Today’s Lecture
Thermodynamics
Heat Transfer
Heat Transfer
1.Conduction.
2. Convection
3. Radiation
Heat Conduction
A rectangular slab of thickness Δx and
with an area A.
The front side of the slab is at a
temperature T; the back side has a
somewhat different temperature,
T+ΔT.
Heat flows from the hotter to the
cooler side of the slab
We are trying to calculate the heatflow rate, the amount of heat flowing
through the slab per unit time,
H = ΔQ/Δt.
We expect H to be proportional to the area, A, of the slab, the
temperature difference, ΔT, between the back and the front and inversely
proportional to the thickness of the slab, Δx.
H should also depend of properties of the material the slab is made of…
Heat Conduction
Bringing all the parts together:
ΔT
H = − kA
Δx
The coefficient k reflects specific
properties of the material of the
slab and is called thermal
Heat flows from the hotter to the
cooler side of the slab
conductivity
H = ΔQ/Δt – heat-flow rate is measured in Joules/second, J/s, or Watts, W.
Thermal conductivity, k, is measured in W/m⋅K.
Increasing Heat-Flow Rate for More
Efficient Heat Transfer.
ΔT
H = − kA
Δx
Cooling and heating radiators:
• thin, highly heat conductive materials (metals) - small Δx
large k;
• large surface area A.
Reducing Heat-Flow Rate for Better
Thermal Insulation.
ΔT
H = − kA
Δx
Thick (large Δx) cavities in house walls
filled with insulating (small k) materials;
Reducing the number of walls (small surface area A).
Thermodynamic Equilibrium
An object can maintain thermal equilibrium by
balancing energy gain and loss. A home is a good
example of thermal energy balance. Energy from
the furnace balances the heat loss, maintaining a
constant temperature.
A home is insulated so that it looses 350W/oC.
What is the temperature inside during a 40
person party with the heat turned off, if it is 12oC
outside and the average power output per person
is 100W?
Assume an inside temperature, T. Then the expression for energy balance is:
Thermal Resistance
ΔT
H = − kA
Δx
Analogous to the
Ohm’s law:
V
I=
R
The current, I = Δq/Δt, amount of charge per unit time, is
analogous to the heat-flow rate, H = ΔQ/Δt, “thermal
current”. The voltage, V, the factor driving the electric current,
is analogous to temperature difference, ΔT, “thermal voltage” .
Continuing the analogy: electric resistance, R, is analogous to
Thermal resistance is introduced as
The electrical conductivity, σ, is analogous to the thermal conductivity, k.
Thermal Resistance
Thermal resistance
is introduced as:
Δx
R=
kA
Heat-flow rate
equation is now:
ΔT
H =−
R
When heat flows through two
materials, it is analogous to two
resistors connected in a series.
To see this we first write down the heat flow
through each slab:
What about this temperature T2?
Thermal Resistance
Thermal resistance
is introduced as:
Δx
R=
kA
Heat-flow rate
equation is now:
ΔT
H =−
R
The temperature T2 is the temperature
required so that at equilibrium the heat
flow is uniform, H1 = H2 = H. This is
analogous to the current being the same
through both resistors connected in
series.
The analogy is complete. The “thermal current” is uniform and the “thermal
voltages” add.
Thermal Resistance
A bar of a gold alloy (Au) is in thermal
contact with a bar of a silver alloy (Ag). Both
bars have the same length, Δx, and cross
sectional area, A. One end of the bar is held
at 30°C while the other is held at 80°C.
Given that the thermal conductivities, k, of
the gold and silver alloys are 300W/(m°C)
and 450W/(m °C) respectively, what is the
temperature at the junction when the heat
flow reaches equilibrium?
Since T3 = 80oC and T1 = 30oC, T2 is given by:
For this junction temperature the heat flow is continuous throughout the bar!
Licking an ice cream, which is
frozen, seems to be OK…
What about the door handles, when it is freezing outside?
Which one would
you prefer?
Cold metals are especially bad because of
their high heat… conductivity
In order to keep your tongue above 0°C you basically
have to heat the whole piece of metal…
Otherwise…
Thermal Conductivities and Specific heats
of Different Materials.
Example
Consider a copper pan that is 1.5mm thick with a surface area of 300cm2.
The pan is heating 2kg of water whose temperature is rising at a rate of
.15K/sec. Find the temperature difference across the thickness of the
bottom of the pan.
Solution: The rate of heat conduction across the bottom of the pan
supplies the heat necessary to raise the temperature of the water.
This temperature difference depends on the material used to make
The pan as well as the thickness of the bottom of the pan, Δx/k.
Heat Transfer
1. Conduction.
1.Convection
2. Radiation
Convection is heat transfer by the bulk
motion of a fluid.
Calculation of convective heat transfer is
complicated due to details of the associated
fluid motion.
Convection generates an active flow and
efficient mixing of the liquid.
Hot and cold liquid is brought in a thermal contact;
it reduces the distance across which the
conduction occurs and increases the contact
area.
ΔT
H = − kA
Δx
Heat Transfer
1. Conduction
1. Convection
2. Radiation
All matter radiates energy via
black-body radiation!
Black Body Radiation
A body is a “black body” if it absorbs all the light that
falls on it. This applies not only to dark objects (black
ink) but also to stars.
In equilibrium an object emits radiation at the same rate
it is absorbed. A perfect absorber is therefore also a
perfect emitter. At room temperature such an object is
black, hence the name.
When the object is not in equilibrium it still radiates at a
rate set by its own temperature. The burner on a stove
starts out black, but when heated it glows red. Since it
still absorbs all the light that hits it, it is still a black body.
Stars are the same!
Black body demonstration here
Black Body Radiation
As the poker gets hotter its color changes from black to red to
white! Yet it is still a black body!
Intensity Distribution for Black
Body Radiation
Max Planck generated an expression for the intensity distribution for
the radiation from a black body as a function of wavelength and
temperature.
I, T 
2hc 2
 5 e hc/k B T − 1
To do this he had to make the assumption
that energies of an oscillator inside a black
body only had discrete values, they were
quantized (Physics 2D)!
E n  nhf and h  6. 626  10 −34 J  s
These oscillators only emitted or absorbed
energy when they changed quantum states.
Blackbody Spectra for Different
Temperatures
By fitting distribution curves for black body radiation Planck was
able to determine the value of the constant that bears his name,
Planck’s constant,
h = 6.632x10-34Js
Blackbody Spectra for Different
Temperatures
Wien’s law states that the peak in the intensity of blackbody
radiation is inversely proportional to the temperature, λmax = a / T. If
you can take derivatives this law is straightforward to derive. The
intensity distribution is:
I, T 
2hc 2
 5 e hc/k B T − 1
It is easier to take the derivatives if we simply change variables
such that x = hc/λkBT. The intensity distribution then becomes:
2k B T 5 x 5
Ix, T 
x
h4 c3 e − 1
By solving a transcendental
equation we find:
dI
dx
xx max
 0 → x max  4. 965
Blackbody Spectra for Different
Temperatures
Wien’s law states that the peak in the intensity of blackbody radiation
is inversely proportional to the temperature, λmax = a / T.
Defining x = hc/λkBT, we found that xmax= 4.965. From this we can
solve for the wavelength that corresponds to the maximum in the
intensity distribution:
 max
−3
hc
2.
90

