ans. the geyser heats the water, raising the temperature...
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
4MarksQuestions
1.Ageyserheatswaterflowingattherateof3.0litresperminutefrom27°Cto77°C.If
thegeyseroperatesonagasburner,whatistherateofconsumptionofthefuelifits
heatofcombustionis J/g?
Ans.Waterisflowingatarateof3.0litre/min.
Thegeyserheatsthewater,raisingthetemperaturefrom27°Cto77°C.
Initialtemperature, =27°C
Finaltemperature, =77°C
∴Riseintemperature,ΔT=
=77–27=50°C
Heatofcombustion=
Specificheatofwater,c=4.2
Massofflowingwater,m=3.0litre/min=3000g/min
Totalheatused,ΔQ=mcΔT
=
=
Rateofconsumption= =15.75g/min
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2.Whatamountofheatmustbesuppliedto kgofnitrogen(atroom
temperature)toraiseitstemperatureby45°Catconstantpressure?(Molecularmassof
.)
Ans.Massofnitrogen,
Riseintemperature,ΔT=45°C
Molecularmassof ,M=28
Universalgasconstant,R=
Numberofmoles,
Molarspecificheatatconstantpressurefornitrogen,
Thetotalamountofheattobesuppliedisgivenbytherelation:
=933.38J
Therefore,theamountofheattobesuppliedis933.38J.
3.Explainwhy
(a)Twobodiesatdifferenttemperatures and ifbroughtinthermalcontactdonot
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necessarilysettletothemeantemperature .
(b)Thecoolantinachemicaloranuclearplant(i.e.,theliquidusedtopreventthe
differentpartsofaplantfromgettingtoohot)shouldhavehighspecificheat.
(c)Airpressureinacartyreincreasesduringdriving.
(d)Theclimateofaharbourtownismoretemperatethanthatofatowninadesertat
thesamelatitude.
Ans.(a)Whentwobodiesatdifferenttemperatures and arebroughtinthermal
contact,heatflowsfromthebodyatthehighertemperaturetothebodyatthelower
temperaturetillequilibriumisachieved,i.e.,thetemperaturesofboththebodiesbecome
equal.Theequilibriumtemperatureisequaltothemeantemperature only
whenthethermalcapacitiesofboththebodiesareequal.
(b)Thecoolantinachemicalornuclearplantshouldhaveahighspecificheat.Thisis
becausehigherthespecificheatofthecoolant,higherisitsheat-absorbingcapacityandvice
versa.Hence,aliquidhavingahighspecificheatisthebestcoolanttobeusedinanuclearor
chemicalplant.Thiswouldpreventdifferentpartsoftheplantfromgettingtoohot.
(c)Whenacarisinmotion,theairtemperatureinsidethecarincreasesbecauseofthe
motionoftheairmolecules.AccordingtoCharles'law,temperatureisdirectlyproportional
topressure.Hence,ifthetemperatureinsideatyreincreases,thentheairpressureinitwill
alsoincrease.
(d)Aharbourtownhasamoretemperateclimate(i.e.,withouttheextremesofheatorcold)
thanatownlocatedinadesertatthesamelatitude.Thisisbecausetherelativehumidityina
harbourtownismorethanitisinadeserttown.
4.Acylinderwithamovablepistoncontains3molesofhydrogenatstandard
temperatureandpressure.Thewallsofthecylinderaremadeofaheatinsulator,and
thepistonisinsulatedbyhavingapileofsandonit.Bywhatfactordoesthepressureof
thegasincreaseifthegasiscompressedtohalfitsoriginalvolume?
Ans.Thecylinderiscompletelyinsulatedfromitssurroundings.Asaresult,noheatis
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exchangedbetweenthesystem(cylinder)anditssurroundings.Thus,theprocessis
adiabatic.
Initialpressureinsidethecylinder=
Finalpressureinsidethecylinder=
Initialvolumeinsidethecylinder=
Finalvolumeinsidethecylinder=
Ratioofspecificheats,Y=1.4
Foranadiabaticprocess,wehave:
Thefinalvolumeiscompressedtohalfofitsinitialvolume.
∴
Hence,thepressureincreasesbyafactorof2.639.
5.InchangingthestateofagasadiabaticallyfromanequilibriumstateAtoanother
equilibriumstateB,anamountofworkequalto22.3Jisdoneonthesystem.Ifthegas
istakenfromstateAtoBviaaprocessinwhichthenetheatabsorbedbythesystemis
9.35cal,howmuchisthenetworkdonebythesysteminthelattercase?(Take1cal=
4.19J)
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Ans.Theworkdone(W)onthesystemwhilethegaschangesfromstateAtostateBis22.3J.
Thisisanadiabaticprocess.Hence,changeinheatiszero.
∴ΔQ=0
ΔW=–22.3J(Sincetheworkisdoneonthesystem)
Fromthefirstlawofthermodynamics,wehave:
ΔQ=ΔU+ΔW
Where,
ΔU=Changeintheinternalenergyofthegas
∴ΔU=ΔQ–ΔW=–(–22.3J)
ΔU=+22.3J
WhenthegasgoesfromstateAtostateBviaaprocess,thenetheatabsorbedbythesystem
is:
ΔQ=9.35cal=9.35 4.19=39.1765J
Heatabsorbed,ΔQ=ΔU+ΔQ
∴ΔW=ΔQ–ΔU
=39.1765–22.3
=16.8765J
Therefore,16.88Jofworkisdonebythesystem.
6.TwocylindersAandBofequalcapacityareconnectedtoeachotherviaastopcock.
Acontainsagasatstandardtemperatureandpressure.Biscompletelyevacuated.The
entiresystemisthermallyinsulated.Thestopcockissuddenlyopened.Answerthe
following:
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(a)WhatisthefinalpressureofthegasinAandB?
(b)Whatisthechangeininternalenergyofthegas?
(c)Whatisthechangeinthetemperatureofthegas?
(d)Dotheintermediatestatesofthesystem(beforesettlingtothefinalequilibrium
state)lieonitsP-V-Tsurface?
Ans.(a)0.5atm
(b)Zero
(c)Zero
(d)No
Explanation:
(a)ThevolumeavailabletothegasisdoubledassoonasthestopcockbetweencylindersA
andBisopened.Sincevolumeisinverselyproportionaltopressure,thepressurewill
decreasetoone-halfoftheoriginalvalue.Sincetheinitialpressureofthegasis1atm,the
pressureineachcylinderwillbe0.5atm.
(b)Theinternalenergyofthegascanchangeonlywhenworkisdonebyoronthegas.Since
inthiscasenoworkisdonebyoronthegas,theinternalenergyofthegaswillnotchange.
(c)Sincenoworkisbeingdonebythegasduringtheexpansionofthegas,thetemperature
ofthegaswillnotchangeatall.
(d)Thegivenprocessisacaseoffreeexpansion.Itisrapidandcannotbecontrolled.The
intermediatestatesdonotsatisfythegasequationandsincetheyareinnon-equilibrium
states,theydonotlieontheP-V-Tsurfaceofthesystem.
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
5MarksQuestions
1.Derivetheequationofstateforadiabaticchange?
Ans.LetP=pressure,V=volumeandT=Temperatureofthegasinacylinderfittedwitha
perfectly
frictionlesspiston.
SupposeasmallamountofheatdQisgiventothesystem.Theheatisspentintwoways:-
1)InincreasingthetemperatureofthegasbylasmallrangedT,atconstantvolume
2)Inexpansionofgasbyasmallvolumedv
So,dQ=CVdT+PdV
Inadiabaticchange,noheatissuppliedfromoutside
So,dQ=O
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CVdt+PdV=O→(1)
Acc.tostandardgasequation
PV=RT
Diffbothsides
PdV+VdP=RdR
RdT=PdV+VdP(dR=OasRisaconstant)
dT=
Usingthisinequationi)
Cv
CVPdV+CVVdP+RPdV=O
(CV+R)PdV+CVVdP=O→2)
As,CP–CV=R
orCP=R+CV
Soequation2)becomes
CPPdV+CVVdP=O
DividingaboveequationbyCVPV
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Integratingbothsides
LogeV+LogeP=constant
Loge +LogeP=constant
LogeP =constant
=antilog(constant)
K=anotherconstant
2.Deriveanexpressionfortheworkdoneduringisothermalexpansion?
Ans.Consideronegrammoleofidealgasinitiallywithpressure,volumeandtemperatureas
P,V,T,LetthegasexpandtoavolumeV2,whenpressurereducestoP2andatthesame
temperatureT
IfA=Areaofcross–sectionofpiston
Force=Pressure×Area
F=PxA
Ifweassumethatpistonmovesadisplacementdx,
theworkdone:→dw=Fdx
dw=P×A×dx
dw=P×dv
TotalworkdoneinincreasingthevolumefromV1toV2
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W=
Since,PV=RT(fromidealgasequation)
P=
W=
W=RT
W=RTLoge
W=RT
W=RTLoge
W=2.3026RTLog10
AsP1V1=P2V2
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SoW=2.3026RTLog10
3.BrieflydescribeaCarnotcycleandderiveanexpressionfortheefficiencyofCarnot
cycle?
Ans.TheconstructionofaheatenginefollowingCarnotcycleis:-
1)Sourceofheat:-ItismaintainedathighertemperatureT1
2)Sinkofheat–ItismaintainedatlowertemperatureT2
3)Workingbase:-Aperfectidealgasistheworkingsubstance.
Theory:-Carnotcycleconsistoffourstages:-
1)Isothermalexpansion
2)Adiabaticexpansion
3)Isothermalcompression
4)Adiabaticcompression.
4.Discussbrieflyenergydistributionofablackbodyradiation.Hencededucewien’s
displacementlaw?
