1. PHYSICS 231 Lecture 28: Thermal conduction
ISOLATION
Remco Zegers
Walk-in hour: Thursday 11:30-13:30 am
Helproom
PHY 231

2. Previously: Phase Change
GAS(high T)
Q=cgasmT
Q=csolidmT
Solid (low T)
liquid 
solid
Gas 
liquid
Q=mLv
liquid (medium T)
Q=mLf
Q=cliquidmT
PHY 231

3. PHY 231

4. How can heat be transferred?
PHY 231

5. Conduction Touching different materials: Some feel cold, others
feel warm, but all are at the same temperature…
PHY 231

6. Thermal conductivity metal T=200C wood T=200C The heat transfer
in the metal is
much faster than
in the wood:
(thermal
conductivity)
T=370C
T=370C
PHY 231

7. Heat transfer via conduction
Conduction occurs if there is a
temperature difference between
two parts of a conducting medium
Rate of energy transfer P
P=Q/t (unit Watt)
P=kA(Th-Tc)/x=kAT/x
k: thermal conductivity
Unit:J/(msoC)
metal k~300 J/(msoC)
gases k~0.1 J/(msoC)
nonmetals~1 J/(msoC)
PHY 231

8. Example A glass window (A=4m2,x=0.5cm)
separates a living room (T=200C)
from the outside (T=0oC). A) What
is the rate of heat transfer through
the window (kglass=0.84 J/(msoC))?
B) By what fraction does it change
if the surface becomes 2x smaller
and the temperature drops to -200C?
A) P=kAT/x=0.84*4*20/0.005=13440 Watt
B) Porig=kAT/x Pnew=k(0.5A)(2T)/x=Porig
The heat transfer is the same
PHY 231

9. Another one. Heat sink Heat reservoir
An insulated gold wire (I.e. no heat lost to the air) is at
one end connected to a heat reservoir (T=1000C) and at the
other end connected to a heat sink (T=200C). If its length
is 1m and P=200W what is its cross section (A)?
kgold=314 J/(ms0C).
P=kAT/x=314*A*80/1=25120*A=200
A=8.0E-03 m2
PHY 231

10. And another Water 0.5L 1000C A=0.03m2 thickness: 0.5cm. 1500C
A student working for his exam feels hungry and starts boiling
water (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= cal
Heat added by the stove: P=kAT/x=1*0.03*50/0.005=
=300 cal
P=Q/t t=Q/P=270000/300=900 s (15 minutes)
He’ll be hungry for a bit longer…
PHY 231

11. Isolation Tc Th inside L1 L2 L3
A house is built with 10cm thick wooden walls and roofs.
The owner decides to install insulation. After installation
the walls and roof are 4cm wood+2cm isolation+4cm wood.
If kwood=0.10 J/(ms0C) and kisolation=0.02 J/(ms0C), by what
factor does he reduce his heating bill?
Pbefore=AT/[0.10/0.10]=AT
Pafter=AT/[0.04/ / /0.10]=0.55AT
Almost a factor of 2 (1.81)!
PHY 231

12. Overview of material for exam: 8
=I (compare to F=ma)
Moment of inertia I: I=(miri2) (kgm2)
: angular acceleration (rad/s2)
I depends on the choice of rotation axis!! F r
PHY 231

13. Chapter 8: Rotational KE
Rotational KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
[mgh+½mv2+½I2]initial= [mgh+½mv2+½I2]final
PHY 231

14. Ch 8: Angular momentum =v/r L=I =mr2v/r =mvr
Conservation of angular momentum
If the net torque equal zero, the
angular momentum L does not change
Iii=Iff
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15. Chapter 9 Young’s modulus.
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16. Chapter 9 Shear modulus
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17. Chapter 9: Bulk modulus
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18. Chapter 9: density
water=1.0x103 kg/m3
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19. Ch. 9 Pascal Enclosed fluid: F1/A1=F2/A2
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the whole
fluid and all the walls of the container that hold the fluid.
PHY 231

20. h: distance between liquid surface and the point where you measure P h
Pressure at depth h
P = P0+ fluidgh
h: distance between liquid surface
and the point where you measure P h P
Buoyant force for submerged object
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 h B w
Buoyant force for floating object
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 231

21. Bernoulli’s equation P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant
The sum of the pressure (P), the kinetic energy per unit
volume (½v2) 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 of
continuity.
PHY 231

22. Fs Fs Fg Surface tension Fs=L
L: contact length between object and liquid
: surface tension Fs Fs  Fg
Units of : N/m=J/m2
Energy per unit surface
PHY 231

23. Poiseuille’s Law How fast does a fluid flow through a tube? R4(P1-P2)
(unit: m3/s)
Rate of flow Q= v/t=
8L
: coefficient of viscosity
PHY 231

24. Ch. 10: Temperature scales
Conversions
Tcelsius=Tkelvin-273.5
Tfahrenheit=9/5*Tcelcius+32
We will use Tkelvin.
If Tkelvin=0, the atoms/molecules
have no kinetic energy and every
substance is a solid; it is called the
Absolute zero-point.
Celsius
Fahrenheit
Kelvin
PHY 231

25. Ch. 10: Thermal expansion L=LoT A=AoT =2 V=VoT =3 
length L
L=LoT
surface
A=AoT =2
volume
V=VoT =3  L0
: coefficient of linear expansion
different for each material
Some examples:
=24E-06 1/K Aluminum
=1.2E-04 1/K Alcohol
T=T0
T=T0+T
PHY 231

26. Boyle & Charles & Gay-Lussac IDEAL GAS LAW
PV/T = nR = Nkb
n: number of particles in the gas (mol)
R: universal gas constant 8.31 J/mol·K
N: number of atoms/molecules
kb:boltzmann’s constant 1.38x10-23 J/K
n=N/NA NA:Avogadro’s constant 6.02x1023
If no molecules are extracted from or added to a system:
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27. Pressure Number of Molecules Mass of 1 molecule
Averaged squared velocity
Number of molecules
per unit volume
Volume
Average translation kinetic energy
PHY 231

28. Average molecular kinetic energy
Ideal gases
Average molecular kinetic energy
Total kinetic energy
rms speed of a molecule
M=Molar mass (kg/mol)
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29. Ch. 11 Heat transfer to an object
The amount of energy transfer Q to an object with mass m
when its temperature is raised by T:
Q=cmT
Change in
temperature
Mass of object
Energy transfer
(J or cal)
Specific heat
(J/(kgoC) or cal/(goC)
PHY 231

30. Calorimetry If we connect two objects with different temperature
energy will transferred from the hotter to the cooler
one 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 231

31. Phase Change GAS(high T) Q=cgasmT Q=csolidmT Solid (low T) liquid 
Q=mLv
liquid (medium T)
Q=mLf
Q=cliquidmT
Make sure to understand: ice -> water -> steam
PHY 231

32. Heat transfer via conduction
Rate of energy transfer P
P=Q/t (unit Watt)
P=kA(Th-Tc)/x=kAT/x
k: thermal conductivity
Unit:J/(msoC)
For several layers:
PHY 231