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CBSEClass11physics

ImportantQuestions

Chapter11

ThermalPropertiesofMatter

5MarksQuestions

1.Answerthefollowing:

(a)Thetriple-pointofwaterisastandardfixedpointinmodernthermometry.Why?

Whatiswrongintakingthemeltingpointoficeandtheboilingpointofwateras

standardfixedpoints(aswasoriginallydoneintheCelsiusscale)?

(b)ThereweretwofixedpointsintheoriginalCelsiusscaleasmentionedabovewhich

wereassignedthenumber0°Cand100°Crespectively.Ontheabsolutescale,oneofthe

fixedpointsisthetriple-pointofwater,whichontheKelvinabsolutescaleisassigned

thenumber273.16K.Whatistheotherfixedpointonthis(Kelvin)scale?

(c)Theabsolutetemperature(Kelvinscale)Tisrelatedtothetemperaturetconthe

Celsiusscaleby

=T-273.15

Whydowehave273.15inthisrelation,andnot273.16?

(d)Whatisthetemperatureofthetriple-pointofwateronanabsolutescalewhoseunit

intervalsizeisequaltothatoftheFahrenheitscale?

Ans.(a)Thetriplepointofwaterhasauniquevalueof273.16K.Atparticularvaluesof

volumeandpressure,thetriplepointofwaterisalways273.16K.Themeltingpointofice

andboilingpointofwaterdonothaveparticularvaluesbecausethesepointsdependon

pressureandtemperature.

(b)Theabsolutezeroor0KistheotherfixedpointontheKelvinabsolutescale.

(c)Thetemperature273.16Kisthetriplepointofwater.Itisnotthemeltingpointofice.The

temperature0°ConCelsiusscaleisthemeltingpointofice.ItscorrespondingvalueonKelvin

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scaleis273.15K.

Hence,absolutetemperature(Kelvinscale)T,isrelatedtotemperaturetc,onCelsiusscaleas:

=T–273.15

(d)LetTFbethetemperatureonFahrenheitscaleand bethetemperatureonabsolute

scale.Boththetemperaturescanberelatedas:

……..(i)

Let bethetemperatureonFahrenheitscaleand bethetemperatureonabsolute

scale.Boththetemperaturescanberelatedas:

………..(ii)

Itisgiventhat:

Subtractingequation(i)fromequation(ii),weget:

Triplepointofwater=273.16K

∴Triplepointofwateronabsolutescale= =491.69

2.TwoidealgasthermometersAandBuseoxygenandhydrogenrespectively.The

followingobservationsaremade:

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Temperature PressurethermometerA PressurethermometerB

Triple-pointofwater

Normalmeltingpointofsulphur

(a)Whatistheabsolutetemperatureofnormalmeltingpointofsulphurasreadby

thermometersAandB?

(b)Whatdoyouthinkisthereasonbehindtheslightdifferenceinanswersof

thermometersAandB?(Thethermometersarenotfaulty).Whatfurtherprocedureis

neededintheexperimenttoreducethediscrepancybetweenthetworeadings?

Ans.(a)Triplepointofwater,T=273.16K.

Atthistemperature,pressureinthermometerA,

Let bethenormalmeltingpointofsulphur.

Atthistemperature,pressureinthermometerA,

AccordingtoCharles'law,wehavetherelation:

=392.69K

Therefore,theabsolutetemperatureofthenormalmeltingpointofsulphurasreadby

thermometerAis392.69K.

Attriplepoint273.16K,thepressureinthermometerB,

Attemperature ,thepressureinthermometerB,

AccordingtoCharles'law,wecanwritetherelation:

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Therefore,theabsolutetemperatureofthenormalmeltingpointofsulphurasreadby

thermometerBis391.98K.

(b)TheoxygenandhydrogengaspresentinthermometersAandBrespectivelyarenot

perfectidealgases.Hence,thereisaslightdifferencebetweenthereadingsofthermometers

AandB.

Toreducethediscrepancybetweenthetworeadings,theexperimentshouldbecarried

underlowpressureconditions.Atlowpressure,thesegasesbehaveasperfectidealgases.

3.Aholeisdrilledinacoppersheet.Thediameteroftheholeis4.24cmat27.0°C.What

isthechangeinthediameteroftheholewhenthesheetisheatedto227°C?Coefficient

oflinearexpansionofcopper= .

Ans.Initialtemperature, =27.0°C

Diameteroftheholeat =4.24cm

Finaltemperature, =227°C

Diameteroftheholeat

Co-efficientoflinearexpansionofcopper,

Forco-efficientofsuperficialexpansion ,andchangeintemperatureΔT,wehavethe

relation:

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But

Changeindiameter= =0.0144cm

Hence,thediameterincreasesby cm.

4.Abrasswire1.8mlongat27°Cisheldtautwithlittletensionbetweentworigid

supports.Ifthewireiscooledtoatemperatureof-39°C,whatisthetensiondeveloped

inthewire,ifitsdiameteris2.0mm?Co-efficientoflinearexpansionofbrass=

;Young'smodulusofbrass= .