Ans.Forablackbody,themonochromaticemittance oftheblackbodyandthe
wavelength oftheradiationemitted.
So,atagiventemperatureofblackbody:→
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a)Theenergyemittedisnotdistributeduniformlyamongstallwavelengths.
b)Theenergyemittedinmaximumcorrespondingtoacertainwavelength andits
fallsoneithersideofit.
Astemperatureofblackbodyisincreased.
a)Thetotalenergyemittedrapidlyincreasesforanygivenwavelength.
b)Thewavelengthcorrespondingtowhichenergyemittedismaximumisshiftedtowards
shorterwavelengthsidei.e, mdecreaseswithriseintemperature
or mT=constant
Thusisthewein’sdisplacementlaw.
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CBSEClass11Physics
Chapter-12(Thermodynamics)
NUMERICALS
1. WhenasystemistakenfromstateAtostateBalongthepathACB,80kcalofheat
flowsintothesystemand30kcalofworkisdone.
a. HowmuchheatflowsintothesystemalongpathADBiftheworkdoneis10k
Cal?
b. WhenthesystemisreturnedfromBtoAalongthecurvedpaththeWorkdoneis
20kcal.Doesthesystemabsorborlibrateheat.
c. IfUA=0andUD=40kcal,findtheheatabsorbedintheprocessAD
Ans.
a. dWADB=+10kcal
Internalenergyispathindependent
dQADB=duACB=50kcal
dQADB=50+10=60kcal.
b.
=-50-20=-70kcal
c. UA=0UD=40kcal
dWADB=10kcal
dWDB=0sincedV=0
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dQAD=40+10=50kcal
2. moleofheliumiscontainedinacontaineratS.T.P.Howmuchheatenergyisneeded
todoublethepressureofthegas,keepingthevolumeConstant?Heatcapacityofgas
is3Jg-1K-1.
Ans.
CV=MCV=12J/molekM→Molecularmass
3. Thevolumeof steamproducedby1gofwaterat 100°C is 1650cm3.Calculate the
changeininternalenergyduringthechangeofstategivenJ=4.2x107ergcal-1g=98
Jcm/s2?
latentheatofsteam=540Cal/g
Ans.Massofwater=1g=10-3kg
Volumeofwater
=1cm3
Changeinvolume=1650-1=1649cm3
dQ=ML=540cal=540×4.2×107erg
P=1atm=76×13.6×981
Du=dQ-pdv=22.68×109-1.67×109
=21.01×109erg.
4. What is the coefficient of performance ( ) of a carnot refrigerator working
between30°CandO°C?
Ans.
5. Calculatethefallintemperaturewhenagasinitiallyat72°Cisexpandedsuddenly
toeighttimesitsoriginalvolume.( =5.3)
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Ans.
6. Asteamengineintakesteamat200°Candafterdoingworkexhaustsitdirectlyinair
at100°Ccalculatethepercentageofheatusedfordoingwork.Assumetheengineto
beanidealengine?
Ans.
7. A perfect Carnot engine utilizes an ideal gas the source temperature is 500K and
sink temperature is 375K. If the engine takes 600kcal per cycle from the Source,
Calculate
a. Theefficiencyofengine
b. Workdonepercycle
c. Heatrejectedtosinkpercycle
Ans.
a. T1=500KT2=375K
Q1=Heatabsorbed=600kcal
=25%
b.
c. w=Q1-Q2Q2=Q1-W=600-150
=450kcal
8. TwocarnotenginesAandBareoperatedinseries.ThefirstoneAreceivesheatat
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900KandrejecttoareservoirattemperatureTK.ThesecondengineBreceivesthe
heat rejected by the first engine and in turn rejects to a heat reservoir at 400 K
calculatethetemperatureTwhen
i. Theefficienciesofthetwoenginesareequal
ii. Theworkoutputofthetwoenginesareequal
Ans.WA=WB
T=650K
T2=900×400
=600K
T1=273KT2=673K
Massofgas=10mole
=-8.4×104Jworkbeingdoneonthegas
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Du=-dw=8.4×104J
9. TenmoleofhydrogenatNTPiscompressedadiabaticallysothatitstemperaturebecome
400°CHowmuchworkisdoneonthegas?whatistheincreaseintheinternalenergyof
thegas
R=8,4Jmol-1K-1 =1.4
10. ThetemperatureT1andT2of thetwoheatreservoirs inan idealcarnotenginebe
1500°Cand500°Crespectively,whichoftheseincreasingT1by100°CordecreasingT2
by100°Cwouldresultinagreaterimprovementintheefficiencyoftheengine.
Ans.
i. T1isincreasedfrom1500°Cto1600°C
T1=1873K
T2remainconstantT2=773k
ii. T1remainconstant1500°C
T1=1500+273=1773k
T2isdecreasedby100i.e.400°C
T2=400+273=673k
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CBSEClass11Physics
Chapter-12(Thermodynamics)
ShortAnswerTypeQuestion(2Marks)
1. AThermosbottlecontainingteaisvigorouslyshaken.Whatwillbetheeffectonthe
temperatureoftea.
Ans.Temperatureofteawillrise.
2. Writetwolimitationofthefirstlawofthermodynamics.
Ans.
(i)itdoesnotgivethedirectionofflowofheat.
(ii)ItdoesnotexplainwhyheatcarnotbespontaneouslyconvertedintoWork.
3. WritetheexpressionsforCv,andCpofagasintermsofgasconstantRand where
Ans. =Cp/Cv
Cp-Cv=R
Cp= Cv
4. NorealenginecanhaveanefficiencygreaterthatofacarnotengineWorking
betweenthesametowtemperatures.Why?
Ans.InCarnotengine,
i. Thereisabsolutelynofrictionbetweenthewallofcylinderandpiston.
ii. Workingsubstanceisanidealgas
Inrealenginetheseconditioncarnotbefulfilled.
5. Whywateratthebaseofawaterfallisslightlywarmerthanatthetop?
Ans.Potentialenergyconvertedintokineticenergy,somepartofkineticenergyis
Convertedintoheat.
6. Whenicemelts,thechangeininternalenergyisgreaterthantheheatsupplied.
Why?
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Ans.dq-du+dw
Du=dq-pdv
Whenicemeltchangeinvolumeisnegative
7. ExplainwhytwoisothermalCurvesneverintersect.
Ans.PV
8. AnidealmonatomicgasistakenroundthecycleABCDAasshown.Calculatethe
workdoneduringthecycle.
Ans.PV
9. Canaroombecooledbyopeningthedoorofrefrigeratorinaclosedroom?
Ans.No,ItaVoiletsSecondslaw.
10. ExplainWhatismeantbyisothermalandadiabaticoperations.
Ans.AdiabaticaProcess-Pressure,volumeandtemperatureofthesystemchangesbut
thereisnoexchangeofheat.
IsothermalProcess-Pressure,volumechangestemperatureremainConstant.
11. TwobodiesatdifferenttemperaturesT,andTifbroughtinthermalcontactdonot
necessarilysettletothemeantemperature(T1+T2)/2Explain?
Ans.Heatflowsfromhighertemperaturetolowertemperatureuntilthetemperature
becomeequalonlywherethethermalcapacitiesoftwobodiesareequal.
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CBSEClass11Physics
Chapter-12(Thermodynamics)
VeryShortAnswerTypeQuestion(1Marks)
1. Whysparkisproducedwhentwosubstancesarestruckhardagainsteachother?
Ans.Workisconvertedintoheat.
2. WhatistheSpecificheatofagasinanisothermalprocess.
Ans.Infinite
3. Onwhatfactors,doestheefficiencyofCarnotenginedepend?
Ans.
4. WhataretwoessentialfeaturesofCarnot'sidealheatengine.
Ans.
(i)Sourceandsinkhaveinfiniteheatcapacities.
(ii)Eachprocessoftheengine'scycleisfullyreversible
5. PlotagraphbetweeninternalenergyUandTemperature(T)ofanidealgas.
Ans.
6. Refrigeratortransfersheatfromcoldbodytoahotbody.Doesthisviolatethe
secondlawofthermodynamics?
Ans.No,ExternalWorkisdone
7. Whatisheatpump?
Ans.Aheatpumpisadevicewhichusesmechanicalworktoremoveheat.
8. Givetwoexampleofheatpump?
Ans.Refrigerator,AirConditioner.
9. Whatisheatengine?
Ans.Heatengineisadevicewhichconvertheatenergyintomechanicalenergy.
10. Whyagasiscooledwhenexpanded?
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Ans.Decreaseininternalenergy.
11. Canthetemperatureofanisolatedsystemchange?
Ans.Yesinanadiabaticprocess
12. Whichoneasolid,aliquidoragasofthesamemassandatthesametemperature
hasthegreatestinternalenergy.
Ans.Gashasgreatestinternalenergyandsolidhasleastinternalenergy.
13. UnderwhatidealconditiontheefficiencyofaCarnotenginebe100%.
Ans.IfthetemperatureofsinkisOK.
14. Whichthermodynamicvariableisdefinedbythefirstlawofthermodynamics?
Ans.Internalenergy.
15. Giveanexamplewhereheatbeaddedtoasystemwithoutincreasingits
temperature.
Ans.Melting.
16. WhatistheefficiencyofcarnotengineWorkingbetweenicepointandsteampoint?
Ans.
17. Twoblocksofthesamemetalhavingmasses5gand10gcollideagainstatargetwith
thesamevelocity.Ifthetotalenergyusedinheatingtheballswhichwillattain
highertemperature?
Ans.Boththeballswillundergothesameriseintemperature.
18. Whatisthespecificheatofagasinanadiabaticprocess.Workisconvertedinto
heat.
Ans.Zero.
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
1MarksQuestions
1.Ifaairisacylinderissuddenlycompressedbyapiston.Whathappenstothe
pressureofair?
