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HeatandTherm

ody

namics

Whyw

ou

ldsome

onedesigninga

pipelineincludethesestrangeloops?

Temperatureand

theZerothLaw

of

Thermody

namics

Zeroth

Law

ofThermody

namics

Ifobjec

tsAandBareseparatelyinthermal

equilibriumwithathirdobjectC,ThenAandB

areine

quilibriumwitheachother

Implication:

Thislaw

isimportantbecausewecannow

definewhattemperatureis

Tempera

tureisapropertythatcandeterminewheth

er

objectsa

reinthermalequilibrium

ThermalEqu

ilibriu

m

y

Wecansaythat

objectsinthermalequilibriumwitheachotherare

atthesa

metemperature

Orobjectswithdifferenttemperaturesarenotintherm

al

equilibriu

m

Wayso

fmeasu

ringtemperatures

1.

Expansionofmaterialsas

temper

aturechanges

Mater

ialsexpand

when

heated.

Mercu

ryandalcohol

Theflowofelectronsina

materialvarywith

temperature

(resistancethermometers)

Resistanceismeasuredas

temperatureisvaried

2.Temperaturedependenceof

pressureatconstantvolume

(gasthermometers)

Pressurechangeswithtemperature

(volumeofGasisconstant)

Pressureiszeroat

T=-2

73.1

5C

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Thermalex

pansion

Linearexpansionobje

ctsincreasesinlengthwhen

temperatureincreases,and

contractswhentemperature

decreases.T~L

Note:Differentmaterialshavedifferent

coefficientsofthermalexpansion

VolumeExpansion

volumeofanobjectchangeswith

temperature

Sample

applicationofthermal

ex

pans

ion

Circuitbreaker.T

heswitchisoffifthetemperatureincreases.

WHYM

ATTEREXPANDS?

Atomsoscillatewithgreateramplitude

whentemperatureincreases

10-

10m

Atordinarytemperature,atomsoscillate

withamplitudeof10-11m

Heat

y

Rememb

erwhenthereistemperaturedifferencebetween

twoobjects,heatflowsfromaregionofhigher

temperaturetoaregionoflowertemperature.

Heat

Heatisdefinedasthetransferofene

rgyacrossthe

boundaryofasystemduetoatempe

raturedifference

betweenthesystemanditssurround

ings

UnitsofHeat

calorie(cal)

theamountofenergytran

sfernecessaryto

raisethetemperatureof1gofwaterfrom

14.5

Cto15.5

C

Joule(J)isalsoaunitofheat

1calorie=4.186J

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Mechanical

equ

ivalentofheat

Heatcapacitya

ndspecificHeat

TheheatcapacityCofa

particularsampleofasubstance

isdefinedastheamountof

energyneededtoraisethe

temperatureofthatsampleb

y1C

Q

=CT

Specificheatistheamountofheatneededtoraisethe

temperatureofaunitmasso

fmaterialbyaunitdegreein

temperature.

Q

=mcT

m=

mass

c=specificheat

T=changeintemperature

Specificheatofsomecommon

materials

Water:

c=1.0cal/g-

0C

Ice:

c=0.5cal/g-

0C

Lead:

c=0.031cal/g-

0C

Latent

Heat

Thechang

einphase(solid,

liquid,gas)ofasubstancedueto

theadditionorremovalofheatwithoutchangingits

temperat

ure Ic

eHeat

Ti=00C

LiquidWater

Tf=00C

y

Latenth

eatoffusion

theheatneededtochangethe

phaseofmatterfromSolidtoLiquid

y

Latenth

eatofvaporization-theheatneededtochange

thephase

ofmatterfromLiquidtoGas

y

Latenth

eatofsolidificationliquidtosolid

y

Latenth

eatofsublimationSolidtogas

Sample

A10gsolidiceinitiallyat-2oCisheated.T

heice

turnedtoliquidwaterandhavea

final

temperatureof10oC.Howmuchheatwasadded?

G

iven:m=10g

Tf=10oC

Ti=-2oC

Q=?

Solution:

Qtotal=Qtempchange

+Q

phasechange

+Qtempchange

=mcice

T+mLf+mcwater

T

=(10g)(0.5cal/g-oC)(0-(-2))+(10g)(80cal/g)+(10g)(1cal/g-oC)(10-0)

=10cal+800cal+100cal

=910cal

y

Note:Seemanualformoresamplecalculations

AdditionorRemovalofHea

tcauses.

y

Temperaturechange

y

Phasechange

y

Expansion

y

Changeininternalenergy

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ModesofHeat

Transfer

y

Conduction

y

Convection

y

Radiation

Condu

ction

Condu

ction

-Theflowofthermalenergydirectlythroughmatter

withoutmotionofthe

materialitself.

-Kineticenergytransferbetweenmoleculesand

atomsthatcomprisestheconductor

Higher

Temperature

Conductor

Lower

Temperature

Condu

ction

Factors

thataffecttherate

ofthermalcondu

ction

1.

Material(thermalconductivity)

a.

Con

ductors(metal)

b.Insu

lators(wood,asbestos,gases)

2.

TemperatureDifference

Obvio

uslythermalenergytransfer(heat)w

ill

onlyo

ccurifthereisdifferenceintempera

ture

3.Thick

nessofthematerialslabandthecross

sectio

nalarea

Factors

thataffecttherate

ofthermalcondu

ction

Conv

ection

y

Convectionisthetransferofheatbymas

smotionofafluid

fromoneregionofspacetoanother.

Radiation

y

Radiationisthetransferofheatbyelectr

omagneticwaves

suchasvisiblelight,

infrared,andultravioletradiation.Does

notneedamediumtopropagate

y

Allobjectsemitcontinuouslyintheform

ofproducedby

thermalvibrationsofthemolecules

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Sample

y

Dewerflask

InthePhilippinesitisco

mmonlyknownastermos

(ThermosBrand)

Law

sofThermody

namics

1.

FirstLawofTherm

odynamics

Q=U+W

2.

SecondLawofThermodynamics

Entropy

HeatEngines

FirstLaw

ofThermody

namics

Statement:

Theneth

eattransferredequalsthechangeinINTERNA

L

ENERGY

ofthesystemandWORKISDONEBYthe

system.

Q=U+W

Q=heat

U=inte

rnalenergy

W=workdone

FirstLaw

ofThermody

namics

y

AstatementofLawofConservationEnergy

(Inthisca

seHeatisconvertedtomechanicalenergy)

Internalenergyincreases

andworkisdoneis

done

bythesystem

Heat

Interna

lEnergy

y

Thesomeofalltheenergiesofthemoleculesandatoms

that

makeup

thesystem

y

Includestranslational,rotationalkineticenergyandthe

potential

energybetweenthemoleculesoratomsthatm

ake

upthesystem

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FirstLaw

ofThermody

namics;

AdiabaticProcess

AdiabaticProcess:NOHEATISADDED

Q=0

Therefore

U

=W

AdiabaticProce

ss

Workdonebythesystem:decreasessystems

internalenergy

(result:LowerT

emperature)

Workdoneintothe

system:Increasessystems

internalenergy

(result:HigherTem

perature)

AdiabaticProce

ss

Ex

ample

250C50

C

-50C

-150C

150C

4km 1km2

km3km 0

km

SecondLaw

ofThermody

namics

Heatwil

lneverofitselfflowfromacoldobjectt

oa

hotobject

y

KelvinPlanckStatementof2ndLawofthermodynamics:

Itisim

possibletoconstructaheatenginethat,

operatinginacycle,producesnoeffectotherth

an

theabso

rptionofenergyfromareservoirandthe

perform

anceofanequal

amount

ofwork

Inshort

Noheatenginecan

completely

convertheatinto

mechanicalwork

Heatengine-device

thatconvertsinternal

energytomechanical

energy

SampleofHeatEngine

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Refrigerator

Idealefficiency

ofaheatengine

y

Example:

Findthe

efficiencyofaheatenginethatabsorbs

2000Jofenergyfromahotreservoirandexhausts

1500Jtoacoldreservoir?

Solution: e=1(Qc/Qh)

=1(1500J/2000J)

=0.25or25%

Entrop

y

Entropy

=fromG

reekwordsmeaningturning

into

=me

asureofhow

muchenergyorheatis

unavailableforconversionintowork

=me

asureoftheamountofdisorder

S=Q/T

Entrop

y

Entropyisa

quantitativemeasureofDISORDER

Restating2ndLawofThermodynamics:

Inanyis

olatedsystem,entropyincreases

Wheneverenergyisfreelytransformsfromoneformtoanother,

thedirectionofthetransformationistowardsastateofgreater

disorder(

greaterentropy)

Implicat

ion:NaturalSystemtendstowardsgreater

disorder

ReadMore

y

http://www.a

llaboutscience.org/laws-o

f-thermodynamics-

faq.htm

y

ConceptualPhysicsbyG.

Hewitt

y

CollegePhysicsbySerwayandVaughn