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
Physics140,Prof.M.Nikolic 3
Weincludeininternalenergy:
• Kine@cenergyoftransla@on,vibra@on,androta@onofmoleculesduetotheirindividualmo@ons
• Poten@alenergyduetointerac@onsbetweenpar@clesinthesystem• Chemicalandnuclearenergy(bindingenergies)
Heat
Physics140,Prof.M.Nikolic 4
Heatisthetransferofenergyfromonebodytoanotherduetothetemperaturedifference.
Calorictheory
• Un@laboutthemid-19thcentury,scien@stsbelievedthatheatwasafluid,called“caloric”anditwasbelievedthatitcouldbetransferredbetweenobjectsbutneithercreatednordestroyed.
• Toheatupanobjectthiscalorichadtoflowintoit.
Kine4ctheory
• JamesPrescoUJouleshowedthatheatisreallyaformofenergy• AllmaUerismadeupofatoms/[email protected]
theymove,thehoUeranobjectwillbe.
Thermalexpansion-solids
Physics140,Prof.M.Nikolic 5
Oscilla@onsatarela@velylowtemperaturehavearela@velysmallamplitude.
Oscilla@onsatarela@velyhightemperaturehavearela@velylargeamplitude.
Athightemperatures,[email protected].
Measuringheat
Physics140,Prof.M.Nikolic 6
• HeatistransferofenergyèitismeasuredinJoules(J)
• InChemistryandBiology,peopleusethecalorie(cal)
Calorieisamountofenergyneededtoraisethetemperatureof1gramofwater1degreeCelsius.
1cal=4.186J
• FoodcalorieorCalorie(Cal)is1000calories 1Cal=1000cal
Calories:1090Cal=1.09millioncal=1.09millionx4.2J=4.6millionJ
Heatcapacityandspecificheat
Physics140,Prof.M.Nikolic 7
SpecificHeat(specificheatcapacity)istheheatcapacityperunitmass.
Unitsofheatcapacity:[J/K]
Heatcapacityistheenergynecessarytochangeanobject’stemperatureby10C.
TQCΔ
=
Unitsofspecificheat:[J/kgK]TmQ
mCc
Δ==
Or Q =mcΔT =mc(Tfinal −Tinitial )
Physics140,Prof.M.Nikolic 8
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
Physics140,Prof.M.Nikolic 9
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
Physics140,Prof.M.Nikolic 10
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
Physics140,Prof.M.Nikolic 11
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
Physics140,Prof.M.Nikolic 12
Whenasubstancechangesphasesenergyistransferredwithoutachangeintemperature.
The“hiddenenergy”requiredtochangethephaseofasubstanceiscalledlatentheat(L).
Fusion Vaporiza@on
Physics140,Prof.M.Nikolic 13
Phasetransi@onsTotalheatinvolvedinthephasetransi@on Q =mL
1. Thelatentheatoffusion(Lf)istheheatperunitmassneededtoproducethesolid-liquidphasetransi@on.
2. Thelatentheatofvaporiza4on(Lv)istheheatperunitmassneededtoproducetheliquid-gasphasetransi@on.
Physics140,Prof.M.Nikolic 14
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
Physics140,Prof.M.Nikolic 15
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
Physics140,Prof.M.Nikolic 16
Heattransfer
Physics140,Prof.M.Nikolic 18
1. Conduc@on:heattransferthroughdirectcontact2. Convec@on:heattransferthroughfluids3. Radia@on:heattransferthroughelectromagne@cradia@on
Thermalconduc@on
Physics140,Prof.M.Nikolic 19
• Throughdirectcontact,heatcanbeconductedfromregionsofhightemperaturetoregionsoflowtemperature.
• Energyistransferredbycollisionsbetweenneighboringatomsormolecules.
Heattransferred
Q ~ A ΔT ⋅ td
P = Qt=κA ΔT
d
Fourierlawofthermalconduc@on
κisthethermalconduc4vity
Thermalconduc@on
Physics140,Prof.M.Nikolic 20
PRAdPT ==Δκ
Thermalresistance:
AdRκ
=
Units:KelvinperWaI[K/W]
k–thermalconduc@vityofmaterialUnits:WaUpermeter@mesKelvin[W/m�K]
Thermalconvec@on
Physics140,Prof.M.Nikolic 21
Theairabovetheradiatorheatsup.
Coldairisallowedtoentertheregionvacatedbythewarmed-upairandiswarmedupinturn.
Interac@onswithslower,colderairmoleculestendstodispersetheinternalenergy,makingtheroomfeelwarm.
Thermalradia@on
Physics140,Prof.M.Nikolic 22
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)
Physics140,Prof.M.Nikolic 25
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
Physics140,Prof.M.Nikolic 26
Visiblelight
Infraredlight
Physics140,Prof.M.Nikolic 27