Ans.Sincethesuddencompressioncausesheatingandriseintemperatureandifthepiston
ismaintainedatsamePositionthenthepressurefallsastemperaturedecreases.
2.Whatistheratiooffindvolumetoinitialvolumeifthegasiscompressed
adiabaticallytillitstemperatureisdoubled?
Ans.SinceforanadiabaticProcess,
PVY=constant
SincePV=RT
P=
So, constant
OrTVy-1=constantT1,V1=InitialtemperatureandInitialVolume
∴T1V1y-1=T2V2y-1T2,V2=FinaltemperatureandFinalvolume.
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SinceT2=2T1(Given)
So,
Since >1, islessthan .
3.WhatistheratioofslopesofP-Vgraphsofadiabaticandisothermalprocess?
Ans.TheslopeofP-Vgraphis
Foranisothermalprocess,(PV=constant)
So,
Foranadiabaticprocess(PVY=constant)
Divide2)by1)
So,theratioofadiabaticslopetoisothermalslopeisY.
4.WhatisthefoundationofThermodynamics?
Ans.Thefoundationofthermodynamicsisthelawofconservationofenergyandthefactthe
heatflowsfromahotbodytoacoldbody.
5.Differentiatebetweenisothermalandadiabaticprocess?
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Ans.
Isothermalprocess
Adiabaticprocess
1) Inthis,temperatureremainsconstant 1) Inthis,noheatisaddedorremoved.
2) Itoccursslowly 2) Itoccurssuddenly.
3)Here,systemisthermallyconductingto
surroundings3)
Here,systemisthermallyinsulatedfrom
surroundings.
4) Stateequation:→PV=constant 4)Stateequation:→PVY=constant.
6.ACarnotenginedevelops100H.P.andoperatesbetween270Cand2270C.Find1)
thermalefficiency;2)heatsupplied3)heatrejected?
Ans.Here,energy=W=100H.P.
=100×746W(1H.P.=746W)
=
Hightemperature,TH=2270C=227+273=500K
Lowtemperature,Th=270C=27+273=300K
1)Thermalefficiency,
2)TheheatsuppliedQHisgivenby:-
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3)TheheatrejectedQLisgivenby:-
or
7.Drawap–vdiagramforisothermalandadiabaticexpansion?
Ans.
8.Statezerothlawofthermodynamics?
Ans.Acc.tothis,whenthethermodynamicsystemAandBareseparatelyinthermal
equilibriumwithathirdthermodynamicsystemC,thenthesystemAandBareinthermal
equilibriumwitheachotheralso.
9.Canagasbeliquefiedatanytemperaturebyincreaseofpressurealone?
Ans.No,agascanbeliquefiedbypressurealone,onlywhentemperatureofgasisbelowits
criticaltemperature.
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10.Canyoudesignheatenergyof100%efficiency?
Ans.Sinceefficiencyofheatengine=1- ,so,efficiencywillbe100%or1ifT2=OKorT1=
α.Sinceboththeseconditionscannotbepracticallyattained,soheatenginecannothave
100%efficiency.
11.Ifairisabadconductorofheat,whydowenotfeelwarmwithoutclothes?
Ans.Thisisbecausewhenwearewithoutclothesaircarriesawayheatfromourbodydueto
convectionandwefeelcold.
12.Abodywithlargereflectivityisapooremitterwhy?
Ans.Thisisbecauseabodywithlargereflectivityisapoorabsorberofheatandpoor
absorbersarepooremitters.
13.Animalscurlintoaball,whentheyfeelverycold?
Ans.Whenanimalscurl,theydecreasetheirsurfaceareaandsinceenergyradiatedvaries
directlytosurfaceareahencelossofheatduetoradiationisalsoreduced.
14.Whyistheenergyofthermalradiationlessthanthatofvisiblelight?
Ans.Theenergyofanelectromagneticwareisgivenby:-E=hf
h=Planck’sconstant;f=frequencyofwave.Sincethefrequencyofthermalradiationisless
thanthatofvisiblelight,theenergyassociatedwiththermalradiationislessthanassociated
withvisiblelight.
15.TworodsAandBareofequallength.EachrodhasitsendsattemperatureT1andT2
(T1>T2).WhatistheconditionthatwillensureequalratesofflowthroughtherodsA
andB?
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Ans.Heatflow,
K=Thermalconductivity
A=Area
T1=Temperatureofhotbody
T2=Temperatureofcoldbody
d=distancebetweenhotandcoldbody.
Q=heatflow
Whentherodshavethesamerateofconduction,
Q1=Q2
K1,K2→Thermalconductivityoffirstandsecondregion
A1,A2→Areaoffirstandsecondregion
or,K1A1=K2A2
or
16.ASphereisatatemperatureof6ook.ItscoolingrateisRinanexternal
environmentof200k.Iftemperaturefallsto400k.WhatisthecoolingrateR1interms
ofR?
Ans.Acc.toStefan’slaw;
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E=constantT4
Also,R1=constant(T24–T1
4)
R=constant(T34–T1
4)
T2=heatofhotjunction=400K
T1=heatofcoldjunction=200K
T3=heatofhotjunction=600K
R1=constant
R1=constant
Divideeq41)&2)
Therefore,
17.Ifthetemperatureofthesunisdoubled,therateofenergyreceivedoneachwill
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increasesbywhatfactor?
Ans.ByStefan’slaw:→
RateofenergyradiatedαT4
T=Temperature
E1=constantT14
E2=constantT24
T1=Initialtemperature
T2=Finaltemperature
T2=2T1
T24=(2)4T1
4
T24=16T1
4
E2=constant(16T14)
E2=16(constantT14)
E2=16E1
18.Onawinternight,youfeelwarmerwhencloudscovertheskythanwhenskyis
clear.Why?
Ans.Weknowthatearthabsorbsheatindayandradiatesatnight.Whenskyiscovered,with
clouds,theheatradiatedbyearthisreflectedbackandearthbecomeswarmer.Butifskyis
cleartheheatradiatedbyearthescapesintospace.
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19.Ifabodyisheatedfrom270Cto9270Cthenwhatwillbetheratioofenergiesof
radiationemitted?
Ans.Since,ByStefan’slaw:→
E=Energyradiated
T=Temperature.
E1,T1 Initialenergyandtemperature
E2,T2 Finalenergyandtemperature.
T1=270C=27+273=300K
T2=9270C=927+273K=1200K.
E=constantT4
So,E1=constantT14
Equatingequation1)&2)
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orE1:E2=1:256
20.Whichhasahigherspecificheat;waterorsand?
Ans.Waterhashigherspecificheatthansandas
,whereT=Temperature,Q=Heat,m=Mass,
C=Specificheat;Sinceforwatertemperatureincreaseslessslowlythansandhencethe
result.
21.Whyislatentheatofvaporizationofamaterialgreaterthanthatoflatentheatof
fusion?
Ans.Whenaliquidchangesintoagas,thereislargeincreaseinthevolumeandalarge
amountofworkhastobedoneagainstthesurroundingatmosphereandheatassociated
withchangefromsolidtogasislatentheatofvaporizationandhencetheanswer.
22.DrawaP–VdiagramforLiquidandgasatvarioustemperaturesshowingcritical
point?
Ans.
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23.Whyistemperaturegradientrequiredforflowofheatfromonebodytoanother?
Ans.Heatflowsfromhighertemperaturetolowertemperature.Therefore,temperature
gradient(i.e.temperaturedifference)isrequiredfortheheattoflowonepartofsolidto
another.
24.WhyareCalorimetersmadeupofmetalonly?
Ans.Calorimetersaremadeupofmetalonlybecausetheyaregoodconductorofheatand
hencetheheatexchangeisquickwhichthebasicrequirementfortheworkingof
calorimeter.
25.Ifabodyhasinfiniteheatcapacity?Whatdoesitsignify?
Ans.Infiniteheatcapacitymeansthattherewillbenochangeintemperaturewhetherheat
istakenoutorgiventothesubstance.
26.Definetriplepointofwater?
Ans.Triplepointofwaterrepresentsthevaluesofpressureandtemperatureatwhichwater
co-existsinequilibriuminallthethreestatesofmatter.
27.StateDulongandpetitlaw?
Ans.Acc.tothislaw,thespecificheatofallthesolidsisconstantatroomtemperatureandis
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equalto3R.
28.Whytheclockpendulumsaremadeofinvar,amaterialoflowvalueofcoefficient
oflinearexpansion?
Ans.TheclockpendulumsaremadeofInverbecauseithaslowvalueofα(co-efficientof
linearexpansion)i.e.forasmallchangeintemperature,thelengthofpendulumwillnot
changemuch.
29.Whydoesthedensityofsolid|liquiddecreaseswithriseintemperature?
Ans.LetP=Densityofsolid|liquidattemperatureT
P1=Densityofsolid|liquidatTemperatureT+∆T
SinceDensity=
So,P= →(1)P1= (2)
V1=VolumeofsolidattemperatureT+∆T
V=VolumeofsolidattemperatureT
Sinceonincreasingthetemperature,solids|liquidsexpandthatistheirvolumesincreases,
sobyequation
i)&2)Densityisinverselyproportionaltovolumes,soifvolumeincreasesonincreasingthe
temperature,Densitywilldecrease.
30.TwobodiesatdifferenttemperaturesT1,andT2arebroughtinthermalcontactdo
notnecessarilysettledowntothemeantemperatureofT1andT2?
Ans.TwobodiesatdifftemperaturesT1andT2wheninthermalcontactdonotsettlealways
attheirmeantemperaturebecausethethermalcapacitiesoftwobodiesmaynotbealways
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equal.
31.Theresistanceofcertainplatinumresistancethermometerisfoundtobe2.56Ωat
00cand3.56Ωat1000c.Whenthethermometerisimmersedinagivenliquid,its
resistanceisobservedto5.06Ω.Determinethetemperatureofliquid?
Ans.Ro=Resistanceat00c=2.56Ω
Rt=ResistanceattemperatureT=1000c=3.56Ω100
Rt=Resistanceatunknowntemperaturet;
Rt=5.06Ω
Since,
t=
=
=
=
t=2500c
32.CalculateCpforair,giventhatCv=0.162calg-1k-1anddensityairatN.T.Pis
0.001293g|cm3?
Ans.Specificheatatconstantpressure=Cp=?
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Specificheatatconstantvolume=Cv=0.162Calg-1k-1
Now,Cp–Cv=
OrCP–Cv=
Cp–Cv=
=
=
=6.8×10-4+2
Cp–Cv=0.068
Cp=0.162+0.068
Cp=0.23Calg-1k-1
33.Developarelationbetweentheco-efficientoflinearexpansion,co-efficient
superficialexpansionandcoefficientofcubicalexpansionofasolid?
Ans.Since,co-efficientoflinearexpansion=α=
∆L=changeinlength
L=length
∆T=changeintemperature
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Similarly,co-efficientofsuperficialexpansion=β=
∆S=changeinarea
S=originalarea
∆T=changeintemperature
Co-efficientofcubicalexpansion,=Y=
∆V=changeinvolume
V=originalvolume
∆T=changeintemperature.
Now,∆L=αL∆T
L+∆L=L+αL∆T
L+∆L=L(1+α∆T)→(1)
SimilarlyV+∆V=V(1+Y∆T)→(2)
AndS+∆S=S(1+β∆T)→(3)
Also,(V+∆V)=(L+∆L)3
V+∆V=
V+∆V=L3
Sinceα2,α3arenegligible,so,
V+γV∆T=V(1+3α∆T)[asL3=V]
So,V+γV∆T=V+V3α∆T
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γV∆T=3α∆T
Y=3α
Similarly,β=2α[usingL2=S(Area)]
So,
34.Calculatetheamountofheatrequiredtoconvert1.00kgoficeat–100cintosteamat
1000catnormalpressure.Specificheatofice=2100J|kg|k.Latentheatoffusionofice=
3.36x105J|kg,specificheatofwater=4200J|kg|k.Latentheatofvaporizationofwater=
2.25x106J|kg?
Ans.(1)Here,heatisrequiredtoraisethetemperatureoficefrom–100cto00c.
So,changeintemperature=∆T=T2-T1=0-(-10)=100c
So,∆Q1=cm∆T
C=specificheatofice
M=Massofice
∆T=100c
∆Q1=2100×1×10=21000J
(2)Heatrequiredtomelttheiceto00cwater:-
∆Q2=mL
L=Latentheatoffusionofice=3.36×105J/kg
m=Massofice
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∆Q2=1×3.36×105J/kg
∆Q2=3.36×105J
∆Q2=336000J
(3)Heatrequiredtoraisethetemperatureofwaterfrom00cto1000c:-
∆T=T2-T1=100-0=1000c
∆Q3=cm∆Tc=specificheatofwater
=4200×1×100
=420,000J
(4)Heatrequiredtoconvert1000cwatertosteamat1000c
∆Q4=mLL=Latentheatofvapourisation=2.25×106J/kg
∆Q4=1×2.25×106J|kg
∆Q4=2250000J
∴TotalHeatrequired=∆Q1+∆Q2+∆Q3+∆Q4
∆Qtotal=21000+336000+420000+2250000
∆Qtotal=3027000J
∆Qtotal=3.027x106J
35.Whyismercuryusedinmakingthermometers?
Ans.Mercuryisusedinmakingthermometersbecauseithaswideandusefultemperature
rangeandhasauniformrateofexpansion.
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36.Howwouldathermometerbedifferentifglassexpandedmorewithincreasing
temperaturethanmercury?
Ans.Ifglassexpandedmorewithincreasingtemperaturethanmercury,thescaleofthe
thermometerwouldbeupsidedown.
37.Showthevariationofspecificheatatconstantpressurewithtemperature?
Ans.
38.Twothermometersareconstructedinthesamewayexceptthatonehasaspherical
bulbandtheotheranelongatedcylindricalbulb.Whichonewillresponsequicklyto
temperaturechange?
Ans.Thethermometerwithcylindricalbulbwillrespondquicklytotemperaturechanges
becausethesurfaceareaofcylindricalbulbisgreaterthantheofsphericalbulb.
39.StateCarnot’sTheorem?
Ans.AccordingtoCarnot’sTheorem,noengineworkingbetweentwotemperaturescanbe
moreefficientthanaCarnot’sreversibleengineworkingbetweenthesametemperatures.
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
2MarksQuestionsPart1
1.AmotorcartyrehasaPressureoffouratmosphereataroomtemperatureof270C.If
thetyresuddenlybursts,calculatethetemperatureofescapinggas?
Ans.Sincethetyresuddenlybursts,thechangetakingplaceisadiabatic,foradiabatic
change:-
Or
Hence,T1=273+27=300K
P1=InitialPressure;P2=finalPressure
So,
So,Puttingtheabovevaluesineq4i)
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Taking1.4Power
W1=-150J→(1)
WorkdonebythegasintheprocessB→Cis:→
Addingequationi)&2)
NetworkdonebythegasinthewholeprocessisW=W1+W2
W=150–70=-22OJ
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T2=201.8K
∴T2=201.8–273=-71.20C
2.HowdoesCarnotcycleoperates?
Ans.ACarnotcycleoperatesafollows:-
1)Itreceivesthermalenergyisothermallyfromsomehotreservoirmaintainedataconstant
hightemperatureTH.
2)Itrejectsthermalenergyisothermallytoaconstantlow–temperaturereservoir(T2).
3)Thechangeintemperatureisreversibleadiabaticprocess.
Suchacycle,whichconsistoftwoisothermalprocessesboundedbytwoadiabaticprocesses,
iscalledCarnotcycle.
3.CalculatetheworkdonebythegasingoingfromtheP-Vgraphofthethermodynamic
behaviorofagasfrompointAtopointBtopointC?
Ans.WorkdonebythegasintheprocessA→Bis
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W1=-(areaundercurveAB)
=-
=-
PAB=500Pa
=5×105N|m2
4.Whydoesabsolutezeronotcorrespondtozeroenergy?
Ans.Thetotalenergyofagasisthesumofkineticandpotentialenergyofitsmolecules.
Sincethekineticenergyisafunctionofthetemperatureofthegas.Henceatabsolutezero,
thekineticenergyofthemoleculesceasesbutpotentialenergyisnotzero.So,absolutezero
temperatureisnotthetemperatureofzeroenergy.
5.StatetheSecondlawofthermodynamicsandwrite2applicationsofit?
Ans.Accordingtosecondlawofthermodynamics,whenacoldbodyandahotbodyare
broughtintocontactwitheachother,heatalwaysfromhotBodytothecoldbody.Also,that
noheatenginethatworksincyclecompletelyconvertsheatintowork.
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Secondlawofthermodynamicsisusedinworkingofheatengineandofrefrigerator.
6.At00Candnormalatmosphericpressure,thevolumeof1gofwaterincreasesfrom
1cm3to1.091cm3onfreezing.Whatwillbethechangeinitsinternalenergy?Normal
atmosphericpressureis1.013x105N|m2andthelatentheatofmeltingoficeis80cal/g?
Ans.Since,heatisgivenoutby1gofwaterinfreezingis
m=Massofwater=1g
Q=-(mLf)Lf=Latentheatofmeltingofice=80cal|g
Duringfreezing,thewaterexpandsagainstatmosphericpressure.Hence,externalwork
done(W)bywateris:-W=P×∆V
P=1.013×105N|m2;∆V=1.091–1=0.091cm3=o.o91×10-6m3
∆V=V2–V1;V2=finalvolume=1.91cm3
V1=Initialvolume=1cm3
So,W=
W=0.0092J
Since,1cal=4.2Jso,
W=
Sincetheworkhasbeendonebyice,itwillbetakenpositive.
Acc.tofirstlawofthermodynamics,
Q=∆∪+W∆∪=changeininternalenergy
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So,∆∪=Q–W
=
∆∪=-80.0022cal
Negativesignindicatesthatinternalenergyofwaterdecreasesonfreezing.
7.TwodifferentadiabaticpathsforthesamegasintersecttwothermalsatT1andT2as
showninP-Vdiagram.Howdoes Comparewith ?
Ans.Now,ABandCDareisothermalsattemperatureT1andT2respectivelyandBCandAD
areadiabatic.
SincepointsAandDlieonthesameadiabatic.
T1VAY-1=T2VDY-1
Also,pointsBandClieonthesameadiabatic,
orT1VBY-1=T2VCY-1
Fromequation1)&2)
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8.Theinternalenergyofacompressedgasislessthanthatoftherarifiedgasatthe
sametemperature.Why?
Ans.Theinternalenergyofacompressedgasislessthanthatofrarifiedgasatthesame
temperaturebecauseincompressedgas,themutualattractionbetweenthemolecules
increasesasthemoleculescomesclose.Therefore,potentialenergyisaddedtointernal
energyandsincepotentialenergyisnegative,totalinternalenergydecreases.
9.ConsiderthecyclicprocessABCAonasample2molofanidealgasasshown.The
temperatureofthegasatAandBare300Kand500Krespected.Totalof1200Jofheatis
withdrawnfromthesample.FindtheworkdonebythegasinpartBC?
