phy 231 1 physics 231 lecture 28: thermal conduction remco zegers walk-in hour: thursday 11:30-13:30...
Post on 21-Dec-2015
218 views
TRANSCRIPT
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 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 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