PHY 2311

PHYSICS 231Lecture 28: Thermal conduction

Remco ZegersWalk-in hour: Thursday 11:30-13:30 am

Helproom

PHY 2312

Previously: Phase Change

GAS(high T)

liquid (medium T)

Solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid

Q=mLf

Q=mLv

PHY 2313

PHY 2314

How can heat be transferred?

PHY 2315

Conduction

Touching different materials: Some feel cold, othersfeel warm, but all are at the same temperature…

PHY 2316

Thermal conductivity

metal wood

T=370CT=370C

T=200C

T=200C

The heat transferin the metal is much faster thanin the wood:(thermal conductivity)

PHY 2317

Heat transfer via conduction

Conduction occurs if there is a temperature difference betweentwo parts of a conducting medium

Rate of energy transfer PP=Q/t (unit Watt)P=kA(Th-Tc)/x=kAT/x

k: thermal conductivityUnit:J/(msoC)

metal k~300 J/(msoC)gases k~0.1 J/(msoC)nonmetals~1 J/(msoC)

PHY 2318

ExampleA glass window (A=4m2,x=0.5cm) separates a living room (T=200C) from the outside (T=0oC). A) Whatis the rate of heat transfer throughthe window (kglass=0.84 J/(msoC))?B) By what fraction does it changeif the surface becomes 2x smallerand the temperature drops to -200C?

A) P=kAT/x=0.84*4*20/0.005=13440 WattB) Porig=kAT/x Pnew=k(0.5A)(2T)/x=Porig

The heat transfer is the same

PHY 2319

Another one.

Heat reservoir Heat sink

An insulated gold wire (I.e. no heat lost to the air) is atone end connected to a heat reservoir (T=1000C) and at theother end connected to a heat sink (T=200C). If its lengthis 1m and P=200W what is its cross section (A)?

kgold=314 J/(ms0C).P=kAT/x=314*A*80/1=25120*A=200A=8.0E-03 m2

PHY 23110

And anotherWater 0.5L1000C

1500C

A=0.03m2 thickness: 0.5cm.

A student working for his exam feels hungry and starts boilingwater (0.5L) for some noodles. He leaves the kitchen when the water just boils.The stove’s temperature is 1500C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (Lv=540 cal/g, kpan=1 cal/(ms0C)

To boil away 0.5L (=500g) of water: Q=Lv*500=270000 calHeat added by the stove: P=kAT/x=1*0.03*50/0.005=

=300 calP=Q/t t=Q/P=270000/300=900 s (15 minutes)He’ll be hungry for a bit longer…

PHY 23111

Isolation

L1L2 L3

insi

de

ThTc

iii

ch

kL

TTA

t

QP

)/(

)(

A house is built with 10cm thick wooden walls and roofs.The owner decides to install insulation. After installationthe walls and roof are 4cm wood+2cm isolation+4cm wood.If kwood=0.10 J/(ms0C) and kisolation=0.02 J/(ms0C), by whatfactor does he reduce his heating bill?

Pbefore=AT/[0.10/0.10]=ATPafter=AT/[0.04/0.10+0.02/0.02+0.04/0.10]=0.55ATAlmost a factor of 2 (1.81)!

PHY 23112

Overview of material for exam: 8

=I (compare to F=ma)Moment of inertia I: I=(miri

2) (kgm2): angular acceleration (rad/s2)

I depends on the choice of rotation axis!!

F F

r r

PHY 23113

Chapter 8: Rotational KE

Rotational KEr=½I2

Conservation of energy for rotating object:

[PE+KEt+KEr]initial= [PE+KEt+KEr]final

[mgh+½mv2+½I2]initial= [mgh+½mv2+½I2]final

PHY 23114

Ch 8: Angular momentum

0L then 0 if

0

00

t

L

t

LL

IL

t

II

tII

Conservation of angular momentumIf the net torque equal zero, theangular momentum L does not change

Iii=Iff

=v/rL=I =mr2v/r =mvr

PHY 23115

Chapter 9 Young’s modulus.

LA

FL

LL

AFY

Y

0

0

0

2

/

/

L/L :strain tensile

](Pa) Pascal[N/mF/A :stress tensile

strain tensile

stress tensile

PHY 23116

Chapter 9 Shear modulus

xA

Fh

hx

AFS

S

/

/

x/h :strainshear

](Pa) Pascal[N/mF/A :stressshear

strainshear

stressshear

2

PHY 23117

Chapter 9: Bulk modulus

pressureP

VV

P

VV

AFB

B

00

0

2

//

/

V/V :strain volume

](Pa) Pascal[N/mF/A :stress volume

strain volume

stress volume

PHY 23118

Chapter 9: density

)/( 3mkgV

M

)4(/

Cwatermaterialspecific o

water=1.0x103 kg/m3

PHY 23119

Ch. 9 Pascal

Pascal’s principle: a change in pressure applied to a fluid that is enclosed is transmitted to the wholefluid and all the walls of the container that hold the fluid.