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Ans.Thechangeininternalenergyduringthecyclicprocessiszero.Therefore,heatsupplied
tothegasisequaltoworkdonebyit,
∴WAB+WBC+WCA=-1200J→(1)
(-vebecausethecyclicprocessistracedanticlockwisethenetworkdonebythesystemis
negative)
TheworkdoneduringtheprocessABis
WAB=PA(VB-VA)=nR(TB-TA)
WAB=2×8.3(500-300)=3320J→2)
R=Universalgasconstant
N=No.ofvolume
Sinceinthisprocess,thevolumeincreases,theworkdonebythegasispositive.
Now,WCA=O( volumeofgasremainsconstant)
∴3320+WBC+O=-1200(Usingequation1)&2)
WBC=-1200–3320
WBC=-4520J
10.Arefrigeratorplacedinaroomat300Khasinsidetemperature264K.Howmany
caloriesofheatshallbedeliveredtotheroomforeach1Kcalofenergyconsumedby
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therefrigerator,ideally?
Ans.Hightemperature,TH=300K
Lowtemperature,Th=264K
Energy=1Kcal.
Co-efficientofperformance,isgivenby:-
Now,COP=
QL=heatrejected
QL=COP×W
QL=
Themechanicalworkdonebythecompressoroftherefrigeratoris:-
W=QH–QL
QH=W+QL
QH=
QH=8.33Kcal
11.Ifthedoorofarefrigeratoriskeptopeninaroom,willitmaketheroomwarmor
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cool?
Ans.Sincearefrigeratorisaheatenginethatoperatesinthereversedirectioni.e.itextracts
heatfromacoldbodyandtransformsittohotbody.Sinceitexhaustmoreheatintoroom
thanitextractsfromit.Therefore,theneteffectisanincreaseintemperatureoftheroom.
12.ThefollowingfigureshowsaprocessABCAperformedonanidealgas,findthenet
heatgiventothesystemduringtheprocess?
Ans.Sincetheprocessiscyclic,thechangeininternalenergyiszero.Therefore,theheat
giventothesystemisequaltoworkdonebyit.Thenetworkdonebythegasintheprocess
ABCAis:-
W=WAB+WBC+WCA
NowWAB=O
DuringthepathBC,temperatureremainsconstant.Soitisanisothermalprocess.So,WBC=
nRT2Loge
DuringtheCA,VαTsothat isconstant.
∴WorkdonebythegasduringthepartCAis:-
WCA=P(V1–V2)
=nR(T1–T2)
=-nR(T2–T1)→Usingequation1)
W=O+nRT2Loge -nR(T2–T1)
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13.Acertaingasatatmosphericpressureiscompressedadiabaticallysothatitsvolume
becomeshalfofitsoriginalvolume.Calculatetheresultingpressure?
Ans.Lettheoriginalvolume,V1=V
∴finalvolumeV2= (∴volumebecomehalf)
InitialpressureP1=o.76mofHgcolumn
FinalpressureP2aftercompression=?
Asthechangeisadiabatic,so
Y= =1.4forair
P2=P1
=0.76×
P2=0.76×(2)1.4
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P2=2mofHgcolumn
P2=hsg
P2=2.672x105N|m2
P2=2×(13.6x103)×9.8
h=heightofHgcolumn
s=Densityofair
g=Accelerationduetogravity
14.Whyisconversionofheatintoworknotpossiblewithoutasinkatlower
temperature?
Ans.Forconvertingheatenergyintoworkcontinuouslyapartofheatenergyabsorbed
fromthesourcehastoberejected.Theheatenergycanberejectedonlytoabodyatlower
temperaturewhichissink,sowerequireasinktoconvertheatintowork
15.Writethesignconventionsfortheheatandworkdoneduringathermodynamic
process?
Ans.1)WhenheatissuppliedtoasystemdQistakenpositivebutwhenheatissuppliedbya
system,dQistakennegative.
2)Whenagasexpands,dwistakenaspositivebutwhenagascompresses,workdoneis
takenasnegative.
16.Doestheworkingofanelectricrefrigeratordefysecondlawofthermodynamics?
Ans.No,itisnotagainstthesecondlaw;thisisbecauseexternalworkisdonebythe
compressororforthistransferofheat.
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17.ACarnotengineabsorb6×105calat2270ccalculateworkdonepercyclebythe
engineifitsinkisat1270c?
Ans.Here,heatabsorbed=Q1=6×105cal.
Initialtemperature=T1=2270c=227+273=500K
Finaltemperature=T2=1270c=127+273=400K
As,forCarnotengine;
Q2=
Q2=4.8×105cal
Q2=Finalheatemitted
Asw=Q1–Q2=6×105–4.8×105
=1.2×105cal
Work=w=1.2×105×4.2J
Dore=5.04×105J
18.Howdoessecondlawofthermodynamicsexplainexpansionofgas?
Ans.Sincefromsecondlaw,.
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dS≥OdS=changeinentropy
Duringtheexpansionofgas,thethermodynamicprobabilityofgasislargerandhenceits
entropyisalsoverylarge.Sinceformsecondlaw,entropycannotdecrease∴followingthe
secondlaw,gasmoleculesmovefromonepartitiontoanother.
19.Whyisithotteratthesamedistanceoverthetopofthefirethaninfrontofit?
Ans.Atapointinfrontoffire,heatisreceivedduetotheprocessofradiationonly,whileata
pointabovethefire,heatreachesbothduetoradiationandconvection.Hencetheresult.
20.Ametalrodoflength20cmanddiameter2cmiscoveredwithanon-conducting
substance.Oneofitendsismaintainedat1000cwhiletheotherisat00c.Itisfoundthat
25goficemeltsin5mincalculatecoefficientofthermalconductivityofmetal?
Ans.Lengthofrod=∆x=20cm=2×10-3m
Diameter=2cm
R=10-2m
Areaofcross-section=πr2
=π(10-2)2
=10-4πsq.m
∆T=T2–T1=100–0=1000c
Massoficemelted=m=25g
Latentheatoffice=80cal/g
Heatconducted,∆Q=mL
=25x80
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=2000cal
=2000×4.2J
∆t=5min=300s
So,
K= =
=
K=1.78J|s|m|0c
K=coefficientofthermalconductivity
21.CalculatethetemperatureinKelvinatwhichaperfectlyblackbodyradiatesatthe
rateof5.67w/cm2?
Ans.E=5.67w|cm2;E=energyradiated
=5.67x107erg|s|cm2
=Stefan’sconstant=5.67×10-5ergs|s|cm2|K4,fromStefan’slaw
E=σT4
T=
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T=
22.Howdoyouexplaintheemissionoflong-wavelengthbytheobjectatlow
temperature?
Ans.SincebyWein’slaw:→
i.etemperatureisinverselyproportionaltothewavelengthso,iftemperatureisless,then
wavelengthwillbelong.Iftemperatureishigh,thenwavelengthwillbeshort.
23.Iftheradiationfromthemoongivesmaximaat =4700A0and =14x10-6m.What
conclusioncanbedrawnfromtheaboveinformation?
Ans.Acc.towien’sdisplacementlaw,
Now,accordingtothequestion, m=4700A0=4700×10-10m
T1=Temperatureofmoon,
T1=
b=2.9×10-3mK
Letthetemperaturecorrespondingto
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So,T2=
T2=
24.Differentiatebetweenconduction,convectionandradiation?
Ans.
Properties Conduction Convection Radiation
1.Material
MediumEssential Essential NotEssential
2. MoleculesDonotleavetheirmean
position
Morebodilyfromone
placetoanother.
Mediumdoesnot
playanypart
3.Transferof
heat
Canbeinanydirection
alonganypartOnlyverticallyupward
Inalldirectionin
straightlines
4.Speedof
transferofheatSlow Rapid
Fastestwiththe
speedoflight.
25.Thetilefloorfeelscolderthanthewoodenflooreventhoughbothfloormaterials
areatsametemperature.Why?
Ans.Thishappensbecausethetileisbetterheatconductorthanwood.Theheatconducted
fromourfoottothewoodisnotconductedawayrapidly.So,thewoodquicklyheatsuponits
surfacetothetemperatureofourfoot.Butthetileconductstheheatawayrapidlyandthus
cantakemoreheatfromourfoot,soitssurfacetemperaturedrops.
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
2MarksQuestionsPart2
26.Aroomhasa4mx4mx10cmconcreteroof(K1=1.26w|m|0C).Atsomeinstant,the
temperatureoutsideis460candradius320c.
1)Calculateamountofheatflowingpersecondintotheroomthroughtheroof.
2)Ifbricks(K2–0.56w|m|0c)ofthickness7.5cmarelaiddownonroof,calculatethe
newrateofheatflowunderthesametemperatureconditions?
Ans.1)Areaofroof=4×4=16m2
Thicknessofroof,x1=10cm=0.1m,
Thermalresistanceoftheroofisgivenby:-
∴Rateofheatflowthroughtheroofis:-
2)Thethermalresistanceofthebrickisgivenby:-
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Theequivalentthermalresistanceoftheroofnowis:→
∴Rateofheatthroughtheroofis:→
27.Abarocopperoflength75cmandabaroflength125cmarejoinedendtoend.Both
areofcircularcross–sectionwithdiameters2cm.Thefreeendsofcopperandsteelare
maintainedat1000cand00crespectively.Thesurfacesofthebarsarethermally
insulated.Whatisthetemperatureofcopper–steeljunction?Thermalconductivityof
copper=9.2x10-2kcal|m|0c|sandthatofsteelis1.1x10-2kcal|m|0c|s?
Ans.l1=lengthsofcopperbarsAB
l2=lengthofsteelbarsBC.
Θ1=temperatureoffreeendsA
Θ2=temperatureoffreeendsC.
Θ=temperatureofcopper–steel.
