Physics 140 Heat -...

Physics140

Heat

Chapter14

Internalenergy

Physics140,Prof.M.Nikolic 2

Theaveragetransla@onalkine@cenergyofasinglepar@cle

kTK23

tr =

Theinternalenergyofasystemisthesumofalltheenergyofallthemoleculesinthesystem.

NkTU23

=Foranidealgas:

Wedonotincludeininternalenergy•  Kine@cenergyoftransla@on,vibra@on,androta@ononmacroscopiclevel•  Poten@alenergyduetoexternalinterac@ons

Internalenergy


Weincludeininternalenergy:

•  Kine@cenergyoftransla@on,vibra@on,androta@onofmoleculesduetotheirindividualmo@ons

•  Poten@alenergyduetointerac@onsbetweenpar@clesinthesystem•  Chemicalandnuclearenergy(bindingenergies)

Heat


Heatisthetransferofenergyfromonebodytoanotherduetothetemperaturedifference.

Calorictheory

•  Un@laboutthemid-19thcentury,scien@stsbelievedthatheatwasafluid,called“caloric”anditwasbelievedthatitcouldbetransferredbetweenobjectsbutneithercreatednordestroyed.

•  Toheatupanobjectthiscalorichadtoflowintoit.

Kine4ctheory

•  JamesPrescoUJouleshowedthatheatisreallyaformofenergy•  AllmaUerismadeupofatoms/[email protected]

theymove,thehoUeranobjectwillbe.

Thermalexpansion-solids


Oscilla@onsatarela@velylowtemperaturehavearela@velysmallamplitude.

Oscilla@onsatarela@velyhightemperaturehavearela@velylargeamplitude.

Athightemperatures,[email protected].

Measuringheat


•  HeatistransferofenergyèitismeasuredinJoules(J)

•  InChemistryandBiology,peopleusethecalorie(cal)

Calorieisamountofenergyneededtoraisethetemperatureof1gramofwater1degreeCelsius.

1cal=4.186J

•  FoodcalorieorCalorie(Cal)is1000calories 1Cal=1000cal

Calories:1090Cal=1.09millioncal=1.09millionx4.2J=4.6millionJ

Heatcapacityandspecificheat


SpecificHeat(specificheatcapacity)istheheatcapacityperunitmass.

Unitsofheatcapacity:[J/K]

Heatcapacityistheenergynecessarytochangeanobject’stemperatureby10C.

TQCΔ

=

Unitsofspecificheat:[J/kgK]TmQ

mCc

Δ==

Or Q =mcΔT =mc(Tfinal −Tinitial )


Exercise:SpecificheatA0.400kgaluminumteakeUlecontains2.00kgofwaterat15.00C.HowmuchheatisrequiredtoraisethetemperatureofthewaterandkeUleto1000C?Thespecificheatofwateris4186J/kgKandaluminumis900J/kgK.

Whatisgiven:mAl=0.4kgmw=2kgTi=15oCTf=100oCcw=4186J/kgKcAl=900J/kgK

)( if TTmcTmcQ −=Δ=

WhenlookingintotemperaturedifferenceyoucankeeptemperatureinCelsius:ΔT=100oC-15oC=85oC=85K

TheheatneededtoraisethetemperatureofthewatertoTfis:

wwww TcmQ Δ= kJ 712K 85K J/kg 4186kg 2w =⋅⋅=Q

TheheatneededtoraisethetemperatureofthealuminumtoTfis:

AlAlAlAl TcmQ Δ= kJ 6.30K 85K J/kg 900kg 4.0Al =⋅⋅=Q

kJ 6.742kJ 6.30kJ 712Alwtotal =+=+= QQQ

Calorimetry


Acalorimeterisadevicethatcanmeasurethespecificheatofsubstances.

•  Itconsistsofaninsulatedcontainer(aluminum)withaknownmassfilledwiththeknownquan@tyofwater

•  Boththewaterandcalorimeterhavethesame(known)ini@altemperature

•  Afertheaddi@onoftheunknownsubstance,thesystemofcalorimeterpluswaterplussubstanceeventuallyreachthesamefinaltemperature.

Ifnoheatentersorleavesthesystem,thenwecansay:

Qcalorimeter +Qwater +Qsubs tance =Qsystem = 0

mcalccal (Tf −Ti )+mwatercwater (Tf −Ti )+msubcsub(Tf −Tsub ) = 0

Specificheatofidealgases


U =32NkBT =

32nRTInternalenergyiftheidealgas

Inaclosedsystem(whenthevolumeofthegasisconstant)èaddedheat=changeofinternalenergy

Q = ΔU = n 32RΔT = nCVΔT

molJ/K 5.12

23

== RCV Molarspecificheatatconstantvolumeofmonoatomicidealgas

molJ/K 8.20

25

== RCVMolarspecificheatatconstantvolumeofdiatomicidealgas


Exercise:Hea@ngthenitrogenAcontainerofnitrogengas(N2)at23oCcontains425Latapressureof3.5atm.If26.6kJofheatareaddedtothecontainer,whatwillbethenewtemperatureofthegas?Whatisgiven:Ti=23oC=23+273=296KVi=425l=425dm3=0.425m3Pi=3.5atm=3.5x1.013x105PaQ=26.6kJ=26600J

Q = nCVΔT ΔT = Tf −Ti =QnCV

Nitrogen(N2)isadiatomicgas:Cv=20.8J/K/mol

Thenumberofmolesnisgivenbytheidealgaslaw:

n = PiViRTi

mol 22.61K 296315.8

m 425.0Pa 10013.15.3 35

=⋅

⋅××=n

K 9.316K 9.20K 296J/K/mol 8.20mol 2.61J 26600K296 =+=

⋅+=+=

Vif nC

QTT

Phasetransi@ons


Whenasubstancechangesphasesenergyistransferredwithoutachangeintemperature.