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Ans.Initialtemperature, =27°C

Lengthofthebrasswireat ,l=1.8m

Finaltemperature, =–39°C

Diameterofthewire,d=2.0mm= m

Tensiondevelopedinthewire=F

Coefficientoflinearexpansionofbrass,α=

Young'smodulusofbrass,Y=

Young'smodulusisgivenbytherelation:

…………..(i)

Where,

F=Tensiondevelopedinthewire

A=Areaofcross-sectionofthewire.

ΔL=Changeinthelength,givenbytherelation:

ΔL=αL …(ii)

Equatingequations(i)and(ii),weget:

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(Thenegativesignindicatesthatthetensionisdirectedinward.)

Hence,thetensiondevelopedinthewireis .

5.Abrassrodoflength50cmanddiameter3.0mmisjoinedtoasteelrodofthesame

lengthanddiameter.Whatisthechangeinlengthofthecombinedrodat250°C,ifthe

originallengthsareat40.0°C?Istherea'thermalstress'developedatthejunction?The

endsoftherodarefreetoexpand(Co-efficientoflinearexpansionofbrass=

,steel= ).

Ans.Initialtemperature, =40°C

Finaltemperature, =250°C

Changeintemperature, =210°C

Lengthofthebrassrodat =50cm

Diameterofthebrassrodat =3.0mm

Lengthofthesteelrodat =50cm

Diameterofthesteelrodat =3.0mm

Coefficientoflinearexpansionofbrass, =

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Coefficientoflinearexpansionofsteel, =

Fortheexpansioninthebrassrod,wehave:

=0.2205cm

Fortheexpansioninthesteelrod,wehave:

=0.126cm

Totalchangeinthelengthsofbrassandsteel,

=0.2205+0.126

=0.346cm

Totalchangeinthelengthofthecombinedrod=0.346cm

Sincetherodexpandsfreelyfrombothends,nothermalstressisdevelopedatthejunction.

6.AnswerthefollowingquestionsbasedontheP-Tphasediagramofcarbondioxide:

(a)Atwhattemperatureandpressurecanthesolid,liquidandvapourphasesof

co-existinequilibrium?

(b)Whatistheeffectofdecreaseofpressureonthefusionandboilingpointof ?

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(c)Whatarethecriticaltemperatureandpressurefor ?Whatistheir

significance?

(d)Is solid,liquidorgasat(a)-70°Cunder1atm,(b)-60°Cunder10atm,(c)15°C

under56atm?

Ans.(a)TheP-Tphasediagramfor isshowninthefollowingfigure.

Cisthetriplepointofthe phasediagram.Thismeansthatatthetemperatureand

pressurecorrespondingtothispoint(i.e.,at-56.6°Cand5.11atm),thesolid,liquid,and

vaporousphasesof co-existinequilibrium.

(b)Thefusionandboilingpointsof decreasewithadecreaseinpressure.

(c)Thecriticaltemperatureandcriticalpressureof are31.1°Cand73atmrespectively.

Evenifitiscompressedtoapressuregreaterthan73atm, willnotliquefyabovethe

criticaltemperature.

(d)ItcanbeconcludedfromtheP-Tphasediagramof that:

(a) isgaseousat-70°C,under1atmpressure

(b) issolidat-60°C,under10atmpressure

(c) isliquidat15°C,under56atmpressure

7.AnswerthefollowingquestionsbasedontheP-Tphasediagramof :

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(a) at1atmpressureandtemperature-60°Ciscompressedisothermally.Doesit

gothroughaliquidphase?

(b)Whathappenswhen at4atmpressureiscooledfromroomtemperatureat

constantpressure?

(c)Describequalitativelythechangesinagivenmassofsolid at10atmpressure

andtemperature-65°Casitisheateduptoroomtemperatureatconstantpressure.

(d) isheatedtoatemperature70°Candcompressedisothermally.Whatchanges

initspropertiesdoyouexpecttoobserve?

Ans.(a)No

(b)Itcondensestosoliddirectly.

(c)Thefusionandboilingpointsaregivenbytheintersectionpointwherethisparallelline

cutsthefusionandvaporisationcurves.

(d)Itdepartsfromidealgasbehaviouraspressureincreases.

Explanation:

(a)TheP-Tphasediagramfor isshowninthefollowingfigure.

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At1atmpressureandat-60°C, liestotheleftof-56.6°C(triplepointC).Hence,itliesin

theregionofvaporousandsolidphases.

Thus,C condensesintothesolidstatedirectly,withoutgoingthroughtheliquidstate.

(b)At4atmpressure, liesbelow5.11atm(triplepointC).Hence,itliesintheregionof

vaporousandsolidphases.Thus,itcondensesintothesolidstatedirectly,withoutpassing

throughtheliquidstate.

(c)Whenthetemperatureofamassofsolid (at10atmpressureandat-65°C)is

increased,itchangestotheliquidphaseandthentothevaporousphase.Itformsaline

paralleltothetemperatureaxisat10atm.Thefusionandboilingpointsaregivenbythe

intersectionpointwherethisparallellinecutsthefusionandvaporisationcurves.