Insteadystate,theheatflowingpersecondthroughtwobarsisthesamei.e
H1=H2
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∴Temperatureofjunction=θ:→
28.TworodsAandBareofequallength.EachrodhasitsendsattemperaturesT1and
T2.WhatistheconditionthatwillensureequalratesofflowofheatthroughtherodsA
andB?
Ans.Sinceθ=
Θ=heatflow
K=co–efficientofthermalconductivity
A=Cross–SectionalArea
Θ1=Temperatureofhotbody
Θ2=Temperatureofcoldbody
X=distancebetweenhotandcoldfaces
t=time
ForrodA:
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Forequalratesofflow, KAAA=KBAB
29.Alayerofice10cmthickisformedonapond.Thetemperatureofairis–100C.
Calculatehowlongitwilltakeforthethicknessoficetoincreaseby1mm.Densityof
ice=1g|cm3;Thermalconductivityofice=0.005Cal|s|cm|0C;Latentheatofice=
80Cal|g?
Ans.Lett=timerequiredtoincreasethethicknessoficeby1mm(=0.1cm)
Massoficerequiredtobeformedis:-
m=VolumexDensity
LetA=Areaofuppersurface
Volume=AreaxThickness
=A×0.1
m=(A×0.1)×1
m=0.1Agram→1)
Now,heatmustflowfromlowersurfacetotheuppersurfaceoficeandfinallyinto
atmosphere.
Θ=heatthatflowsoutofpondintoatmosphere.
=Latentheatofice
m=Massofice
k=co–efficientofthermalconductivity
A=Cross–sectionalArea
t=time
x=Distancebetweenhotandcoldsurface
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θ1=temperatureofhotsurface
θ2=temperatureofcoldsurface
∴θ=mL;
Θ=0.1×A×80(Usingequation1)
Θ=8ACal→2)
But
Now,x=10cm,
K=0.005cal|cm|∆|0C
Θ1–θ2=0–(-10)=100C
30.TwoconductingslabsofthermalconductivitiesK1andK2arejoinedasshowninthe
figure.Thetemperatureoftheendsofslabareθ1andθ2(θ1>θ2).Findthefinal
temperatureof(θm)?
Ans.Letθ1=temperatureofhotslab
Θ2=temperatureofcoldslab
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K1=Co–efficientofthermalconductivityofhotslab
K2=Co–efficientofthermalconductivityofcoldslab
Θm=finaltemperature
d=Distanceb/whotandcoldsurface
A=Areaofcross–section
t=time
Now,sinceissteadystate,therateofheattransferinboththeslabsissamei.e
Θ1–θm=becausefirstheatflowsfromθ1tothejunction
Θ2–θm=thenheatflowsfromjunctiontosecondsurface
So,
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So,
31.Theendsofthetworodsofdifferentmaterialswiththeirthermalconductivities,
radiiofcross–sectionandlengthintheratio1:2aremaintainedatthesame
temperaturedifference.Iftherateofflowofheatthroughthelargerrodis4cal|s,
whatistherateofflowthroughtheshorterrod?
Ans.K1=thermalconductivityoffirstregion
K2=thermalconductivityofsecondregion
r1=radiusofcrosssectionoffirstregion
r2=radiusofcross–sectionofsecondregion
l1–lengthoffirstregion
l2=lengthofsecondregion
Θ1=heatflowoffirstregion
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Θ2=heatflowofsecondregion
Now,
Also,
and
Now,weknow,
So,Let
Now,Divideeq41)&2)
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Since
32.Whatarethermalradiation?Writeitspropertiesofthermalradiation?
Ans.Theradiantenergyemittedbeabodysolelyonaccountofitstemperatureiscalled
thermalradiation.
PropertiesofthermalRadiation:-
1)Theytravelthroughvacuum
2)Theyobeylawsofrefraction
3)Theycanberefracted
4)Theytravelwiththespeedoflight
5)Theydonotheatthemediumthroughwhichtheypasses.
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6)Theyexhibitphenomenaofinterference,diffractionandpolarization.
33.Anindirectlyheatedfilamentisradiatingmaximumenergyofwavelength2.16x10-
7m.Findthenetamountofheatenergylostpersecondperunitarea,thetemperature
ofsurroundingairis130C.Givenb=2.88x10-3mk,σ=5.77x10-8J|s|m2|k4?
Ans.ByWien’sLaw:-
Theproductofwavelength atwhichmaximumenergyisemittedandtheabsolute
temperature(T)oftheblackbodyisalwaysconstant.
i.e T=constant=b→(1)
b=Wien’sconstant=2.9×10-3mK
Now,
T=13333.3K
Now,Temperatureofsurrounding,To=13+273=286K.
Netamountofheatenergylostpersecondperunitarea:-
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E=1.824×108J/s/m2
34.Animalsintheforestfindshelterfromcoldinholesinthesnow.Why?
Ans.Animalsintheforestfindshelterfromcoldinholesinthesnowbecausesnowhas
trappedair(asinicethereisnoair)so,itactsasaheatinsulator.Therefore,thesnow
preventsthetransmissionofheatfromthebodyoftheanimaltotheoutside.
35.Abrassboilerhasabaseareaof0.15m2andthickness1.0cm.Itboilswateratthe
rateof6kg|minwhenplacedonagasstove,Estimatethetemperatureofthepartof
flameincontactwiththeboiler.Thermalconductivityofbrass=109J|s|m|0C,heatof
vaporizationofwater=2256J|g?
Ans.Rateofboilingofwateris=6.0Kg/min
=
=100g/s
∴Rateatwhichheatissuppliedbytheflametowateris:-
m=Rateofboilingofwater
L=heatofvaporizationofwater
θ=mL
=
Θ=225600J/s
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Now,T2=Temperatureofcoldjunction=1000C
T1=Temperatureofhotjunction
T2=Temperatureofcoldjunction
t=time
x=Distanceb/whotandcoldjunction
Now,x=1.0cm=1.0X10-2m
K=109J|s|m|0C
A=0.15m2
t=1s
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T1=237.980C
36.Howdoyouexplainheatingofroomsbasedonprincipleofconvection?
Ans.Convectionistheprocessbywhichheatistransmittedfromonepointtoanotherdueto
themovementofheatedparticlesofthesubstance.
Duringheatingoftheroombyaheater,theairmoleculesinimmediatecontactwithheater
areheatedup,theyacquiresufficientenergyandriseupward.Thecoolairparticlesnearto
theroofaredenseandmoredownandinturnitisheatedandthemovesupwards.Henceby
themovementofheatedairparticles,theentireroomheatsup.
37.Ifforagas, =0.67thenwhichgasisthis:-monatomic,diatomicandtriatomic?
Ans.Sinceforanidealgas,CP–CV=R→1)
CP=Specificheatatconstantpressure
CV=Specificheatatconstantvolume
R=UniversalGasConstant
Andgiven
or
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Andweknow,that =1.67isformonatomicgas;Sothegasismonatomicinquestion.
38.A50gleadbullet,specificheat0.02cal|g|0Cisinitiallyat300C.Itisfixedvertically
upwardwithaspeedof840m|sandonreturningtothestartinglevelstrikesacakeof
iceat00C.Howmuchiceismelted?Assumethatallenergyisspentinmeltingiceonly?
Ans.Speedofbullethittingtheice=V=840m|s
Heatproducedduetokineticenergyofthebullet:-=
Now,m=Massofbullet=
Hence
Now,heatgivenbybulletduetotemperaturedifference=mc
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m=Massofbullet
c=Specificheatofbullet
Q2=InitialTemperature
Q1=FinalTemperature
Totalheatgivenbybullet=4200+30=4230Cal.
Now,entireheatofbulletisusedinmeltingtheiceonly,LetM=MassofIcethatmelted
L=Latentheatofice
Hence
39.Agasmixtureconsistsof2molesofoxygenand4molesofargonattemperatureT.If
weneglectallvibrationmodes,findthetotalenergyofthesystem?
Ans.LetNA=Avogadro’sNumber
No.ofdegreesoffreedomofO2molecule(diatomic)=5
No.ofdegreesoffreedomof2molesofoxygen=2NA×5=10NA
No.ofdegreesoffreedomof4molesofargon(monatomic)=4NA×3
=12NA( 3=degreesoffreedom)
Totaldegreesoffreedomofmixture=10NA+12NA=22NA→1)
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Energyassociatedwitheachdegreeoffreedom|molecule=
Totalenergyofmixture=22NA× ( Usingequation1)
40.ShowthatCP-CV=RWhere[CP=specificheatatconstantpressure;CV=specific
heatatconstantvolumeandR=UniversalGasconstant]foranidealgas?
Ans.Now,Letfirstheatthegasatconstantvolumeandtemperatureincreasesby So,
Sincevolumeremainsthesame,hencenoworkisheatingthegasthenaccordingtolawof
conservationofenergy,theentireheatsuppliedgoesintoraisingtheinternalenergyand
hencethetemperatureofthegas.
Now,
∴∆U=increaseintheinternalenergyofthegasLetheatthegasatconstantpressureandifthetemperatureofthegasincreasesby∆Tbuthereexternalworkisdonetoexpandthegas
hence
But
Now,formidealgasequation:→
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Subtractingequation3)fromequation4)
Put
or
41.Howdoyoujustifythatwhenabodyisbeingheatedatmeltingpoint,the
temperatureremainsConstant?
Ans.Whenabodyisbeingheatedbelowthemaltingpoint,theheatsuppliedincreasesthe
potentialaswellasthekineticenergyofthemolecules.Duetotheincreaseinthekinetic
energyofthemoleclues,thetemperatureincreases.Butatmeltingpoint,heatgoes,to
increaseonlythepotentialenergyofmoleculesandhencethetemperatureremainsthe
same.
42.DrawandexplainaP–Tdiagramforwatershowingdifferentphases?
Ans.