Enclosed fluid: F1/A1=F2/A2

PHY 23120

P = P0+ fluidghh: distance between liquid surface and the point where you measure P

P0

P

h

B = fluidVobjectg = Mfluidg = wfluid

The buoyant force equals the weight of the amount of water that can be put in the volume taken by the object.If object is not moving: B=wobject object= fluid

Pressure at depth h

Buoyant force for submerged object

Buoyant force for floating objecthB

w

The buoyant force equals the weight of the amount of water that can be put in the part of the volume of the object that is under water.objectVobject= waterVdisplaced h= objectVobject/(waterA)

PHY 23121

Bernoulli’s equation

P1+½v12+gy1=

P2+½v22+gy2

P+½v2+gy=constant

The sum of the pressure (P), the kinetic energy per unit volume (½v2) and the potential energy per unit volume (gy)is constant at all points along a path of flow.

Note that for an incompressible fluid:A1v1=A2v2

This is called the equation ofcontinuity.

PHY 23122

Surface tension

Fs=LL: contact length between object and liquid: surface tension

Fs Fs

Fg

Units of : N/m=J/m2

Energy per unit surface

PHY 23123

Poiseuille’s Law

How fast does a fluid flowthrough a tube?

Rate of flow Q= v/t=R4(P1-P2)

8L(unit: m3/s)

: coefficient of viscosity

PHY 23124

Ch. 10: Temperature scales

ConversionsTcelsius=Tkelvin-273.5Tfahrenheit=9/5*Tcelcius+32

We will use Tkelvin.

If Tkelvin=0, the atoms/moleculeshave no kinetic energy and everysubstance is a solid; it is called theAbsolute zero-point.

Kelvin

Celsius Fahrenheit

PHY 23125

Ch. 10: Thermal expansion

L=LoT

L0

L

T=T0T=T0+T

A=AoT =2

V=VoT =3

length

surface

volume

Some examples:=24E-06 1/K Aluminum=1.2E-04 1/K Alcohol

: coefficient of linear expansion different for each material

PHY 23126

Boyle & Charles & Gay-LussacIDEAL GAS LAW

PV/T = nR = Nkb

n: number of particles in the gas (mol)R: universal gas constant 8.31 J/mol·KN: number of atoms/moleculeskb:boltzmann’s constant 1.38x10-23 J/Kn=N/NA NA:Avogadro’s constant 6.02x1023

If no molecules are extracted from or added to a system:

2

22

1

11 constant T

VP

T

VP

T

PV

PHY 23127

Pressure

2

2

1

3

2vm

V

NP

Number of Molecules

Volume

Mass of 1 molecule

Averaged squared velocity

Average translation kinetic energy

Number of moleculesper unit volume

PHY 23128

M

RT

m

Tkvv

nRTTNkE

Tkvm

brms

Bkin

B

33

2

3

2

32

3

2

1

2

2

Average molecular kinetic energy

Total kinetic energy

rms speed of a moleculeM=Molar mass (kg/mol)

Ideal gases

PHY 23129

Ch. 11 Heat transfer to an object

Q=cmT

Energy transfer(J or cal)

Specific heat(J/(kgoC) or cal/(goC)

Mass of object

Change in temperature

The amount of energy transfer Q to an object with mass m when its temperature is raised by T:

PHY 23130

Calorimetry

If we connect two objects with different temperatureenergy will transferred from the hotter to the coolerone until their temperatures are the same. If the system is isolated:

Qcold=-Qhot

mcoldccold(Tfinal-Tcold)=-mhotchot(Tfinal-Thot)

the final temperature is: Tfinal=

mcoldccoldTcold+mhotchotThot

mcoldccold+mhotchot

PHY 23131

Phase Change

GAS(high T)

liquid (medium T)

Solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid

Q=mLf

Q=mLv

Make sure to understand: ice -> water -> steam

PHY 23132

Heat transfer via conduction

Rate of energy transfer PP=Q/t (unit Watt)P=kA(Th-Tc)/x=kAT/x

k: thermal conductivityUnit:J/(msoC)

iii

ch

kL

TTA

t

QP

)/(

)(For several layers: