 (area under the curve) (00 cm )(00 kpa) (00 0 m )(.0 0 Pa) 40 J hus the work done on the gas s 40 J. Assess: he area under the curve s negatve because the gas s compressed. hus, the envronment does postve work on the gas to compress t. P.5. Prepare: For a gas n a sealed contaner (n s constant) we use Equaton.4. In ths case we want to solve or, p V but we get to cancel the volumes snce they are the same. We need to take the usual steps o convertng temperatures to the absolute scale and pressures rom gauge pressures to absolute pressures. 0.00C 7 K p 55.9 kpa 0. kpa 57. KPa p 65. kpa 0. kpa 66.4 KPa p p 57. KPa 66.4 KPa 7 K 89 K 6 C Assess: We expected the nal temperature to be hgher than the ntal temperature as the pressure rose. he answer s a reasonable real-le temperature.

-4 Chapter P.6. Prepare: We are gven the volume and gauge pressure at the bottom o the sea: b g Vb.0 cm and p.5 atm. he absolute pressure s: pb pb g atm.5 atm. Snce the process s sothermal, rom the deal gas law, s constant. When the bubble reaches the surace, ts pressure s atm. Snce s constant, we can equate ts value at the bottom o the sea wth ts value at the top as ollows: (.5 atm)(.0 cm ) (.0 atm)v t hs equatons can be solved to yeld: Vt.5 cm. (b) As the bubble rses t expands. I no heat were exchanged wth the sea, t would cool by adabatc expanson. Snce t does not cool down, we know heat s lowng nto the bubble. Assess: We can understand why heat lows nto the bubble we realze that as the bubble expands t does work on ts surroundngs and does ths work at the expense o ts thermal energy. he sea responds to the loss n thermal energy o the bubble by gvng t heat. You can magne the bubble dong an nntesmal amount o work, whch causes ts temperature to decrease nntesmally, whch n turn causes heat to low rom the sea to the slghtly cooler bubble untl they have the same temperature agan. P.7. Prepare: We are gven the ntal volume, temperature and pressure o the gas; 0 C 9 K and p.0 atm, as well as the nal volume and temperature; 0 C 6 K. We can use the deal gas law. that We can rewrte the deal gas law as s constant, that s, hs equaton, n turn, can be solved or p. V 4.0 m, V mand nr. Snce the number o moles o gas s constant, t ollows p V (.0 atm)(4.0 m )(6 K) V ( m )(9 K) p 0.0 atm Assess: hs s n agreement wth what we know, that pressure decreases wth ncreasng alttude. P.8. Prepare: Equaton.9 apples. able. has coecents o expanson or varous materals. Note that the alumnum and the steel have derent coecents o lnear thermal expanson. he total expanson n the rod wll be the expanson n the steel added to the expanson o the alumnum. L L L ( L ) ( L ) alumnum steel steel steel alumnum alumnum 6 6 5 ( 0 K )(0.00 m)(0 K) ( 0 K )(0.0 m)(0 K).0 0 m Assess: hs s a very small expanson, as expected. P.9. Prepare: We are gven much o the data needed n Equaton.9 except the coecent o lnear 6 expanson or steel, whch we look up n able.. L 0.7 mm, K, 0 K. steel

hermal Propertes o Matter -5 We want to know the orgnal length, so we solve Equaton.9 or that quantty. L (0.7 mm) m L 4.7 m 6 (0 K )( K) 000 mm Assess: hs seems to be a reasonable answer n the realm o daly le, about the wdth o a room. It would have taken an alumnum beam only about hal that long to produce the same L under the same. P.0. Prepare: he gaps must be at least the length o the thermal expanson n the rals. Equaton.9 and able. apply. he expanson o a sngle ral durng ths temperature change s L L 6 ( 0 K)( m)(50 K 6 K) 4.9 0 m he rals should be lad wth gaps o about hal a centmeter. Assess: hs result makes sense. Gaps on real tracks are about ths sze or larger. P.. Prepare: We need to rearrange Equaton. to gve the ractonal volume expanson (the percentage expanson s then just 00 tmes that ractonal expanson). 6 Look up Al 690 K n able.. 0 K. V V 6 (69 0 K )(0 K) 0.008 0.8% Assess: he expanson s small, but greater than t would have been or steel. Whle 0C s a larger temperature swng than we mght see on a daly bass, there are stuatons (engnes, etc.) where there are sgncant temperature derences, and people who desgn and buld precse thngs must take such expansons nto account. P.. Prepare: he heat needed to change an object s temperature by s Q Mc. he mass o the ce cube s 0.00 kg and ts specc heat rom able.4 s c ce 090 J/(kg K). Q (0.00 kg)(090 J/(kg K))(4 K 7 K),500 J hus, the energy removed rom the ce block s,500 J. Assess: he negatve sgn wth Q means loss o energy because removal o heat rom the ce reduces ts thermal energy and ts temperature. P.. Prepare: he mass o the mercury s M 0 g.0 0 kg, the specc heat rom able.4 c mercury 40 J/kg K, the bolng pont b 57C, and the heat o vaporzaton rom able.5 L V.96 0 5 J/kg. We wll use Equaton. or obtanng heat needed to rase ts temperature to bolng and Equaton.4 or obtanng heat needed to bol mercury nto vapors at the bolng temperature. Note that heatng the mercury at ts bolng pont changes ts thermal energy wthout a change n temperature. We also note that s the same whether we calculate t n Kelvns or n C, so we don t have to convert C nto K. he heat requred or the mercury to change to the vapor phase s the sum o two steps. he rst step s he second step s Q Mc mercury (.0 0 kg)(40 J/(kg K))(57C 0C) 944 J Q ML V (.0 0 kg)(.96 0 5 J/kg) 590 J he total heat needed s 6870 J, whch wll be reported as 6900 J. Assess: More energy s needed to vaporze mercury (86%) than to warm t to ts bolng temperature (4%), as we would expect. P.4. Prepare: he mass o the mercury s 0 g.0 0 kg and ts specc heat rom able.4 s c Hg 40 J/kg K. he mass o the water s 0 g.0 0 kg and ts specc heat rom able.4 s c water 490 J/kg K. We wll use Equaton. to obtan the needed heats. (a) he heat needed to change the mercury s temperature s

-6 Chapter Q 00 J Q McHg 5.7 K 5.7C, whch wll be reported as 6C. Mc (0.00 kg)(40 J/(kg K)) Hg (b) he amount o heat requred to rase the temperature o the same amount o water by the same number o degrees s Q Mc water (0.00 kg)(490 J/(kg K))(5.7 K) 000 J Assess: Q s drectly proportonal to c water and the specc heat or water s much hgher than the specc heat or mercury. hs explans why Q water Q mercury. P.5. Prepare: We ll compute the energy necessary to evaporate.5 L and then dvde by an hour to get the rate. One lter o water has a mass o one klogram. he rate s the energy dvded by the tme. 5 6 Q ML (.5 kg)(40 J/K) 8.4 0 J 6 Q 8.40 J h rate 00 W t h 600 s Assess: hat s an mpressve power output, but necessary to keep cool n tropcal clmates. Notce that the value gven or L at body temperature s derent rom the one gven n able.5 or standard temperature (0C). P.6. Prepare: We are asked to nd the tme t takes or the allgator s temperature to rse rom 5 C to 0 C, that s by 5 K. It s absorbed at a rate o 00 W. We use the mammalan specc heat: c 400 J/(kg K). We wll need to use the denton o power, whch n ths case can be wrtten as P Q / t, as well as the ormula or heat absorbed n connecton wth an ncrease n temperature, Q mc. he tme needed or the allgator to reach ts nal temperature can be obtaned by solvng the power equaton: t Q / P. hs, coupled wth the ormula or heat gves us the ollowng: Q mc (00 kg)(400 J/(kg K))(5 K) t 5,500 s P P 00 W hs works out to a tme o (5,500 s)( hr)/(600 s) or 7 hr. Assess: hs seems lke a long tme. But consder that a cup o water o about / 5 kg takes about mn to heat n a 600 W mcrowave oven. he allgator s about 500 tmes more massve and the temperature ncrease o the allgator s about 5 tmes less. So we mght expect the allgator to take around 00 mn or about hr to heat up. he derence between and the actual value o 7 comes rom the derent wattages n the two cases as well as the derent specc heats. P.7. Prepare: Water n the body s converted to water vapor. Equaton. apples. Each breath converts 5 mg o water to water vapor. he heat requred or ths s 5 5 Q ML v (.50 kg)(40 J/kg) 60 J At breaths/mn, there are 0. breaths/s. Multplyng by the above result o 60 J/breath we obtan that P J/s or P W s the rate o heat loss. Convertng to Calores/day, we have P Calore 60 s 60 mn 4 h W 50 Calore 490 J mn h d Assess: hs seems reasonable snce t s a small racton o a person s daly calorc ntake o about 000 calores. P.8. Prepare: We need to nd the temperature ncrease and we are gven the metabolc power P 000 W, the tme o exercse t 0 mn (0 mn)(60 s)/( mn) 800 s, and the man s mass

hermal Propertes o Matter -7 m 70 kg. We need the ormula or power, whch n ths case s P E / t th and Equaton.0 or heat absorbed. Here the heat produced by the exercse goes to ncreasng the thermal energy o the man, so we can wrte the ollowng: Eth mc. For the specc heat, we use the mammalan specc heat rom able.4: c 400 J/kg K. Frst we wll nd the thermal energy produced by the exercse by solvng the power equaton. We can solve the equaton th E 6 th P t (000 W)(800 s).8 0 J E mc or. 6 (.8 0 J) Eth / mc 7.56 K (70 kg)(400 J/(kg K)) o two sgncant gures, hs temperature ncreases by 7.6 K. Assess: Even though he only exercsed or 0 mn, the man s temperature has ncreased by 7.6 C or 9F (7.56 C) 4 F. hs would brng a temperature o 98.6 F up to about F, whch would be very dangerous. 5C P.9. Prepare: he water rst needs to be rased rom 0C to 00C (assumng normal atmospherc pressure), and then changed rom lqud to gas at the bolng temperature. We ll use Equaton. Q Mc or the rst term and Equaton. Q ML or the second. We look up the specc heat o water n able.4 as ollows: c 490 J/(kg K). We look up the heat o vaporzaton o water n able.5: L v.6 0 5 J/kg. We are gven M 0.0 kg. 00 0 80 K. Q Mc ML 5 v (0.0 kg)(490 J/(kg K))(80 K) (0.0 kg)(.6 0 J/kg),000 J 0,000 J 60,000 J Assess: hat s qute a ew joules, but notce that the vast majorty s needed to change the water rom lqud to sold rather than rase t rom 0C to 00C. Pay attenton to the sgncant gures. he terms that have actors multpled together each end up wth two sgncant gures, but when those terms are added, the ten-thousand s place s the last sgncant gure (see actcs Box., rule ). P.40. Prepare: We have a thermal nteracton between the copper pellets and the water. he ntal temperatures o both copper pellets and the water are known and we denote the common nal temperature by. he specc heats o copper and water rom able.4 are as ollows: c c 85 J/(kg K) and c w 490 J/ (kg K). Whle the mass o copper pellets s M c 0.00 kg, the mass o water s obtaned rom M w V: M w (00 ml)(0 6 m / ml) (000 kg/m ) 0.0 kg. We can determne the common nal temperature usng Equatons. and.5. he conservaton o energy equaton Q c Q w 0 s M c ( 00C) M c ( 0C) 0 J c c w w Solvng ths equaton or the nal temperature gves M c (00C) M c (0C) (0.00 kg)(85 J/(kg K))(00C) (0.0 kg)(490 J/(kg K)(0C) c c w w Mccc M wcw (0.00 kg)(85 J/(kg K) (0.0 kg)(490 J/(kg K) 8 C he nal temperature o the water and the copper s 8 C. Assess: Due to the large specc heat o water compared to copper and three tmes more water n our system, we expected a small temperature ncrease.

-8 Chapter P.4. Prepare: A thermal nteracton between the copper block and the water leads to a common nal temperature denoted by. he ntal temperatures o both the copper block and the water are known. he specc heats o copper and water rom able.4 are as ollows: c c 85 J/(kg K) and c w 490 J/(kg K). Whle the mass o the water s known, we can determne the mass o the copper block usng Equatons. and.5. he conservaton o energy equaton Q copper Q water 0 J s M copperc copper ( copper) M waterc water( water ) 0 J Both the copper and the water reach the common nal temperature 5.5C. hus M copper (85 J/(kg K)(5.5C 00C).0 kg(490 J/(kg K))(5.5C 0C) 0 J Mcopper 0.0 kg Assess: Due to the large specc heat o water compared to copper, a smaller value obtaned or the mass o the copper block s reasonable. P.4. Prepare: We have a thermal nteracton between the alumnum pan and the water that leads to the common nal temperature denoted by. he ntal temperature o the alumnum pan s unknown, but that o the water s known. he specc heats o alumnum and water rom able.4 are c Al 900 J/(kg K) and c w 490 J/(kg K). he mass o the alumnum pan s M Al 0.750 kg and the mass o water s M water 0 kg. We can determne the ntal temperature o the pan usng Equatons. and.5. he conservaton o energy equaton Q Al Q water 0 J s M Al c Al ( Al ) M water c water ( water ) he pan and water reach a common nal temperature 4.0C (0.750 kg)(900 J/(kg K))(4.0C Al) (0 kg/m )(490 J/(kg K))(4.0C 0.0C) (675.0 J/K)(4.0C ) 67,600 J 0 J 70C Al Assess: Due to () the large specc heat o water compared to alumnum and () the large mass o water compared to alumnum, an ncrease o temperature o 4C needed hgh ntal temperature o the pan. P.4. Prepare: We have a thermal nteracton between the metal sphere and the mercury that leads to the common nal temperature 99.0C. he ntal temperatures o the metal sphere and the mercury are known. he specc heat o mercury rom able.4 s c Hg 40 J/(kg K). he mass o the metal sphere s M metal 0.500 kg and the mass o water s M Hg 4.08 kg. Our strategy s to determne c metal usng Equatons. and.5. he conservaton o energy equaton Q metal Q Hg 0 J s M metalc metal( metal ) M Hg c Hg ( Hg ) 0 J (0.500 kg)c metal(99c 00C) (4.08 kg)(40 J/(kg K))(99C 0C) 0 J We nd that cmetal 449 J/(kg K). he metal s ron. P.44. Prepare: hs s a basc calormetry problem. We are gven the mass and ntal temperature o the alumnum mal 00 g and Al 0 C, as well as the mass and ntal temperature o the coee mc 500 g and c 85 C. We obtan the specc heats rom able.4, accordng to whch cal 900 J/(kg K) and c 490 J/(kg K). c Al he basc calormetry equaton becomes Al Al Al c c c 0 m c m c whch we can solve or the nal temperature:

hermal Propertes o Matter -9 (0.00 kg)(900 J/kg)( 0 C) (0.500 kg)(490 J/(kg K))( 85 C) 0 J (80 J/K) 74000 J 0 76 C Assess: Notce that even though the alumnum was comparable n mass to the coee, ts temperature change o 96 C was ar more than the temperature change o the coee, 9 C. hs s a result o water havng a much hgher specc heat than metal. P.45. Prepare: Problem-Solvng Strategy. wll be useul. Assume the heat s transerred entrely rom the coee to the ce snce the contaner s well-nsulated. All the ce melts and s converted to water at 0C, otherwse the letover ce would contnue to cool the coee. he coee transers addtonal heat to the water to rase ts temperature to 0C. he ce s at ts meltng temperature when t s put n the coee. he heat requred to melt the ce s Q melt M ce L. Once the ce melts, heat goes nto the water that results n rasng ts temperature to 0C. he heat requred s Qwater Mcec water( ) Mcec water( 0C) Mcec water. he temperature o the coee decreases as a result o the ce meltng and the heatng o the resultng water. he mass o the coee s Mcoee V. he densty o the coee s 000 kg/m. he volume o the coee s m V 00 ml (000 L).000 m 000 L 4 hree sgncant gures have been assumed n the volume o coee. Energy conservaton gves M L M c Vc ( ) 0 ce ce water water Where s the ntal temperature o the coee. Solvng or the mass o ce requred, M 4 Vcwater( ) (000 kg/m )(.000 m )(490 J/(kg K))( 50 K) ce 5 L cwater (.0 J/kg)+(490 J/(kg K))(0 K) 9. 0 kg It takes 9 g o ce to cool the coee. Assess: About hal the mass o the coee n ce has to be added to cool the coee, as we mght expect. P..46. Prepare: Heat wll leave the patent and enter the ce, meltng t. We need the ormula or heat absorbed n meltng Q m L, and the ormula or the heat lost by the patent ce ce Q mpcp. For the specc heat, we wll use the value or mammals rom able.4, cp 400 J/kg K. We need to solve or m. ce he calormetry equaton s as ollows: m m 5 ce(. 0 J/kg) (60 kg)(400 J/(kg K))(9 C 40 C) 0 J (400 J/(kg K)) (60 kg)( K) 0.6 kg 60 g (.0 J/kg) ce 5 It takes 60 g o ce to reduce the ever by C. Assess: Snce the rato o the mammalan specc heat to the latent heat o uson o ce s about p K, 00 ollows that a rule o thumb s that the amount o ce needed to cool a person by C s one hundredth o ther body mass. hs s seen to be the case here. P.47. Prepare: We wll use Equatons.6 and.7 to nd the heat needed at constant volume and constant pressure. Snce these equatons nvolve the number o moles o the gas, we wll calculate t rom the mass o the gas and ts molar mass. From able.6, C P 0.8 J/(mol K) and C V.5 J/(mol K). Note that the change n temperature on the Kelvn scale s the same as the change n temperature on the Celsus scale. (a) he atomc mass number o argon s 40. hat s, M mol 40 g/mol. he number o moles o argon gas n the contaner s t

-0 Chapter he amount o heat s whch wll be reported as J. (b) For the sobarc process Q nc P becomes M.0 g n 0.05 mol M 40 g/mol mol Q nc V (0.05 mol)(.5 J/(mol K))(00C).5 J.5 J (0.05 mol)(0.8 J/(mol K)) 60 C P.48. Prepare: he heatng processes are sobarc (n part (a)) and sochorc (n part (b)). O s a datomc deal gas. We wll use Equatons.6 and.7 to nd the heat needed at constant volume and constant pressure. Snce these equatons nvolve the number o moles o the gas, we wll calculate t rom the mass o the gas and ts molar mass. From able.6, C P 9. J/(mol K) and C V 0.9 J/(mol K). Note that the change n temperature on the Kelvn scale s the same as the change n temperature on the Celsus scale. (a) he number o moles o oxygen s For the sobarc process, whch wll be reported as 9 J. (b) For the sochorc process, M.0 g n 0.05 mol M g/mol mol Q nc P (0.05 mol)(9. J/(mol K))(00C) 9. J Q nc V 9. J (0.05 mol)(0.9 J/(mol K)) 40 C P.49. Prepare: From the rst law o thermodynamcs, Q E th W, where W s the work done by the gas. W 0 at constant volume, so, usng Equaton.6, E th Q nc V. From able.6, the value o C V or a monatomc gas s.5 J/(mol K) (whch s equal to R/). For a datomc gas, we take C V to be 0.8 J/(mol K) (whch s equal to 5R/). For a monatomc gas, Eth ncv.0 J (.0 mol)(.5 J/(mol K)) 0.0800 C or 0.080 K P.50. Prepare: he heatng s an sochorc process. From Equaton.6, Q nc V, and rom able.6, the value o C V or oxygen s 0.9 J/(mol K) and or helum s.5 J/(mol K). he number o moles n o each gas s obtaned rom the mass o the gas and ts molar mass. he number o moles o helum s For the sochorc processes, QHe ncv (0.50 mol)(.5 J/ (mol K )) Because Q Q, He O M.0 g n 0.50 mol M 4 g/mol mol M QO nc V (0.9 J/ (mol K )) g/mol M (0.50 mol)(.5 J/(mol K)) (0.9 J/(mol K)) M 9.6 g g/mol P.5. Prepare: Please reer to the ollowng gure. he work done by a gas s equal to the area under the graph between V and V. he work done by a gas s postve when V > 0, negatve when V < 0, and zero when V 0. W 0, W Area (I) Area (II), and W Area (II). hus, the net work done s equal to Area (I). hat s, the work done by the gas per cycle s the area nsde the closed p-versus-v curve. We also need to convert the unts o pressure rom atm to Pa usng the converson: atm.0 0 5 Pa.

hermal Propertes o Matter - W by gas he area nsde the trangle s 5 6 6.00 Pa 6 ( atm atm)(6000 m 000 m ) atm (4000 m ) 4 J atm P.5. Prepare: Please reer to the next gure. he work done by a gas s equal to the area under the graph between V and V. he work done by a gas s postve when V > 0, negatve when V < 0, and zero when V 0. W = I + III, W = II + IV, and W = III IV. hus, the net work done s equal to I + II. hat s, the work done by the gas per cycle s the area nsde the closed p-versus-v curve. We also need to convert the unts o pressure rom atm to Pa usng the converson atm.0 0 5 Pa and the unts o volume rom cm to m usng cm 0 6 m. he work done by the gas per cycle s the area enclosed wthn the curve. We have (60 J) 60 J ( p 00 kpa)(800 cm 00 cm ) 600 0 m max max 6 p.00 Pa p.00 Pa 00 kpa 5 5 max P.5. Prepare: he rate o conducton across a temperature derence s gven n Equaton.. Q ka t L where A 4.0 m5.5 m m s the area, L 0.08 m s the thckness o the loorng, and k 0.W/(m K) s the thermal conductvty o wood gven n able.7. 9.6C 6.C.4 C. Q ka t L 0.08 m (0. W/(m K))( m ) (.4 C) 80 J/s 80 W Assess: 80 W s about as much as a dozen ncandescent lght bulbs. In the wnter when you are tryng to keep the room warm ths energy s beng wasted; you could do drastc thngs lke ncrease the thckness o the wood,