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1)Thel-Vcurverepresentthosepointswheretheliquidandvapourphasesarein
equilibrium.
2)Thes–lcurverepresentthepointswherethesolidandliquidphasesexistinequilibrium.
3)Thes–visthesublimationcurvewhereasolidchangesintovapourphasewithout
passingthroughtheliquidstage
4)Triplepoint→Intersectionofthreecurvesisthetriplepoint.Itrepresentsauniquetemperatureandpressureanditisonlyatthispointthatthethreephasescanexisttogether
inequilibrium.
43.Fromwhatheightshouldapieceoficefallsothatitcompletelymelts?Onlyone–
quarterofheatproducedisabsorbedbytheice.Givenlatentheatoficeis3.4×105J|
Kgandaccelerationduetogravity,g=10m|s2?
Ans.Letm=Massofpieceofice
h=heightfromwhichitfalls.
∴LossofPotentialenergy=mgh
ThePotentialenergyoficeisconvertedintoheat.
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Sincetheiceabsorbsonlyone–quarterofthis,
∴Heatabsorbedbyice,
IfLJoules|Kgisthelatentheatofice,them
Equating1)&2)forQ
44.Agascanhaveanyvalueofspecificheatdependinguponhowheatingiscarriedout.
Explain?
Ans.Ifm=Massofgas
Q=heatsupplied
=Changeintemperature
thenspecificheatofgas,
1)Letgasiscompressedsuddenly,Sonoheatissuppliedfromoutside(i.e.Q=0)butthe
temperatureofthegasinthegasincreasesduetocompression,
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2)Letthegasisheatedinawaythatthetemperatureisconstant(∆T=O)then,
Hence,dependinguponconditionsofheating.ThevalueofCwillbedifferent.
45.A0.20Kgaluminumblockat800Cisdroppedinacoppercalorimeterofmass0.05Kg
containing200cm3ofethylalcoholat200C.Whatisthefinaltemperatureofthe
mixture?GivenDensityofethylalcohol=0.81g|cm3;specificheatofethylalcohol=
0.6cal|g|0C;specificheatofcopper=0.094cal|g|0C,specificheatofAl=0.22cal|g
|0C?
Ans.Letθ0C=finaltemperatureofthemixture.
Massofethylalcohol=volume×Density
=200×0.81
=162g
HeatlostbyAluminumblock=MassXspecificheatXfallintemperature
Heatgainedbytheethylalcoholandcalorimeter=(Massofethylalcohol×specificheat×
changeinTemperature)+Massofcoppercalorimeter×specificheatXchangein
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Temperature
But,Heatgained=HeatLost
So,fromequation1)&2)
46.WhyisthereadifferenceinthespecificheatcurveasgivenbyDelong’spetitlaw
andtheexperimentalresultatlowtemperatures?
Ans.Now,fromDulong&Petitlaw,thespecificheatisindependentoftemperaturebutitis
experimentallyseenthatspecificheatatlowertemperaturesisdirectlyproportionaltothe
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cubeoftemperatures.Theabovedependenceisbecauseofthefactthattheparticlesinthe
crystaloscillateasiftheyarecoupledQuantumHarmonicOscillator.
47.SpecificheatofArgonatconstantPressureiso.125cal|g|Kandatconstant
volumeis0.075cal|g|K.CalculatethedensityofargonatN.T.P.GiventhatJ=4.2J|
cal?
Ans.SpecificheatatconstantandPressure,CP=0.125cal|g|K
CP=0.125×4.2×1000J|Kg|K
CP=525J|Kg|K→1)
Specificheatatconstantvolume,CV=0.075cal|g|K
Thegasconstant,rforIkgofgasisgivenby:-
Normalpressure=P=hPg=0.76×13600×9.8=101292.8N|m2
NormalTemperature=T=273K.
SupposeV=Volumeofargoninm3atN.T.P.
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∴DensityofArgon,
48.HowisheatlossreducedinCalorimeter?
Ans.1)Heatlossduetoradiationisreducedbypolishinginnerandoutersurfacesofthe
Calorimeter.
2)Heatlossduetoconductionisreducedbyfillingthespacebetweenthecalorimeterand
insulatingjacketwithpoorconductorofheat.
3)Heatlossduetoconvectionisdonebyusingainsulatinglid.
49.Whatiscriticaltemperature?Howwillyoudifferentiatebetweenagasanda
vapourdependingoncriticaltemperature?
Ans.Thetemperatureabovewhichagasconnotbeliquefied,nomatterhowgreatthe
pressureiscalledcriticaltemperature.Ifthesubstanceliesabovethecriticaltemperature
thenitfallsinthegaseousregion.Ifthesubstanceliesbelowthecriticaltemperaturethanit
fallsinthevapourstage.
50.IfforhydrogenCP–CV=aandforoxygenCP–CV=bwhereCP&CVreferto
specificheatatconstantpressureandvolumethenwhatistherelationbetweenaand
b?
Ans.ForH2,CP–CV=a
CP=Specificheatatconstantpressure
CV=SpecificheatatconstantVolume
ForO2=CP–CV=b
Andr=
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M=Molecularweight
I=Mechaniccalequivalentofheat
Now,weknowthat,CP–CV=r
So,for
fromequation1)
2a=
fromequation2)
32b=
Equatingaboveequationsfor
2a=32b
a=16b
51.Aballisdroppedonafloorfromaheightof2cm.Afterthecollision,itrisesuptoa
heightof1.5m.Assumingthat40%ofmechanicalenergylostgoestothermalenergy
intotheball.Calculatetheriseintemperatureoftheballinthecollision.Specificheat
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capacityoftheballis800J/k.Takeg=10m/s2
Ans.Initialheight=h1=2m
Finalheight=h2=1.5m
Sincepotentialenergy=mechanicalenergyforabodyatrestasK.E=0
Mechanicalenergylost=
=
=
=5J
Now(mechanicalenergylost)×40%=heatgainedbyball
∆T=2.5×10-30C
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
2MarksQuestionsPart3
52.Athermometerhaswrongcalibration.Itreadsthemeltingpointoficeas–100C.It
reads600Cinplaceof500C.Whatisthetemperatureofboilingpointofwateronthe
scale?
Ans.Lowerfixedpointonthewrongscale=-100C.
Let‘n’=no.divisionsbetweenupperandlowerfixedpointsonthisscale.IfQ=readingon
thisscale,then
Now,C=IncorrectReading=600C
Q=CorrectReading=500C
So,
n=140
Now,
On,theCelsiusscale,Boilingpointofwateris1000C
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So,
Q=1300C
53.Writetheadvantagesanddisadvantagesofplatinumresistancethermometer?
Ans.AdvantagesofPlatinumResistancethermometer:-
1)Highaccuracyofmeasurement
2)Measurementsoftemperaturecanbemadeoverawiderangeoftemperaturei.e.from–
2600Cto12000C.
→DisadvantagesofPlatinumResistancethermometer:-
1)HighCost
2)Requiresadditionalequipmentsuchasbridgecircuit,Powersupplyetc.
54.Ifthevolumeofblockofmetalchangesby0.12%whenitisheatedthrough200C.
Whatistheco-efficientoflinearexpansionofthemetal?
Ans.Theco-efficientofcubicalexpansionyofthemetalisgivenby:-
Here,
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∴Co-efficientoflinearexpansionofthemetalis:-
55.Thedensityofasolidat00Cand5000Cisintheratio1.027:1.Findtheco-efficientof
linearexpansionofthesolid?
Ans.Densityat00C=SO
Densityat5000C=S500
Now,SO=S500
Where,Y=Co-efficientofvolumeexpansion
∆T=Changeintemperature
∆T=Changeintemperature
∆T=FinalTemperature–Initialtemperature
∆T=500-00C
∆T=5000C
Or
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Now,Co-efficientoflinearexpansion(α)isrelatedtoco-efficientofvolumeexpansion(Y)as
:-
56.IfoneMoleofamonatomicgasismixedwith3molesofadiatomicgas.Whatisthe
molecularspecificheatofthemixtureatconstantvolume?
Ans.For,amonatomicgas,Specificheatatconsentvolume=CV1= ;R=UniversalGas
Constant
No.ofmolesofmonatomicgas=n1=1mole
No.ofmolesofdiatomicgas=n2=3moles.
For,diatomicgas,specificheatatconstantvolume
Applying,conservationofenergy.
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LetCV=Specificheatofthemixture;
R=UniversalGasconstant
57.Calculatethedifferencebetweentwoprincipalspecificheatsof1gofheliumgasat
N.T.P.GivenMolecularweightofHelium=4andJ=4.186J/calandUniversalGas
constant,R=8.314J/mole/K?
Ans.MolecularweightofHelium=M=4
UniversalGasConstant,R=8.31J|mole|K
CP=specificheatatconstantPressure
CV=specificheatatconstantVolume
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Now, for1moleofgas.
WhereR=UniversalGasConstant=8.31J|mole|K
J=4.186J|cal
M=MolecularweightofHelium=4
58.Whydoesheatflowfromabodyathighertemperaturetoabodyatlower
temperature?
Ans.Whenabodyathighertemperatureisincontactwithabodyatlowertemperature,
moleculewithmorekineticenergythatareincontactwithlessenergeticmoleculesgiveup
someoftheirkineticenergytothelessenergeticones.
59.Aoneliterflaskcontainssomemercury.ITisfoundthatatdifferenttemperatures,
thenvolumeofairinsidetheflaskremainsthesame.Whatisthevolumeofmercuryin
theflask?Giventheco-efficientoflinearexpansionofglass=9×10-6/0Candco-
efficientofvolumeexpansionofmercury=1.8×10-4/0C
Ans.Itisgiventhatvolumeofairintheflaskremainsthesameatdifferenttemperature.This
ispossibleonlywhentheexpansionofglassisexactlyequaltotheexpansionofmercury,
Co-efficientofcubicalexpansionofglassis:-
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Co-efficientofcubicalexpansionofmercuryis:→
Volumeofflask,V=1liter=1000cm3.