The“hiddenenergy”requiredtochangethephaseofasubstanceiscalledlatentheat(L).

Fusion Vaporiza@on


Phasetransi@onsTotalheatinvolvedinthephasetransi@on Q =mL

1.  Thelatentheatoffusion(Lf)istheheatperunitmassneededtoproducethesolid-liquidphasetransi@on.

2.  Thelatentheatofvaporiza4on(Lv)istheheatperunitmassneededtoproducetheliquid-gasphasetransi@on.


Exercise:Phasetransi@onsYoujustdroppeda500-gbarofsilverinanicebucketfilledwith100goficeandattemperatureof-15oC.Ifallicemelted,whatwastheini@altemperatureofsilver?

silverQQ += ice0

Theheatflowsfromtheobjectwithhighertemperaturetotheobjectwithlowertemperatureun@ltheyareatsametemperatures→heatflowsfromthesilverintoice:Qsilver<0andQice>0

Theheatthatflowsintoiceisthesumofa)  Theheattoraisetemperatureoficefrom-15oCto0oC

b)  Theheattomelttheiceat0oC

miceciceΔTice = 0.1 kg ⋅2.05 kJ/kg K ⋅ (0− (−15)) K = 3.07 kJ

miceLf = 0.1 kg ⋅333.7 kJ/kg = 33.4 kJ

Qice =miceciceΔTice +miceLf

Qice = 3.07 kJ + 33.4 kJ = 36.47 kJ

cice=2.05kJ/(kgK)Lf=333.7kJ/kg(table14.4)cs=0.24kJ/(kgK)ms=500g=0.5kgmice=100g=0.1kg


Exercise:Phasetransi@ons

273 K −Ti =−36.47 kJ

0.5 kg ⋅0.24 kJ/kg K= −152 K

Tf −Ti =−Qice

mscs

Ti = 425 K

TheheatthatflowsoutofsilveristhesumistheheattolowerthetemperatureofsilverfromTito0oC

Qs =mscsΔTs =mscs (Tf −Ti ) = −Qice

Youjustdroppeda500-gbarofsilverinanicebucketfilledwith100goficeandattemperatureof-15oC.Ifallicemelted,whatwastheini@altemperatureofsilver?

cice=2.05kJ/(kgK)Lf=333.7kJ/kg(table14.4)cs=0.24kJ/(kgK)ms=500g=0.5kgmice=100g=0.1kg

Phasediagrams-water


Heattransfer


1.  Conduc@on:heattransferthroughdirectcontact2.  Convec@on:heattransferthroughfluids3.  Radia@on:heattransferthroughelectromagne@cradia@on

Thermalconduc@on


•  Throughdirectcontact,heatcanbeconductedfromregionsofhightemperaturetoregionsoflowtemperature.

•  Energyistransferredbycollisionsbetweenneighboringatomsormolecules.

Heattransferred

Q ~ A ΔT ⋅ td

P = Qt=κA ΔT

d

Fourierlawofthermalconduc@on

κisthethermalconduc4vity

Thermalconduc@on


PRAdPT ==Δκ

Thermalresistance:

AdRκ

=

Units:KelvinperWaI[K/W]

k–thermalconduc@vityofmaterialUnits:WaUpermeter@mesKelvin[W/m�K]

Thermalconvec@on


Theairabovetheradiatorheatsup.

Coldairisallowedtoentertheregionvacatedbythewarmed-upairandiswarmedupinturn.

Interac@onswithslower,colderairmoleculestendstodispersetheinternalenergy,makingtheroomfeelwarm.

Thermalradia@on


Allbodiesemitelectromagne@c(EM)radia@on.

Therateofenergyemissionis(Stefan’sLaw): P = eσ AT 4

where•  Aisthesurfaceareaoftheemivngbody•  Tisitstemperature•  σ=5.670×10-8W/m2K4istheStefan-Boltzmannconstant

•  eisemissivity•  e=0foraperfectreflector(perfectlywhiteobject)•  e=1foraperfectabsorber(perfectlyblackobject)


Exercise:Radia@onAtungstenfilamentinalampisheatedtoatemperatureof2.6x103Kbyanelectriccurrent.Thetungstenhasemissivityof0.32.Whatisthesurfaceareaofthefilamentifthelampdelivers40Wofpower?

Whatisgiven:T=2600Ke=0.32P=40Wσ=5.67x10-8W/m2K4

P = eσ AT 44Te

PAσ

=

25428 m 108.4

K 2600K W/m1067.532.0 W40 −

−×=

⋅×⋅=A

TheGreenhouseEffect


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Physics 140 Heat -...

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