- Chapter or smpler, cheaper thngs lke cover the loor wth carpet, whch has a much smaller k. In the summer you mght be grateul to have ths energy conducted rom the room the subloor can stay at a cooler temperature. P.54. Prepare: he bottom o the nteror o the kettle s the same temperature o the bolng water, 00C. Equaton. can be used. Solvng Equaton. or the temperature derence, Q L t ka (400 W/(m K)) (0. m) he bottom o the kettle s only 0. K hotter than the nteror. Assess: hs result makes sense..00 m (800 W). 0 K P.55. Prepare: he rate o energy loss by radaton s gven by Equaton.. Q e A 4 t We are gven e 0.0, 700C 97 K, A 6 (.0 cm.0 cm) 4 cm 0.004 m. he textbook gves Stean s constant as 5.67 0 8 W/(m K 4 ). Q t e A 4 8 4 4 (0.0)(5.67 0 W/(m K ))(0.004 m )(97 K) 4 W Assess: 700C s qute hot, so the cube radates a reasonable amount o energy, but even at 700C the radaton s mostly nrared, not vsble. I we double the (absolute) temperature the total radaton would ncrease by a actor o 6 (due to the 4 ) and also a greater porton o the radaton would be n the vsble range. P.56. Prepare: emssvty s.0. Q t Equaton. apples. For maxmum possble radated power, assume the e A 4 8 4 4 Assess: hs result makes sense or such a small sphere. (.0)(5.67 0 W/(m K ))4 (0.05 m) (7 K) 8.6 W P.57. Prepare: he rate o net energy loss by radaton s gven by Equaton.4. Qnet t e A( ) 4 4 0 where 0 s the termperature o the surroundngs. We are gven 0C 0 K, 0 0C 6 K, and A 0.00 m. We are told to assume the emssvty o seal skn s the same as human skn; the text gves ths vaule as e 0.97. 8 4 he textbook gves Stean s constant as 5.670 W/(m K ). Q t e A 0 (0.97)(5.670 W/(m K ))(0.00 m )[(0 K) (6 K) ] 6.0 W net 4 4 8 4 4 4 Assess: 6 W sn t a lot, but t s sucent to cool the seal when the surroundngs are very cool. I there were no thermal wndows the seal would have dculty regulatng ts temperature.

hermal Propertes o Matter - P.58. Prepare: o use the ormula or the rate o emsson o radated energy we need to know the area o the panels. Snce they emt radaton rom both sdes, the area s gven by the ollowng: A LW (.6 m)(.8 m).0 m. he temperature s 79 K. Snce these panels are desgned to emt radaton, they probably have an emssvty close to so we wll use e. We are ready to nd the power usng Equaton.. P Q t 4 8 4 4 e A ()(5.67 0 W/(m K ))(.0 m )(79 K) 4500 W Assess: hs seems reasonable snce ths s about ve tmes the power emtted by a typcal person, 900 W, and the plates have an area about seven tmes a typcal value or a person, m. he act that the power s not qute seven tmes as great s owng to the somewhat cooler temperature o the panels. P.59. Prepare: he rate o energy loss by radaton s gven by Equaton.. Q e A 4 t We are gven e 0., 500C 77 K, and Q/ t 60 W. We are asked to nd A. 8 4 he textbook gves Stean s constant as 5.670 W/(m K ). Solve the equaton or A. Q/ t 60 W A 4.70 m e 4 8 4 4 (0.)(5.670 W/(m K ))(77 K) 4 Assess: We knew that lght bulb laments have a small surace area, so we are not concerned to get a small answer. he unts work out properly. P.60. Prepare: Snce you are lyng on the ground, your back does not emt radaton or absorb radaton rom the sky. We mght guess that a lttle more than hal o a person s area s o the ground. A typcal person s surace area rom the book s.8 m. So we wll use m or the area n contact wth the ar. We wll use e 0.97. he temperature o your clothng s c 0 K and the temperature o the sky s s K. he net rate that your body loses energy to the sky s gven by Equaton.. Qnet t e A( ) (0.97)(5.67 0 W/(m K ))( m )((0 K) ( K) ) 00 W 4 4 8 4 4 4 c s Assess: hs s about three tmes the net rate o heat loss you would experence you were n a room at room temperature (see Example.). hs s true despte the lower surace area here. he bg derence s that heat s absorbed by your body much slower n ths case. P.6. Prepare: reat the gas n the sealed contaner as an deal gas and use Equaton.4. (a) From the deal gas law equaton nr, the volume V o the contaner s nr (.0 mol)(8. J/(mol K))[(7 0) K] V 0.050 m 5 p.00 Pa

-4 Chapter (b) he beore-and-ater relatonshp o an deal gas n a sealed contaner (constant volume) s p V (7 0) K p p (.0 atm). atm (7 0) K Assess: Note that gas-law calculatons must use n kelvns and pressure must be n Pa. P.6. Prepare: reat the ar n the compressed-ar tank as an deal gas and use Equaton. to nd n. We wll, however, need to convert pressure and temperature to SI unts usng atm.0 0 5 Pa and (n K) (n C) 7. Also note that r h s the volume o a cylnder. (a) From the deal-gas law nr, the number o moles n s 5 p( r h) (50 atm)(.0 0 Pa/atm)[ (0.075 m) (0.50 m)] n 55. mol R R (8.J/(mol K))[(7 0) K] whch wll be reported as 55 mol. (b) At SP, the deal-gas law yelds nr (55. mol)(8. J/(mol K))(7 K) V. m 5 p.00 Pa Assess: he volume o the compressed ar tank s (r )h 8.84 0 m. he volume at SP s 40 tmes the volume o the tank. hat s, the ar s compressed 40 tmes compared to SP values, whch does not look unreasonable. P.6. Prepare: s he gauge pressure o the gas s reat the helum gas n the sealed cylnder as an deal gas. he volume o the cylnder V r h (0.05 m) (0.0 m).56 0 m 5 atm.00 Pa 0 ps 8.69 0 5 Pa 4.7 ps atm so the absolute pressure o the gas s 8.69 0 5 Pa.0 0 5 Pa 9.8 0 5 Pa. he temperature o the gas s (7 0) K 9 K. (a) he number o moles o the gas n the cylnder s So, the number o atoms s (b) he mass o the helum s 5 (9.80 Pa)(.560 m ) n 0.898 mol R (8.J/(mol K))(9 K) N nn A (0.898 mol)(6.00 mol ) 5.4 0 atoms M nm mol (0.898 mol)(4 g/mol).6 g.59 0 kg P.64. Prepare: We are gven a mass o helum that s to be used to ll balloons o a known volume. We need to nd the mass o helum that goes nto a balloon. hs we can do by rst ndng the number o moles rom the deal gas law. For the temperature, we wll use 0 C or 9 K and or the pressure we wll use atm 5 or.0 0 Pa. We can rewrte the deal gas law as n / R to obtan the ollowng: 5 (.0 0 Pa)(0.00 m ) n 0.46 mol (8.5 J/(mol K))(9 K) Helum has an atomc mass o 4 so a mole o helum has a mass o 4 g. hus each balloon holds a mass o helum gven by the ollowng:

hermal Propertes o Matter -5 m (0.46 mol)(4 g/mol).66 g Fnally, the number o balloons that can be lled s ound by dvdng ths number nto the total mass o helum n the tank. 0 g N 8.66 g You can ll 8 balloons wth the helum. Assess: he value we got or the number o moles o helum, 0.4 mol, s reasonable snce a mole o deal gas at SP s about L and n ths problem we have 0 L o helum, or about hal as much. P.65. Prepare: We nd the heat whch must be suppled to the ar rom Equaton.5 or constant pressure processes. he value o C s gven n able.6: Cp 0.8 J/(mol K). he number o moles o ar p nhaled can be obtaned rom the deal gas law. We also nd the nal volume rom the deal gas law and subtract the ntal volume to obtan V. (a) Solvng the deal gas law or the number o moles usng the ntal volume and temperature, we obtan the ollowng: n 5 (.0 0 Pa)(4.00 m ) R (8.5 J/(mol K))(7 K) 0.4 mol he amount o heat whch must be suppled s Q ncp (0.4 mol)(9. J/(molK))(7 K) 44 J 40 J. s (b) From the deal gas law as stated n Equaton.4, the nal volume o the ar s gven by the ollowng: V V he ncrease n volume s V V V 0.4 L. (.0 L)(0 K) 7 K Assess: We could have estmated the change n volume usng Equaton.8, V V, but ths would not.4 L have been as accurate snce depends on temperature (t s gven or temperature vares rom 0C to 7 C. 0 C n able.) and here the P.66. Prepare: he carbon doxde n the cube s an deal gas and we wll use the deal gas Equaton. at SP wth n M/M mol. Usng the deal gas equaton nr MR nr V p pm he molar mass o CO s 44 g/mol or 0.044 kg/mol. hus, (0,000 kg)(8. J/(mol K))(7 K) V 5090 m 5 (.0 0 Pa)(0.044 kg/mol) he length o the cube s L (V) / 7 m. Assess: he length o the cube s large, but so s the mass o the carbon doxde emsson. P.67. Prepare: We wll use the deal gas Equaton.4 and assume that the volume o the tre and that o the ar n the tre s constant. hat s, the gas undergoes an sochorc (constant-volume) process. Because the gas equaton needs absolute rather than gauge pressure, a gauge pressure o 0 ps corresponds to an absolute pressure o (0 ps) (4.7 ps) 44.7 ps. Usng the beore-and-ater relatonshp o an deal gas or an sochorc process, mol p p 7 45 (44.7 ps) 49.4 ps p p 7 5

-6 Chapter Your tre gauge wll read a gauge pressure p 49.4 ps 4.7 ps 4.7 ps, whch s to be reported as 5 ps. Assess: A 5 ps ncrease n gauge pressure due to an ncrease n temperature by 0C s reasonable. P.68. Prepare: For a gas n a sealed contaner (n s constant) we use Equaton.4. p V In our constant volume case V and V cancel. he 5 ps s the gauge pressure. We take the usual steps o convertng the gauge pressure to absolute pressure and convertng the temperatures to absolute temperatures. Solve the top equaton or p. p p 5 ps 4.7 ps 49.7 ps 0C 9 K 0C 7 K 7 K 49.7 ps 46. ps p 9 K hs answer s the absolute nal pressure. o report the answer as a gauge pressure, we subtract atm 4.7 ps. 46.ps 4.7ps =.6ps, whch should be reported to two sgncant gures as ps. Assess: We expected the pressure to decrease wth decreasng temperature. On the absolute scale the temperature ddn t decrease a lot, however, so nether dd the pressure. We dd not need to convert all the pressure data to SI unts because we wanted the answer n the same unts as the orgnal data, and the ormula shows the rato n any unts (usng absolute pressure, not gauge pressure) s the rato o the absolute temperatures. P.69. Prepare: he gas s pressure does not change, so ths s an sobarc process. We wll use Equaton.4 wth p p. he trple pont o water s 0.0C or 7.6 K, so 7.6 K. Because the pressure s a constant, V 68 ml (7.6 K) V 000 ml V V 447.44 K 74C P.70. Prepare: Assume that the compressed ar n the cylnder s an deal gas. he volume o the ar n the cylnder s a constant. We wll use Equaton.4 to calculate the new pressure n atm and compare t wth the maxmum pressure (n atm) o the compressed gas that the cylnder can wthstand. Usng the beore-and-ater relatonshp o an deal gas, p V V K p p (5 atm) V V 9K V 04 atm where we have converted to the Kelvn temperature scale. Because the pressure does not exceed 0 atm, the compressed ar cylnder does not blow. P.7. Prepare: he rgd sphere s volume does not change, so ths s an sochorc process. he ar s assumed to be an deal gas accordng to Equaton.4. (a) When the valve s closed, the ar nsde s at p atm and 00C. he beore-and-ater relatonshp o an deal gas n the closed sphere (constant volume) s (7 0) K p p (.0 atm) 0.7 atm (7 00) K (b) Dry ce s CO. Coolng the sphere to 78.5C gves (7 78.5) K p p (.0 atm) 0.5 atm 7 K

hermal Propertes o Matter -7 Assess: Under constant-volume condtons, pressure decreases wth decrease n temperature, as we would have expected. (b) he nal pressure s 0.5 atm. P.7. Prepare: Please reer to Fgure P.7. Snce the process s occurrng at constant pressure, t s sobarc and the work done by a gas s the area under the p-versus-v curve. he gas s compressng, so the work done by the gas s negatve or the work done on the gas s postve. he work done by gas n an sobarc process s Substtutng nto ths equaton, W p( V V ) 80 J (000 Pa)( V V ) V.00 m 00 cm 4 Assess: he work done on a gas durng compresson s postve. P.7. Prepare: Please reer to Fgure P.7. he work done by a gas s the area under the p-versus-v curve. he gas s expandng, so the work done by the gas s postve or the work done on the gas s negatve. he area under the curve s the area o the rectangle and trangle. he work done by the gas s as ollows: 60 J 6 6 (00 0 m )(00 0 Pa) (00 0 m )(00 0 Pa) hus, the work done on the gas s W 60 J. Assess: he envronment does negatve work on the gas as t expands, so we expected the work to be negatve. P.74. Prepare: Assume the gas s an deal gas. We wll use Equaton., convert all quanttes to SI unts, and reer to Fgure P.74 or ntal and nal pressure and volume. (a) We can nd the temperatures drectly rom the deal gas law. 6 (.0 atm0,00 Pa/atm)(000 cm 0 m /cm ) nr (0.0 mol)(8. J/(mol K)) 66 K 9 C 6 (.0 atm0,00 Pa/atm)(000 cm 0 m /cm ) nr (0.0 mol)(8. J/(mol K)) 66 K 9 C (b), so ths s an sothermal process. (c) A constant volume process has V V. Because p p, restorng the pressure to ts orgnal value means that p p. From the deal gas law, p V 66 K 098 K 85C p V he temperature to be reported s thus 80C. P.75. Prepare: Assume CO gas s an deal gas and Equaton. s applcable. he molar mass or CO s M mol 44 g/mol, so a 0 g pece o dry ce s 0.7 mol. hs becomes 0.7 mol o gas at 0C. (a) Wth V 0,000 cm 0.00 m and 0C 7 K, the pressure s p nr (0.7 mol)(8. J/(mol K))(7 K) 5.56 0 4 Pa 0.509 atm V 0.00 m or 0.5 atm to two sgncant gures. (b) From the sothermal compresson, that s, at constant temperature, p 0.509 atm V V (0.00 m ).700 m 700 cm p.0 atm From the sobarc compresson, that s at constant pressure, V 000 cm V 700 cm (7 K) 6K C