(d)If isheatedto70°Candcompressedisothermally,thenitwillnotexhibitany

transitiontotheliquidstate.Thisisbecause70°Cishigherthanthecriticaltemperatureof

.Itwillremaininthevapourstate,butwilldepartfromitsidealbehaviouraspressure

increases.

8.Achildrunningatemperatureof101°Fisgivenanantipyrin(i.e.amedicinethat

lowersfever)whichcausesanincreaseintherateofevaporationofsweatfromhis

body.Ifthefeverisbroughtdownto98°Fin20min,whatistheaveragerateofextra

evaporationcaused,bythedrug?Assumetheevaporationmechanismtobetheonly

waybywhichheatislost.Themassofthechildis30kg.Thespecificheatofhuman

bodyisapproximatelythesameasthatofwater,andlatentheatofevaporationof

wateratthattemperatureisabout580cal .

Ans.Initialtemperatureofthebodyofthechild, =101°F

Finaltemperatureofthebodyofthechild, =98°F

Changeintemperature,ΔT= °C

Timetakentoreducethetemperature,t=20min

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Massofthechild,m=30kg=

Specificheatofthehumanbody=Specificheatofwater=c

=1000cal/kg/°C

Latentheatofevaporationofwater,L=580cal

Theheatlostbythechildisgivenas:

=50000cal

Let bethemassofthewaterevaporatedfromthechild'sbodyin20min.

Lossofheatthroughwaterisgivenby:

=

∴Averagerateofextraevaporationcausedbythedrug

9.Abrassboilerhasabaseareaof0.15 andthickness1.0cm.Itboilswateratthe

rateof6.0kg/minwhenplacedonagasstove.Estimatethetemperatureofthepartof

theflameincontactwiththeboiler.Thermalconductivityofbrass= ;

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Heatofvaporisationofwater= .

Ans.Baseareaoftheboiler,A=0.15

Thicknessoftheboiler,l=1.0cm=0.01m

Boilingrateofwater,R=6.0kg/min

Mass,m=6kg

Time,t=1min=60s

Thermalconductivityofbrass,K=

Heatofvaporisation,L=

Theamountofheatflowingintowaterthroughthebrassbaseoftheboilerisgivenby:

…………(i)

Where,

=Temperatureoftheflameincontactwiththeboiler

=Boilingpointofwater=100°C

Heatrequiredforboilingthewater:

θ=mL…(ii)

Equatingequations(i)and(ii),weget:

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=

=137.98°C

Therefore,thetemperatureofthepartoftheflameincontactwiththeboileris137.98°C.

10.Explainwhy:

(a)abodywithlargereflectivityisapooremitter

(b)abrasstumblerfeelsmuchcolderthanawoodentrayonachillyday

(c)anopticalpyrometer(formeasuringhightemperatures)calibratedforanideal

blackbodyradiationgivestoolowavalueforthetemperatureofaredhotironpiecein

theopen,butgivesacorrectvalueforthetemperaturewhenthesamepieceisinthe

furnace

(d)theearthwithoutitsatmospherewouldbeinhospitablycold

(e)heatingsystemsbasedoncirculationofsteamaremoreefficientinwarminga

buildingthanthosebasedoncirculationofhotwater

Ans.(a)Abodywithalargereflectivityisapoorabsorberoflightradiations.Apoor

absorberwillinturnbeapooremitterofradiations.Hence,abodywithalargereflectivityis

apooremitter.

(b)Brassisagoodconductorofheat.Whenonetouchesabrasstumbler,heatisconducted

fromthebodytothebrasstumblereasily.Hence,thetemperatureofthebodyreducestoa

lowervalueandonefeelscooler.

Woodisapoorconductorofheat.Whenonetouchesawoodentray,verylittleheatis

conductedfromthebodytothewoodentray.Hence,thereisonlyanegligibledropinthe

temperatureofthebodyandonedoesnotfeelcool.

Thus,abrasstumblerfeelscolderthanawoodentrayonachillyday.

(c)Anopticalpyrometercalibratedforanidealblackbodyradiationgivestoolowavalue

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fortemperatureofaredhotironpiecekeptintheopen.

Blackbodyradiationequationisgivenby:

Where,

E=Energyradiation

T=Temperatureofopticalpyrometer

=Temperatureofopenspace

=Constant

Hence,anincreaseinthetemperatureofopenspacereducestheradiationenergy.

Whenthesamepieceofironisplacedinafurnace,theradiationenergy,

(d)Withoutitsatmosphere,earthwouldbeinhospitablycold.Intheabsenceofatmospheric

gases,noextraheatwillbetrapped.Alltheheatwouldberadiatedbackfromearth's

surface.

(e)Aheatingsystembasedonthecirculationofsteamismoreefficientinwarminga

buildingthanthatbasedonthecirculationofhotwater.Thisisbecausesteamcontains

surplusheatintheformoflatentheat(540cal/g).