LetVmCm3bethevolumeofmercuryintheflask.
Expansionofflask=ExpansionofMercury
∴VolumeofMercury,
60.Arefrigeratoristomaintaineatableskeptinsideat9°C.Ifroomtemperatureis36°
C,calculatethecoefficientofperformance.
Ans.Temperatureinsidetherefrigerator, =9°C=282K
Roomtemperature, =36°C=309K
Coefficientofperformance=
=10.44
Therefore,thecoefficientofperformanceofthegivenrefrigeratoris10.44.
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61.Asteamenginedelivers ofworkperminuteandservices of
heatperminutefromitsboiler.Whatistheefficiencyoftheengine?Howmuchheatis
wastedperminute?
Ans.Workdonebythesteamengineperminute,W=
Heatsuppliedfromtheboiler,H=
Efficiencyoftheengine=
Hence,thepercentageefficiencyoftheengineis15%.
Amountofheatwasted=
= =
Therefore,theamountofheatwastedperminuteis .
62.Athermodynamicsystemistakenfromanoriginalstatetoanintermediatestateby
thelinearprocessshowninFig.(12.13)
ItsvolumeisthenreducedtotheoriginalvaluefromEtoFbyanisobaricprocess.
CalculatethetotalworkdonebythegasfromDtoEtoF
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Ans.TotalworkdonebythegasfromDtoEtoF=AreaofΔDEF
AreaofΔDEF=
Where,
DF=Changeinpressure
=
=300N/
FE=Changeinvolume
= =3.0m3
AreaofΔDEF= =450J
Therefore,thetotalworkdonebythegasfromDtoEtoFis450J.
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CBSEClass11physics
ImportantQuestions
Chapter12
Thermodynamics
3MarksQuestions
1.CalculatetheworkdoneduringtheisothermalProcess?
Ans.Letanidealgasisallowedtoexpandveryslowlyatconstanttemperature.Letthe
expandsfromstateA(P1,V1)tostateB(P2,V2)
TheworkbythegasinexpandingfromstateAtoBis
Foridealgas,PV=NRT
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orP=
Use2)ini)
W=
Sincen,RandTareconstantso,
W=
Wisothermal=nRTLogeV
Wisothermal=
Wisothermal–nRTLoge
Wisothermal=2.303nRTLog10
IfM=MolecularMassofgasthenfor1gramofidealgas,
Wisothermal=2.303
Wisothermal2.303rTLog10
r=Gasconstantfor1gmofanidealgas,
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SinceP1V1=P2V2
SoWisothermal=2.303rTlog10
2.FivemolesofanidealgasaretakeninaCarnotengineworkingbetween1000Cand
300C.Theusefulworkdonein1cycleis420J.Calculatetheratioofthevolumeofthegas
attheendandbeginningoftheisothermalexpansion?
Ans.Hightemperature,TH=1000C=100+273=373K
Lowtemperature,TL=300C,=30+273=303K
Amountofthegas,n=5moles
Usefulworkdonepercycle,W=QH-QL
Now,W=420J
So,QH–QL=420J→1)
Now,
OrQH=
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QL=1818J
or,QH-QL=420J
QH–1818=420J
QH=420+1818=2238J
WhenthegasiscarriedthroughCarnotcycle,theheatabsorbedQHduringisothermal
expansionisequaltotheworkdonebygas.
V1–InitialVolume
V2=FinalVolume,
Inisothermalexpansion,
QH=2.303nRTHLog10
2238=2.303×5×8.4×373Log10
Log10
Log10
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3.Deducetheworkdoneinthefollowingcompletecycle?
Ans.1)WorkdoneduringtheprocessfromAtoB=WAB
WAB=areaABKLA(∴becauseareaunderp-vcurvegivesworkdone)
=areaof∆ABC+areaofrectangle
=
BC=KL=4-1=3l=3x10-3m3
AC=4-2=2N|m2
LC=2-0=2N|m2
WAB=
=3×10-3+6×10-3
WAB=9×10-3J
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Sincegasexpandsduringthisprocess,henceWAB=9×10-3J
2)WorkdoneduringtheprocessfromBtoC(compression)isWBC=-areaBCLK
(-vebecausegascompressesduringBC)
=-KL×LC
WBC=-3×10-3×2
=-6×10-3J
3)WorkdoneduringtheprocessfromCtoA:-
Asthereisnochangeinvolumeofgasinthisprocess,WCA=O
So,networkdoneduringthecompletecycle=WAB+WBC+WCA
=9×10-3-6×10-3+0
Networkdone=3×10-3J
4.Onekilogrammoleculeofagasat400kexpandsisothermallyuntilitsvolumeis
doubled.Findtheamountofworkdoneandheatproduced?
Ans.Initialvolume,V1=V
Finalvolume,V2=2V
InitialtemperatureT=400k
Findtemperature=400k(∴processisisothermal)
Gasconstant,R=8.3kJ|mole|k=8.3x10-3J|mole|k
Workdoneduringisthermalprocess=w=2.3026RTLog10
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W=2.3026×8.3×10-3×400×log10
W=2.3026×8.3×10-3×400×Log10(2)
W=2.3016J
IfHistheamountofheatproducedthan,
5.CalculatedifferenceinefficiencyofaCarnotenergyworkingbetween:-
1)400Kand350K
2)350Kand300K
Ans.Efficiencyofheatengine=n=1-
T2=finaltemperature
T1=Initialtemperature
1)400Kand350K,
T2=350,T1=400
n=1-
=
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n1=
2)350Kand300K
T2=300K;T1=350K
n1=1-
=1-
=
n1=
Changeinefficiency=n2–n1=14.3%-12.5%=1.8%
6.HowdoyouderiveNewton’slawofcoolingfromStefan’slaw?
Ans.Acc.toNewton’slawofcooling,therateoflossofheatofaliquidisdirectlyproportional
tothedifferenceintemperatureoftheliquidandthesurrounding,providedthedifference
intemperatureisverysmall.
LetabodybemaintainedatTK.LetTobethetemperatureofthesurroundings.LetT≫To.
Therewillbelossofheatbethebody
Acc.toStefan’slaw,amountofheatenergylostpersecondperunitareaofthebodyis
E=ϵσ
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σ=Stefan’sconstant
ε=Emissivityofthebodyandsurroundings
E=
IncaseofNewton’scooling,T≈To
E=εσ
E=εσ
Hence,→HencetheNewton’slawofcooling
Eα
7.Definethetermsreflectance,absorptanceandtransmittance.Howaretheyrelated?
Ans.1)Reflectance–Ratioofamountofthermalradiationsreflectedbythebodyinagiven
timetototalamountofthermalradiationsincidentonbodyItisrepresentedbyr,2)
Absorptance–istheratiooftheamountofthermaltothetotalamountofthermalradiations
incidentonit.Itisrepresentedbya
3)Transmittance–Itistheratiooftheamountofthermalradiationstransmittedbybodyin
agiventimetothetotalamountofthermalradiationsincidentonthebodyinagiventime.It
isrepresentedbyt.
LetQ=Amountoftheradiationsincidentbythebodyinagiventime
Q1=Amountofthermalradiationsreflectedbythebodyinagiventime.
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Q2=Amountofthermalradiationsabsorbedbythebodyinagiventime.
Q3=Amountofthermalradiationstransmittedbythebodyinagiventime,
∴Bydefinition,
New,r+a+t=
R+a+t=
R+a+t=1
Ift=0
A=1–r
thatisgoodreflectorsarebadabsorbers
8.IfhalfmoleofheliumiscontainedinacontaineratS.T.P.Howmuchheatenergyis
neededtodoublethepressureofthegas,keepingthevolumeofthegasconstant?Given
specificheatofgas=3J|g|K.
Ans.NumberofmolesofHeliumgas=
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SpecificheatofHeliumgas=
Molecularweight=M=4
Temperature,T1=273K.
∴Molarspecificheatatconstantvolume=CV=MCV
CV=4×3
CV=12J|mol|K
Since,Volumeisconstant,PαTor =Constant
∴
Or
P2=FinalPressure=2P
P1=InitialPressure=P
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Now,Heatrequired,
Heatrequired=1638J
9.Calculatetheamountofheatnecessarytoraisethetemperatureof2molesofHEgas
from200Cto500Cusing:-
1)Constant–VolumeProcess2)ConstantPressureProcess
Herefor,He;CV=1.5RandCP=2.49R
Ans.1)Theamountofheatrequiredforconstant–volumeprocessis:-
Here,n=2moles,CV=1.5R=1.5X8.314J|mol|0C
T2=finalTemperature
T1=InitialTemperature
2)Theamountofheatrequiredforconstant–Pressureprocessis:-
Here,n=2moles,
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Sincethetemperatureriseissameinboththecases,thechangeininternalenergyisthe
samei.e.748J.However,inconstant–pressureProcessexcessheatissuppliedwhichisused
intheexpansionofgas.
10.Anelectricheatersuppliesheattoasystematarateof100W.Ifsystemperforms
workatarateof75Joulespersecond.Atwhatrateistheinternalenergyincreasing?
Ans.Heatissuppliedtothesystematarateof100W.
∴Heatsupplied,Q=100J/s
Thesystemperformsatarateof75J/s.
∴Workdone,W=75J/s
Fromthefirstlawofthermodynamics,wehave:
Q=U+W
Where,
U=Internalenergy
∴U=Q–W
=100–75
=25J/s
=25W
Therefore,theinternalenergyofthegivenelectricheaterincreasesatarateof25W.
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