-8 Chapter or 0C to two sgncant gures. (c) Assess: When volume and pressure appear as a rato o beore and ater values, we do not have to convert them to SI unts. emperature, however, must always be converted to Kelvn n the gas equaton. P.76. Prepare: Please reer to the ollowng gure. he work done n expandng the swm bladder s gven by W p V. o use ths, we need to nd the pressure at the lower depth. he temperature and number o moles o gas reman constant as the sh descends, so rom the deal gas law, s constant, that s or, rearrangng actors, p. Snce the rato o the nal volume to the ntal volume s 60% and the ntal V (.0 atm) V pressure was.0 atm, we can say that V (0.60) V and p 5.0 atm. V (0.60) V We also are gven: 4 V 5.0 0 m, rom whch t ollows that V 4 (0.60) V.0 0 m. Ater descendng, the sh pumps gas nto ts swm bladder to go rom the present volume o 4 4.0 0 m back to ts orgnal volume o 5.0 0 m. he work done s as ollows: W atm Assess: he unts work out to joules snce 5.00 Pa 4 4 (5.0 atm) (5.0 0 m.0 0 m ) 00 J Pa N/m.

hermal Propertes o Matter -9 P.77. Prepare: he rate o heat transer o solar energy s Q/t, whch s equal to solar power. hat s, the heat absorbed s (solar power) t whch s also equal to Mc water accordng to Equaton.. From the gven normaton, we can then easly nd t. he area o the garden pond s A (.5 m) 9.65 m and the mass o water n the pond s 5.9 0 kg. he water absorbs all the solar power, whch s (400 W/m )(9.65 m ) 7854 W hs power s used to rase the temperature o the water. hat s, Q (7854 W) t Mc (5900 kg)(490 J/(kg K))(0 K) t,476 s 8.7 h water Assess: It s a common experence that a pool o water takes a ew hours to warm up. A value o 8.7 h to heat 5900 kg water by 0C s reasonable. P.78. Prepare: Snce the compresson s adabatc, there s no heat low and the rst law o thermodynamcs says that Eth W where W s the work done on the gas. So we need to nd the change n thermal energy rom Equaton.0. Combnng the two equatons mentoned above, we have the ollowng: W Eth nr (0.00 mol)(8.5 J/(mol K)(50 C 0 C) 5 J Assess: We were able to use degrees Celsus n the above calculaton snce we were subtractng two temperatures. P.79. Prepare: Snce ths s an adabatc process, there s no heat low and so the rst law o thermodynamcs says that Eth W. Now W n ths context s work done on the gas. In ths problem, the thermal energy decreases so that Eth and W are negatve. A negatve value o W means that work s done by the gas. We can use Equaton.0 to nd the change n thermal energy. Combnng the two equatons mentoned above gves us the ollowng: W Eth nr (0.5 mol)(8.5 J/(mol K)( 0 C 0 C) 56 J he work done by the gas s 56 J. Assess: he reducton n temperature s made because the gas s dong work n expandng and the uel t uses n dong that work s ts own thermal energy. As the avalable thermal energy goes down, the temperature also goes down. P.80. Prepare: We wll need the ormula or gravtatonal potental energy, Ug mgh, as well as a ormula whch relates a change n temperature to a change n thermal energy. Equaton.4 gves the heat needed to change the temperature o an object. In ths case, the heat produced by the body s metabolsm s turned nto thermal energy so we wrte Eth mc. We wll use the mammalan specc heat rom able.4.

-0 Chapter (a) Susan s change n potental energy s gven by the ollowng equaton: U g mg h (68 kg)(9.8 m/s )(59 m) 900 J 9000 J (b) I ths ncrease n potental energy represents 5% o the total energy used by her body, then she has used our tmes ths much energy, or 57 kj 60 kj. (c) 75% o the energy used s wasted and goes to thermal energy so Eth (0.75)(57 kj) 8 kj. We can solve the equaton E mc or to obtan th Eth 80 J mc (68 kg)(400 J/(kg K)) 0.5 K Assess: hs ncrease s about 0.5 C whch can be converted to 0.9 F. hs doesn t seem lke much, but on the other hand, ths exercse was less strenuous than t mght seem. he energy spent, 60 kj can be converted to about 8 detary Calores. he same amount o energy would be spent n walkng about two ths o a mle. P.8. Prepare: here are two nteractng systems: the nuclear reactor and the water. he heat generated by the nuclear reactor s used to rase the water temperature. For the closed reactor water system, energy conservaton per second requres Q Q reactor Q water 0 J. he heat rom the reactor n t s s Q reactor = 000 MJ =.0 0 9 J and we wll use Equaton. or Q water. he heat absorbed by the water s Q m c m (490 J/(kg K)( K) water water water water.00 J m (490 J/(kg K)( K) 0 J m.980 kg 9 4 water water Each second,.98 0 4 kg o water s needed to remove heat rom the nuclear reactor. hus, the water low per mnute s 4 kg 60 s.980.4 0 6 kg/mn s mn P.8. Prepare: From able.4, the metabolc power o a 68 kg cyclst s 480 W. We assume that 5% o ths goes to propellng the cyclst and the other 75%, or 60 W, becomes heat whch serves to evaporate perspraton. Equaton.6 gves the heat needed to evaporate a lqud. From the dscusson ollowng able.5, a good value or 6 the latent heat o vaporzaton o sweat s L.4 0 J/kg. We solve Equaton.6 or m and get v m Q / L. hs s the mass o perspraton whch heat Q Q could evaporate. We know 60 W and we can combne ths wth the precedng equaton to obtan the ollowng: m Q / L Q / t 60 J/s t t L v 4.50 kg/s 0.54 kg/hr 6 v.4 0 J/kg v

hermal Propertes o Matter - Assess: A klogram o water has a volume o about one lter, so ths s about hal a lter per hour. A value on the order o one lter per hour seems reasonable. P.8. Prepare: he entre knetc energy o the car goes nto heatng the our brakes. Equaton. apples. able.4 gves the specc heat o ron, c 449 J/(kg K). We are gven the mass o each brake dsk, m 4.0 kg and the mass o the car M 00 kg. he speed o the car s v 60 mph, whch we convert to SI unts (keepng one extra sgncant gure n the ntermedate calculaton). 0.447 m/s v 60 mph 6.8 m/s mph Settng the knetc energy equal to the heat gong nto the our brakes gves Now solve or the temperature ncrease. Q 4mc Mv Mv (00 kg)(6.8 m/s) 60 K 8 mc 8 (4.0 kg)(449 J/(kg K)) Assess: hs s a sgncant rse n temperature, but brakes really do heat up qute a bt. he knetc energy o the car must go somewhere. One other opton s to let the translatonal knetc energy o the car be transormed nto rotatonal knetc energy o a lywheel, so that t can be recovered ater the lght turns green. Careul examnaton o the unts shows they cancel properly. P.84. Prepare: We need Equaton.6 whch gves the amount o heat need to evaporate a mass o lqud. We are gven the rate that heat must be absorbed by the water as ollows: Q/ t 500 W. We can use 6 the latent heat o vaporzaton gven n the dscusson ollowng able.5, L.4 0 J/kg. Solvng Equaton.6 or the mass gves us: answer m Q / L Q / t 500 J/s 600 s t t L v 6 v.4 0 J/kg hr v v m Q / L and dvdng both sdes by t gves our.8 kg/hr Assess: hs works out to.8 L/hr, whch seems reasonable. Durng vgorous exercse, humans sweat a couple o lters per hour. An elephant at rest has about twce the metabolc rate o a human exercsng vgorously (usng the data rom able.4), so we would expect a value o around 4 L/hr. P.85. Prepare: Use the conservaton o energy. Frst the gravtatonal potental energy o the water s converted nto knetc energy, and then, on mpact, nto thermal energy. We ll set the change o gravtatonal potental energy (whch decreases, hence the negatve sgn) equal to the change n thermal energy (whch ncreases). Solve or. w Mgh Mcw Mgh gh (9.8 m/s )( 0 m) 0.0 K Mc c 490 J/(kg K) w he answer s small enough to not be notceable, whch probably jbes wth your experence. Assess: I t s not obvous that the mass o the water should cancel (that s, the temperature rse s the same or any amount o water dropped rom 0 m), keep thnkng about t untl t makes sense; each lttle chunk o water starts wth some gravtatonal potental energy, whch changes the thermal energy (and rases the temperature) o that lttle chunk. hat happens or each chunk, so the temperature goes up the same amount no matter how much water you drop. P.86. Prepare: Changng sold lead at 0C to lqud lead at ts meltng pont ( m 8C) requres two steps: rasng the temperature to m and then meltng the sold at m to a lqud at m. We wll use Equatons. and. and work n SI unts. he values o the specc heat and the heat o uson are gven n ables.4 and.5.