11.Abodycoolsfrom80°Cto50°Cin5minutes.Calculatethetimeittakestocoolfrom

60°Cto30°C.Thetemperatureofthesurroundingsis20°C.

Ans.AccordingtoNewton'slawofcooling,wehave:

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Where,

Temperatureofthebody=T

Temperatureofthesurroundings= =20°C

Kisaconstant

Temperatureofthebodyfallsfrom80°Cto50°Cintime,t=5min=300s

Integratingequation(i),weget:

…….(ii)

Thetemperatureofthebodyfallsfrom60°Cto30°Cintime=t'

Hence,weget:

………….(iii)

Equatingequations(ii)and(iii),weget:

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Therefore,thetimetakentocoolthebodyfrom60°Cto30°Cis10minutes.

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CBSEClass11Physics

Chapter-11ThermalPropertiesofMatter

NUMERICALS

1. Howmuchpressurewillamanofweight80kgfexertonthegroundwhen(i)heis

lyingand(2)heisstandingonhisfeet.Givenareaofthebodyofthemanis0.6m2

andthatofhisfeetis80cm2.

Ans.(i)WhenmanislyingP= =1.307×103N/m-2

(ii)WhenmanisstandingthenA=2×80cm2=160×10-4m2

2. Themanualofacarinstructstheownertoinflatethetyrestoapressureof200kpa.

(a)Whatistherecommendedgaugepressure?(b)Whatistherecommended

absolutepressure(c)if,aftertherequiredinflationofthetyres,thecarisdrivento

amountainpeakwheretheatmosphericpressureis10%belowthatatsealevel,

whatwillthetyregaugeread?

Ans.

(a)PressureInstructedbymanual=Pg=2OOKPa

(b)AbsolutePressure=101kP+200kPa=301kPa

(c)AtmountainpeakPa'is10%less

Pa’=90kPa

IfweassumeabsolutepressureintyredoesnotchangeduringdrivingthenPg=P-P’a=

301-90=211kPa

Sothetyrewillread211kPa,pressure.

3. Twostarsradiatemaximumenergyatwavelength,3.6x107mrespectively.Whatis

theratiooftheirtemperatures?

Ans.ByWien'sDisplacementLaw

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=4:3

4. Findthetemperatureof149°Fonkelvinscale.

Ans.

T=286K

5. Ametalpieceof50gspecificheat0.6cal/g°Cinitiallyat120°Cisdroppedin1.6kgof

waterat25°C.Findthefinaltemperatureormixture.

Ans.

C2=1cal/gm°c

∴50×0.6×(120-θ)=1.6×103×1×(θ-25)Θ=26.8°C

6. Aironringofdiameter5.231mistobefixedonawoodenrimofdiameter5.243m

bothinitiallyat27°C.Towhattemperatureshouldtheironringbeheatedsoastofit

therim(Coefficientoflinearexpansionofironis1.2x105k-1?

Ans.

5.243=5.231[1+1.2×10-5(T-300)]

T=191+300=491K=218°C

7. 100goficeat0°Cismixedwith100gofwaterat80°C.Theresultingtemperatureis

6°C.Calculateheatoffurionofice.

Ans.

m1c1(80-6)=m2L+m2c2(6-0)

100×1×74=100L+100×1×6

L=(1×74)-6

=68cal/g

8. Calculateheatrequiredtocovert3kgofwaterat0°Ctosteamat100°CGivenspecific

heatcapacityofH20=4186Jkg-1k-1andlatentheatofstem=2.256x106J/kg

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Ans.HeatrequiredtoconvertH2Oat0°toH2Oat100=m1c1t

=3×4186×1OO

=1255BOOJ

HeatrequiredtoconvertH2Oat100°Ctosteamat100°Cis=mL

=3×2.256x106

=676BOOOJ

Totalheat=80.238OOJ

9. Givenlengthofsteelrod15cm;ofcopper10cm.Theirthermalconductivitiesare

50.2j-sm-1k-1and385j-s-m-1k-1respectively.Areaofcrossectionofsteelis

doubleofareaofCopperrod?

Ans.

A1=2A2

T=317.43K=44.43°C

10. Abodyattemperature94°Ccoolsto86°Cin2min.Whattimewillittaketocool

from82°Cto78°C.Thetemperatureofsurroundingis20°C.

Ans.

…….1

…..2

dividingeq.(1)byeq(2)

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11. Abodyre-emitsalltheradiationitreceives.Findsurfacetemperatureofthebody.

Energyreceivedperunitareaperunittimeis2.835Watt/m2and =5.67x10-8W

m-2K-4.

Ans.

=85k

12. Atwhattemperaturetheresistanceofthermometerwillbe12%moreofits

resistanceat0°C(giventemperaturecoefficientofresistanceis2.5×10-3C-1?

Ans.

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CBSEClass11Physics

Chapter-11ThermalPropertiesofMatter

SHORTANSWERTYPEQUESTIONS(2MARKS)

1. Onahotday,acarisleftinsunlightwithallwindowsclosed.Explainwhyitis

considerablywarmerthanoutside,aftersometime?

Ans.Glasstransmits50%ofheatradiationComingfromahotsourcelikeSunbutdoes

notallowtheradiationfrommoderatelyhotbodiestopassthroughit.

2. Namethesuitablethermometerstomeasurethefollowingtemperatures

(a)-100°C(b)80°C(c)780°C(d)2000°C

Ans.

(a)Gasthermometer;

(b)Mercurythermometer;

(c)Platinumresistancethermometer;

(d)Radiationpyrometer.

3. Theearthwithoutitsatmospherewouldbeinhospitablycold.Why?

Ans.Duetogreen-houseeffect,thepresenceofatmospherepreventsheatradiations

receivedbyearthtogoback.Intheabsenceofatmosphereradiationwillgobackatnight

makingthetemperatureverylowandinhospitable.

4. TheCoolantusedinanuclearplantshouldhavehighspecificheat.Why?

Ans.So,thatitabsorbsmoreheatwithcomparativelysmallchangeintemperatureand

extractslargeamountofheat.

5. Asphere,acubeandadiscmadeofsamematerialandofequalmassesheatedto

sametemperatureof200°C.Thesebodiesarethenkeptatsamelowertemperature

inthesurrounding,whichofthesewillcool(i)fastest,(ii)slowest.Explain.

Ans.Rateofenergyemissionisdirectlyproportionaltoareaofsurfaceforagivenmass

ofmaterial.Surfaceareaofsphereisleastandthatofdiscislargest.Thereforecoolingof

(i)discisfastestand(ii)sphereisslowest.

6.

a. WhypendulumclocksgenerallygofasterinwinterandslowinSummer?

b. Whythebrakedrumsofacarareheatedwhenitmovesdownahillatconstant

speed?

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Ans.

a. Whenthecarmovesdownhill,thedecreaseingravitationalpotentialenergyis

convertedintoworkagainstforceoffrictionbetweenbrakeshoeanddrumwhich

appearsasheat.

b. Timeperiodofpendulum=

InwinterIbecomesshortersoitstimeperiodreducessoitgoesfaster.Insummer

increasesresultinginincreaseintimeperiodsotheclockgoesslower.

7. TheplotsofintensityversuswavelengthforthreeblackbodiesattemperatureT1,T2

andT3respectivelyareshown.

Arrangethetemperatureindecreasingorder.Justifyyouranswer.

Ans.

∴fromWeindisplacementlaw

T1>T3>T2

8. Thetriplepointofwaterisastandardfixedpointinmodernthermometry.Why?

Whymeltingpointoficeorboilingpointofwaternotusedasstandardfixedpoints.

Ans.Themeltingpointoficeaswellastheboilingpointofwaterchangeswithchangein

pressure.Thepresenceofimpuritiesalsochangesthemeltingandboilingpoints.

Howeverthetriplepointofwaterhasauniquetemperatureandisindependentof

externalfactors.Itisthattemperaturesatwhichwater,ice&watervapourco-existthatis

273.16Kandpressure0.46cmofHg.

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CBSEClass11Physics

Chapter-11ThermalPropertiesofMatter

SHORTANSWERTYPEQUESTION(3MARKS)

1. AbigsizeballoonofmassMisheldstationaryinairwiththehelpofasmallblockof

massM/2tiedtoitbyalightstringsuchthatbothfloatinmidair.Describethe

motionoftheballoonandtheblockwhenthestringiscut.Supportyouranswer

withcalculations.

Ans.

Whentheballoonisheldstationaryinair,theforcesactingonitgetbalance

Upthrust=Wt.ofBalloon+Tensioninstring

U=Mg+T

FortheSmallblockofmass floatingstationaryinair

WhenthestringiscutT=0,thesmallblockbeginstofallfreely,theballoonrisesupwith

anacceleration"asuchthat

U—Mg=Ma

intheupwarddirection.

2. Describethedifferenttypesofthermometerscommonlyused.Givetherelation

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betweentemperatureondifferentscales.Givefourreasonsforusingmercuryina

thermometer.

3. Tworodsofdifferentmetalsofcoefficientoflinearexpansion and and

initiallengthI1andI2respectivelyareheatedtothesametemperature,Find

relationin andI2suchthatdifferencebetweentheirlengthsremain

Constant.

Ans.

Giventhatthedifferenceintheirlengthremainconstant

4. Explaintheprincipleofplatinumresistancethermometer.

5. Asteelrailoflength5mandareaofcrosssection40cm2ispreventedfrom

expandingwhilethetemperaturerisesby10°C.Givencoefficientoflinear

expansionofsteelis1.2×10-5k-1.Explainwhyspaceneedstobegivenforthermal

expansion.

Ans.TheCompressivestrain

Thermalstress

whichcorrespondstoanexternalforce

Aforceofthismagnitudecaneasilybendtherails,henceitisimportanttoleavespace

forthermalexpansion.