- Chapter he equaton or the total heat s Q Q Q 000 J Mc lead ( ) ML 000 J M(8 J/( kg K )(8 0) K M(0.5 0 5 000 J J/kg) M 5.5 g (64,44 J/kg) he maxmum mass o lead you can melt wth 000 J o heat s 5.5 g. P.87. Prepare: Heatng the materal ncreases ts thermal energy. Please reer to Fgure P.87. he materal melts at 00C and undergoes a sold-lqud phase change. he materal s temperature ncreases rom 00C to 500C. Bolng occurs at 500C and the materal undergoes a lqud-gas phase change. We wll use Equatons. and. to determne the specc heat and the heat o vaporzaton o the lqud. (a) In the lqud phase, the specc heat o the lqud can be obtaned as ollows: Q Mc (b) he latent heat o vaporzaton s Q c M 0 kj 8 J/(kg K) 0.00 kg 00 K Q 40 kj Lv M (0.00 kg) 5.0 0 J/kg Assess: he values obtaned are o the same order o magntude as n ables.4 and.5 or a ew materals. P.88. Prepare: here are three nteractng systems: alumnum, copper, and ethyl alcohol. he alumnum, copper, and alcohol orm a closed system, so Q Q Al Q Cu Q eth 0 J. We wll use Equaton. or each o the three systems n the above equaton. he specc heats o the three materals are gven n able.4. We must work n SI unts, so have mass n kg and temperature n Kelvns. he mass o the alcohol s M eth V (790 kg/m )(50 0 6 m ) 0.095 kg. Expressed n terms o specc heats and usng the act that, the Q 0 J condton s Substtutng nto ths expresson, MAlcAl Al MCucCuCu Methcetheth 0 J (0.00 kg)(900 J/(kg K)(98 K 47 K) (0.00 kg)(85 J/(kg K)(98 K ) (0.095 kg)(400 J/(kg K)(98 K 88 K) 575 J (7.7 J/K)(98 ) 948 J 0 J 6.6 K 56 C P.89. Prepare: here are two nteractng systems: alumnum and ce. he system comes to thermal equlbrum n our steps: () the ce temperature ncreases rom 0C to 0C, () the ce becomes water at 0C, () the water temperature ncreases rom 0C to 0C, and (4) the cup temperature decreases rom 70C to 0C. We wll use Equatons. and.. he specc heats and the heat o uson are gven n ables.4 and.5. he alumnum and ce orm a closed system, so Q Q Q Q Q 4 0 J. hese quanttes are Q M c (0.00 kg)(090 J/(kg K)(0 K) 090 J ce ce Q M L ce ce water he Q 0 J equaton now becomes he soluton to ths s M Al 0.97 kg. 5 (0.00 kg)(. 0 J/kg),00 J Q M c (0.00 kg)(490 J/(kg K)(0 K) 880 J Q M c M (900 J/(kg K)( 50 K) (45,000 J/kg) M 4 Al Al Al Al 4,770 J (45,000 J/kg)M Al 0 J P.90. Prepare: We have a thermal nteracton between the thermometer and the water. Equaton. wll be used or both the thermometer and water along wth the conservaton o energy equaton. he conservaton o energy equaton Q thermo Q water 0 J s

hermal Propertes o Matter - M thermo c thermo ( thermo ) M waterc water( water ) 0 J he thermometer slghtly cools the water untl both have the same nal temperature 7.C. hus (0.050 kg)(750 J/(kg K)(7.C 0.0C) (0.00 kg)(490 J/(kg K)(7.C ) 90 J 88 (J/K)(7.C ) 0 J 7.5C Assess: he thermometer reads 7.C or a real temperature o 7.5C. hs s reasonable. water P.9. Prepare: here are two nteractng systems: coee (.e., water) and ce. Changng the coee temperature rom 90C to 60C requres our steps: () rase the temperature o ce rom 0C to 0C, () change ce at 0C to water at 0C, () rase the water temperature rom 0C to 60C, and (4) lower the coee temperature rom 90C to 60C. We wll use Equatons. and.. he specc heats o water and ce and the heat o uson o ce are gven n ables.4 and.5. For the closed coee-ce system, Q Q Q ( Q Q Q ) ( Q ) 0 J ce coee 4 ce ce ce ce ce ce ce water ce ce 6 4 coee coee (000 m )( water Q M c M (090 J/(kg K)(0 K) M (4,800 J/kg) Q M L M (0,000 J/kg) Q M c M (490 J/kg)(60 K) M (5,400 J/kg) Q M c he Q 0 J equaton thus becomes water 000 kg/m )(490 J/(kg K)( 0 K) 7,000 J M ce (4,800 0,000 5,400) J/kg 7,70 J 0 J Mce 0.0605 kg 6g Assess: 6 g s the mass o approxmately one ce cube. P.9. Prepare: We wll use the rst law o thermodynamcs to descrbe ths sobarc process. W gas s negatve because the gas s compressed. Compresson transers energy nto the system, that s, work done on the gas s postve. Also, 00 J o heat energy s transerred out o the gas, that s Q 60 J. he rst law o thermodynamcs s E W Q p V Q (4.0 0 5 Pa)(00 600) 0 6 m 00 J 60 J th gas hermal energy ncreases by 60 J. P.9. Prepare: reat helum as an deal gas. he process n part (a) s sochorc and the process n part (b) s sobarc. Intally V (0.0 m) 0.0080 m 8.0 L and 9 K. Helum has an atomc mass number A 4, so g o helum s n M/M mol 0.75 mole o helum. We wll use the deal gas Equatons. and.4. (a) We can nd the ntal pressure rom the deal-gas law as ollows: nr (0.75 mol)(8. J/(mol K)(9 K) p 8 kpa.5 atm V 0.0080 m Heatng the gas wll rase ts temperature. A constant volume process has Q nc V, so Q 000J 07 K nc (0.75 mol)(.5 J/(mol K) V hs rases the nal temperature to 400 K. Because the process s sochorc, p p 400 K p p (.5 atm).atm 9 K (b) he ntal condtons are the same as n part (a), but now Q nc P. hus, Q 000 J 64.K nc (0.75 mol)(0.8 J/(mol K) P