6. Define(i)Specificheatcapacity(ii)Heatcapacity(iii)Molarspecificheatcapacityat

ConstantpressureandatConstantVolumeandWritetheirunits.

7. Plotagraphoftemperatureversustimeshowingthechangeinthestateoficeon

heatingandhenceexplaintheprocess(withreferencetolatentheat)

8. Whatistheeffectofpressureonmeltingpointofasubstance?Whatisregelation.

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Giveapracticalapplicationofit.

9. Whatistheeffectofpressureontheboilingpointofaliquid.Describeasimple

experimenttodemonstratetheboilingofH2Oatatemperaturemuchlowerthan

100°C.Giveapracticalapplicationofthisphenomenon.

10. Stateandexplainsthethreemodesoftransferofheat.Explainhowthelossofheat

duetothesethreemodesisminimisedinathermosflask.

11. DefineCoefficientofthermalconductivity.Twometalslabsofsameareaofcross-

section,thicknessd1andd2havingthermalconductivitiesK1andK2respectively

arekeptincontact.DeduceexpressionforequivalentthermalConductivity.

Ans.Definitionofcoefficientofthermalconductivity.

Insteadystatetheheatpassinginunittimethroughtherodremainsamethatis

wherekisthecoefficientofthermalconductivity

AlsoT1-T2=(T1-T)+(T-T2)

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CBSEClass11Physics

Chapter-11ThermalPropertiesofMatter

VERYSHORTANSWERTYPEQUESTIONS(1MARK)

1. Whatwillbetheeffectofincreasingtemperatureon(i)angleofcontact(ii)surface

tension.

Ans.Angleofcontactincreaseswithincreaseoftemperaturewhilesurfacetension

generallydecreaseswithincreaseoftemperature

2. Explaindowhydetergentshavesmallangleofcontact?

Ans.Detergentsshouldhavesmallangleofcontactsothattheyhawelowsurfacetension

andgreaterabilitytoWetasurface.Furtheras

issmallCos willbelargesohi.e.penetrationwillbehigh.

3. WhydoesmercurymotWetglass?

Ans.MercurydoesnotwetglassbecauseoflargercohesiveforcebetweenHg-Hg

moleculesthantheadhesiveforcesbetweenmercury-glassmolecules.

4. Whyendsofaglasstubebecomeroundedonheating?

Ans.Whenglassisheated,itmelts.Thesurfaceofthisliquidtendstohaveaminimum

area.Foragivenvolume,thesurfaceareaisminimumforasphere.Thisiswhytheends

ofaglasstubebecomeroundedonheating.

5. StateWein'sdisplacementlawforblackbodyradiation.

6. StateStefanBoltzmannlaw.

7. Nametwophysicalchangesthatoccuronheatingabody.

Ans.Volumeandelectricalresistance.

8. Distinguishbetweenheatandtemperature.

9. Whichthermometerismoresensitiveamercuryorgasthermometer?

Ans.GasthermometerismoresensitiveascoefficientofexpansionofGasismorethan

mercury.

10. Metaldischasaholeinit.Whathappenstothesizeoftheholewhendiscisheated?

Ans.Expansionisalwaysoutward,thereforetheholesizeincreasedonheating.

11. Nameasubstancethatcontractsonheating.

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Ans.Ice

12. Agasisfreetoexpandwhatwillbeitsspecificheat?

Ans.Infinity

13. WhatisDeby'stemperature?

Ans.Thetemperatureabovewhichmolarheatcapacityofasolidsubstancebecomes

Constant.

14. Whatistheabsorptivepowerofaperfectlyblackbody?

Ans.One

15. Atwhattemperaturedoesabodystopradiating?

Ans.AtoK.

16. IfKelvintemperatureofanidealblackbodyisdoubled,whatwillbetheeffecton

energyradiatedbyit?

Ans.

17. Inwhichmethodofheattransferdoesgravitynotplayanypart?

Ans.Inconductionandradiation

18. GiveaplotofFahrenheittemperatureversusCelsiustemperature.

Ans.

19. Whybirdsareoftenseentoswelltheirfeatherinwinter?

Ans.Whenbirdsswelltheirfeathers,theytrapairinthefeather.Airbeingapoor

conductorpreventslossofheatandkeepsthebirdwarm.

20. Abrassdiscfitssnuglyinaholeinasteelplate.Shouldweheatorcoolthesystemto

loosenthediscfromthehole.

Ans.Thetemp.coefficientoflinearexpansionforbrassisgreaterthanthatforteel.On

coolingthediscshrinkstoagreaterextentthanthehole,andhencebrassdiscgets

lossened.

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CBSEClass11physics

ImportantQuestions

Chapter11

ThermalPropertiesofMatter

2MarksQuestions

1.Acopperblockofmass2.5kgisheatedinafurnacetoatemperatureof500°Cand

thenplacedonalargeiceblock.Whatisthemaximumamountoficethatcanmelt?

(Specificheatofcopper=0.39J ;heatoffusionofwater=335J ).

Ans.Massofthecopperblock,m=2.5kg=2500g

Riseinthetemperatureofthecopperblock,Δθ=500°C

Specificheatofcopper,C=0.39J

Heatoffusionofwater,L=335J

Themaximumheatthecopperblockcanlose,Q=mCΔθ

=2500 0.39 500

=487500J

Let gbetheamountoficethatmeltswhenthecopperblockisplacedontheiceblock.

Theheatgainedbythemeltedice,Q=

Hence,themaximumamountoficethatcanmeltis1.45kg.

2.A'thermacole'iceboxisacheapandefficientmethodforstoringsmallquantitiesof

cookedfoodinsummerinparticular.Acubicaliceboxofside30cmhasathicknessof

5.0cm.If4.0kgoficeisputinthebox,estimatetheamountoficeremainingafter6h.

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Theoutsidetemperatureis45°C,andco-efficientofthermalconductivityofthermacole

is .[Heatoffusionofwater= ]

Ans.Sideofthegivencubicalicebox,s=30cm=0.3m

Thicknessoftheicebox,l=5.0cm=0.05m

Massoficekeptintheicebox,m=4kg

Timegap,t=6h=6 60 60s

Outsidetemperature,T=45°C

Coefficientofthermalconductivityofthermacole,K=

Heatoffusionofwater,L=

Letm'bethetotalamountoficethatmeltsin6h.

Theamountofheatlostbythefood:

Where,

A=Surfaceareaofthebox=

But

Massoficeleft=4–0.313=3.687kg

Hence,theamountoficeremainingafter6his3.687kg.

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CBSEClass11physics

ImportantQuestions

Chapter11

ThermalPropertiesofMatter

3MarksQuestions

1.Thetriplepointsofneonandcarbondioxideare24.57Kand216.55Krespectively.

ExpressthesetemperaturesontheCelsiusandFahrenheitscales.

Ans.KelvinandCelsiusscalesarerelatedas:

–273.15…(i)

CelsiusandFahrenheitscalesarerelatedas:

…(ii)

Forneon:

=24.57K

=24.57–273.15=–248.58°C

Forcarbondioxide:

=216.55K

∴ =216.55–273.15=–56.60°C

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2.TwoabsolutescalesAandBhavetriplepointsofwaterdefinedtobe200Aand350B.

Whatistherelationbetween ?

Ans.TriplepointofwateronabsolutescaleA, =200A

TriplepointofwateronabsolutescaleB, =350B

TriplepointofwateronKelvinscale, =273.15K

Thetemperature273.15KonKelvinscaleisequivalentto200AonabsolutescaleA.

=

200A=273.15K

Thetemperature273.15KonKelvinscaleisequivalentto350BonabsolutescaleB.

=

350B=273.15

istriplepointofwateronscaleA.

istriplepointofwateronscaleB.

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Therefore,theratio isgivenas4:7.

3.Asteeltape1mlongiscorrectlycalibratedforatemperatureof27.0°C.Thelengthof

asteelrodmeasuredbythistapeisfoundtobe63.0cmonahotdaywhenthe

temperatureis45.0°C.Whatistheactuallengthofthesteelrodonthatday?Whatis

thelengthofthesamesteelrodonadaywhenthetemperatureis27.0°C?Coefficientof

linearexpansionofsteel= .

Ans.LengthofthesteeltapeattemperatureT=27°C,l=1m=100cm

Attemperature =45°C,thelengthofthesteelrod, =63cm

Coefficientoflinearexpansionofsteel,α=

Let betheactuallengthofthesteelrodandl'bethelengthofthesteeltapeat45°C.

=100.0216cm

Hence,theactuallengthofthesteelrodmeasuredbythesteeltapeat45°Ccanbecalculated

as:

=63.0136cm

Therefore,theactuallengthoftherodat45.0°Cis63.0136cm.Itslengthat27.0°Cis63.0cm.

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4.A10kWdrillingmachineisusedtodrillaboreinasmallaluminiumblockofmass

8.0kg.Howmuchistheriseintemperatureoftheblockin2.5minutes,assuming50%

ofpowerisusedupinheatingthemachineitselforlosttothesurroundings.Specific

heatofaluminium=0.91J .

Ans.Powerofthedrillingmachine,P=10kW=

Massofthealuminumblock,m=8.0kg=

Timeforwhichthemachineisused,t=2.5min=2.5 60=150s

Specificheatofaluminium,c=0.91J

Riseinthetemperatureoftheblockafterdrilling= T

Totalenergyofthedrillingmachine=

=

=

Itisgiventhatonly50%ofthepowerisuseful.

Usefulenergy,

But

Therefore,in2.5minutesofdrilling,theriseinthetemperatureoftheblockis103°C.

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5.Inanexperimentonthespecificheatofametal,a0.20kgblockofthemetalat150°C

isdroppedinacoppercalorimeter(ofwaterequivalent0.025kg)containing150cm3of

waterat27°C.Thefinaltemperatureis40°C.Computethespecificheatofthemetal.If

heatlossestothesurroundingsarenotnegligible,isyouranswergreaterorsmaller

thantheactualvalueforspecificheatofthemetal?