-4 Chapter Now the nal temperature s 57 K. Because the process s sobarc, whch s to be reported as 9.8 L. V V 57 K V V (0.0080 m ) 0.00975 m 9.75 L 9 K P.94. Prepare: Please reer to Fgure P.94. he monatomc gas s an deal gas, whch s subject to sobarc and sochorc processes. We wll use the deal gas Equaton. and Equatons.5 and.6. (a) For ths sobarc process, p 4.0 atm, V 800 0 6 m, p 4.0 atm, and V 600 0 6 m. he temperature o the gas s obtaned rom the deal-gas equaton as where n 0.0 mol. Also, hus, the heat requred or the process s nr 90 K 6 V 6000 m 6 780 K V 8000 m Q nc ( ) (0.0 mol)(0.8 J/(mol K)(90 K) 8 J P whch s 80 J n two sgncant gures. hs s heat transerred to the gas. (b) For the sochorc process, V V 600 0 6 m, p 4.0 atm, p.0 atm, and 780 K. can be obtaned rom the deal gas equaton as ollows: he heat requred or the process s.0 atm ( p/ p) (780 K) 90 K 4.0 atm Q nc ( ) (0.0 mol)(.5 J/mol K)(90 K 780 K) 488 J V whch s 490 J n two sgncant gures. Because o the negatve sgn, ths s the amount o heat removed rom the gas. (c) he change n the thermal energy o the gas s E ( Q Q ) ( W W ) 8J 488J W 0J 4 J p V th 4 J (4.0.0 0 5 Pa)(600 0 6 m 800 0 6 m ) 0 J Assess: hs result was expected snce. P.95. Prepare: he gas s assumed to be an deal gas that s subjected to sobarc and sochorc processes. Please reer to Fgure P.95. We wll use SI unts, the deal gas Equaton., and Equatons.5 and.6. (a) he ntal condtons are p.0 atm 04,000 Pa, V 00 cm.0 0 4 m, and 00C 7 K. he number o moles o gas s n 4 (04,000 Pa)(.0 0 m ) 9.80 mol R (8. J/(mol K)(7 K) At pont we have p p.0 atm and V 00 cm V. hs s an sobarc process, so V V V (7 K) 9 K V he gas s heated to rase the temperature rom to. he amount o heat requred s Q nc whch s to be reported as 50 J. hs amount o heat s added durng process. P (9.8 0 mol)(0.8 J/mol K)(9 K 7 K) 5 J

hermal Propertes o Matter -5 (b) Pont returns to 00C 7 K. hs s an sochorc process, so Q nc V (9.8 0 mol)(.5 J/mol K)(7 K 9 K) 9.5 J whch s to be reported as 9 J. hs amount o heat s removed durng process. P.96. Prepare: he heat engne ollows a closed cycle. Please reer to Fgure P.96. (a) he work done by the gas per cycle s the area nsde the closed p-versus-v curve. We get W out 6 (00 kpa 00 kpa)(600 cm 00 cm ) (00 0 Pa)(400 0 m ) 40 J he heat exhausted s QC 80 J 00 J 80 J. So, the heat extracted rom the hot reservor s Q H 80 J 40 J 0 J. (b) he thermal ecency o the engne s or 0. to two sgncant gures. W Q out H 40J 0 J 0.5 P.97. Prepare: Whle the Prepare, Solve, Assess method o problem solvng s meant to dscourage blnd huntng or an equaton and then pluggng and chuggng wthout understandng, we do recognze n ths case that the equaton relatng the rate o thermal energy transer due to conducton and the thermal conductvty (the thng we want) s gven by Equaton.. Q ka = t L We pause to see what aects Q/ t. I the area s larger, then the rate o transer s greater. I the materal s thcker (larger L), then the rate o transer s less. I the temperature derence between the two sdes s greater, then the rate o transer s greater. All o ths makes sense, so we proceed. We are gven L 0.050 m, A.0 m, and 55 K. We also convert the rate o transer to SI unts as ollows: Q Cal 000 cal 4.9 J h mn 5 9 W t h Cal cal 60 mn 60 s Snce we want the thermal conductvty o goose down, we solve the equaton or k. Q L 0.050 m k (9 W) 0.06 W/(m K) t A.0 m 55 K Assess: hs answer s at the lower end o the range lsted n able.7 or ur and eathers, whch s as we expect, snce goose down s well known as a good nsulator. he one number that may not be qute realstc s the 5C nsde the bag; ths s 95F, whch would be warmer than cozy. P.98. Prepare: Assume all the nsulaton comes rom the ar layer. Assume the value or the thermal conductvty o ar n able.7 s a good approxmaton or these condtons. Equaton. apples. Usng Equaton., Q ka t L 0.05 m (0.06 W/ (m K)(.m ) (4 C ( 0 C)) 6 W Assess: hs s a large amount o power, even though ar has the lowest thermal conductvty o all the materals n able.7. P.99. Prepare: We ll use Equaton. n a rato or the two plates. Frst, solve the equaton or.

-6 Chapter Q = t e A We are gven the rates o energy transer or the two plates. Snce they are the same, and snce they are both copper, ther emssvtes are the same. A 4 A 6 A. /4 Q /4 /4 /4 [( t )( e A )] 6 Q /4 [( t )( e A )] A A A A /4 6 Plate s twce as hot as plate on the absolute temperature scale. Assess: Plate has 6 tmes the area o plate, but because plate s twce as hot as plate, they radate energy at the same rate because o the 4 n the ormula. P.00. Prepare: Heat loss by conducton can be calculated wth Equaton.. Equaton. apples to heat lost by radaton. (a) he dead layer o ar separates and nsulates you rom the ar n the room. he conducton through the ar layer s Q ka (0.06 W/ (m K)(.8 m ) (9 K) 84 W t L 0.005 m (b) he heat lost through radaton s gven by Equaton.. Body temperature s 4 7 07 K. he temperature o the walls s 7 7 90 K. Qnet t e A( ) (.97)(5.67 0 W / (m K )(.8 m )((07 K) (90 K) ) 80 W 4 4 8 4 4 4 0 (c) he heat lost to radaton s greater. (d) I the person s metabolzng ood at a rate o 55 W, he eels chlly because he s producng heat at a rate o 55 W and losng heat at a rate o 84 W 80 W or 60 W. Assess: Puttng some clothes on would decrease the heat lost by radaton, convecton, and conducton. P.0. Prepare: Volume thermal expanson ndcates that we should use Equaton.. V V We want to know the change n depth, whch s gven by d V/A. Smlarly the ntal volume n the ormula s gven by V Ad. We wll assume that the temperatures gven (about 7C) are close enough to 0C that we can use the coecent o volume expanson or water gven n able., 0 0 6 K. We are also gven d 500 m and.00 K. V V Ad 6 d d (00 K )(500 m)(.00 K) 0.05 m A A A he correct choce s A. Assess: he answer o cm may not seem lke much, but t could aect low-lyng coastal areas. he reason the area cancels out s because each square centmeter o water wll ncrease n depth by the same amount or a gven temperature change. P.0. Prepare: We can use Equaton.. able.4 lsts the specc heat o water. he mass o the water n the top layer s M d A. he area o the top level o the oceans s 8 0 m 4 A (.60 km ).60 m km he heat requred to change the temperature by one degree Celsus s Q Mc d Ac 7.5 0 J 4 (000 kg/m )(500 m)(.6 0 m )(490 J/(kg K)( K)

hermal Propertes o Matter -7 o one sgncant gure, ths s 0 4 J.he correct choce s A. Assess: hs result makes sense. he densty and specc heat o water are relatvely large. P.0. Prepare: Careully examne Fgure.0a. At C the volume o a mole o water decreases wth ncreasng temperature. Because the graph has a negatve slope at C, the volume would decrease n gong rom C to C. he correct choce s C. Assess: It s unusual or a substance to have a negatve coecent o expanson, but the act that water does at temperatures close to reezng s bologcally mportant. P.04. Prepare: Conducton s the transer o heat drectly through physcal materal. Radaton s energy transer through electromagnetc waves. Evaporaton transers energy through removal o molecules wth hgh thermal energy. Snce there s no mxng n the warmer surace water, convecton does not happen eectvely. he correct choce s B. Assess: Note that the water s heated by the lght rom the sun, so t s heated through radaton.