Ans.Massofthemetal,m=0.20kg=200g

Initialtemperatureofthemetal, =150°C

Finaltemperatureofthemetal, =40°C

Calorimeterhaswaterequivalentofmass,m'=0.025kg=25g

Volumeofwater,V=

Mass(M)ofwaterattemperatureT=27°C:

150 1=150g

Fallinthetemperatureofthemetal:

Specificheatofwater, =4.186J/g/°K

Specificheatofthemetal=C

Heatlostbythemetal, …(i)

Riseinthetemperatureofthewaterandcalorimetersystem:

Heatgainedbythewaterandcalorimetersystem:

…(ii)

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Heatlostbythemetal=Heatgainedbythewaterandcolorimetersystem

mCΔT=(M+m') ΔT'

200 C 110=(150+25) 4.186 13

Ifsomeheatislosttothesurroundings,thenthevalueofCwillbesmallerthantheactual

value.

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CBSEClass11physics

ImportantQuestions

Chapter11

ThermalPropertiesofMatter

4MarksQuestions

1.Theelectricalresistanceinohmsofacertainthermometervarieswithtemperature

accordingtotheapproximatelaw:

Theresistanceis101.6 atthetriple-pointofwater273.16K,and165.5 atthe

normalmeltingpointoflead(600.5K).Whatisthetemperaturewhentheresistanceis

123.4 ?

Ans.Itisgiventhat:

R= …(i)

Where,

aretheinitialresistanceandtemperaturerespectively

RandTarethefinalresistanceandtemperaturerespectively

αisaconstant

Atthetriplepointofwater, =273.15K

Resistanceoflead, =101.6

Atnormalmeltingpointoflead,T=600.5K

Resistanceoflead,R=165.5

Substitutingthesevaluesinequation(i),weget:

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Forresistance, =123.4

Where,Tisthetemperaturewhentheresistanceofleadis123.4

2.Alargesteelwheelistobefittedontoashaftofthesamematerial.At27°C,theouter

diameteroftheshaftis8.70cmandthediameterofthecentralholeinthewheelis8.69

cm.Theshaftiscooledusing'dryice'.Atwhattemperatureoftheshaftdoesthewheel

slipontheshaft?Assumecoefficientoflinearexpansionofthesteeltobeconstantover

therequiredtemperaturerange: = .

Ans.Thegiventemperature,T=27°CcanbewritteninKelvinas:

27+273=300K

OuterdiameterofthesteelshaftatT, =8.70cm

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DiameterofthecentralholeinthewheelatT, =8.69cm

Coefficientoflinearexpansionofsteel, =

Aftertheshaftiscooledusing'dryice',itstemperaturebecomes .

Thewheelwillslipontheshaft,ifthechangeindiameter,Δd=8.69–8.70

=–0.01cm

Temperature ,canbecalculatedfromtherelation:

=

0.01=

=95.78

∴ =204.21K

=204.21–273.16

=–68.95°C

Therefore,thewheelwillslipontheshaftwhenthetemperatureoftheshaftis-69°C.

3.Givenbelowareobservationsonmolarspecificheatsatroomtemperatureofsome

commongases.

GasMolarspecificheat( )

(calmol K )

Hydrogen 4.87

Nitrogen 4.97

Oxygen 5.02

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Nitricoxide 4.99

Carbonmonoxide 5.01

Chlorine 6.17

Themeasuredmolarspecificheatsofthesegasesaremarkedlydifferentfromthosefor

monatomicgases.Typically,molarspecificheatofamonatomicgasis2.92cal/molK.

Explainthisdifference.Whatcanyouinferfromthesomewhatlarger(thantherest)

valueforchlorine?

Ans.Thegaseslistedinthegiventablearediatomic.Besidesthetranslationaldegreeof

freedom,theyhaveotherdegreesoffreedom(modesofmotion).

Heatmustbesuppliedtoincreasethetemperatureofthesegases.Thisincreasestheaverage

energyofallthemodesofmotion.Hence,themolarspecificheatofdiatomicgasesismore

thanthatofmonatomicgases.

Ifonlyrotationalmodeofmotionisconsidered,thenthemolarspecificheatofadiatomic

Withtheexceptionofchlorine,alltheobservationsinthegiventableagreewith .

Thisisbecauseatroomtemperature,chlorinealsohasvibrationalmodesofmotionbesides

rotationalandtranslationalmodesofmotion.

4.Thecoefficientofvolumeexpansionofglycerinis .Whatisthe

fractionalchangeinitsdensityfora30°Criseintemperature?

Ans.Coefficientofvolumeexpansionofglycerin, =

Riseintemperature, =30°C

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Fractionalchangeinitsvolume=

Thischangeisrelatedwiththechangeintemperatureas:

Where,

m=Massofglycerine

=Initialdensityat

=Finaldensityat

Where,

=Fractionalchangeindensity

∴Fractionalchangeinthedensityofglycerin=

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