10
1


m
T
x max k B T
For our Sun the wavelength for the maximum intensity is ~502nm.
Hence the surface temperature of our Sun is found from:
T⊙
−3
2.
90

10

 5780K
−9
502  10
In general the intensity distribution of the black body radiation
from a star yields the surface temperature of that star!
Black Body Radiation
Also From Planck’s intensity distribution it can shown by integrating
the distribution over all wavelengths that the power radiated by a
black body is given by the
Stephan Boltzmann law:
P  eAT 4
Here e is the emissivity which is 1 for a perfect black body. A is the
surface of the black body, T its absolute temperature, and σ is the
Stephan Boltzmann constant which is
found from the integration is given by:
  5. 67  10 −8
For a star of radius r this
expression becomes:
P  4r 2 T 4
From these relations and knowledge of the total power output
(luminosity) of the radiation from the Sun, we can determine its size.
Size of the Sun
The total radiated power
(luminosity) of the Sun is:
P ⊙  3. 86  10 26 W
From Wien’s Law we found that he surface temperature is
T = 5780K. From the Stephan-Boltzmann law we can find the
surface area and/or the radius of the Sun from:
P ⊙  e ⊙ A⊙ T 4⊙  4R 2⊙ T 4⊙
R⊙
 12
T⊙
P⊙
1

4
5780 2
3. 86  10 26
4  5. 67  10 −8
R ⊙  697  10 6 m  697, 000km
So from the study its black body radiation, astronomers can first
determine the temperature of the Sun’s surface, and knowing its
power output determine its size!
Black Body Radiation
In equilibrium an object emits radiation at the same rate
it is absorbed. A perfect emitter is therefore a perfect
absorber. At room temperature this object is black,
hence the name.
When the object is not in equilibrium it still radiates at a
rate set by its own temperature. It now absorbs at a rate
set by the temperature of its surroundings and
Energy Balance
Incident sunlight with flux at
Earth’s surface, S=960W/m2.
Outgoing radiation
(predominantly infrared)
Average temperature
of the Earth is T=14oC?
Natural gases, CO2 and H2O, absorb
radiation, a green house effect.
Nuclear Winter
If a nuclear war resulted in 8%
less solar energy from reaching
the Earth. What would happen
to the 287K average temp?
Energy Balance
Since a blackbody radiates via T4, if
the incident energy is reduced by .92
Global Warming
The opposite of nuclear winter
is happening. The Earth is
experiencing Global Warming.
The increase of CO2 in the
atmosphere is increasing the
green house effect.
To turn around this effect will
require an enormous effort on
all of the world’s citizens!
Thermal Expansion
The change in volume or length of a substance with temperature is
characterized the coefficient of volume or length expansion, which are
given by:
As an example consider filling up your tank with gasoline in the morning
when T=15oC vs the afternoon when T=30oC. For gasoline β = 9.5x10-4K-1.
Assume that your tank has a maximum volume of V=60L, then
At today’s gasoline prices that is approximately $1.00 for each fill up!
Thermal Expansion
There is a relationship between the coefficient of volume and length
expansion. Consider a material whose volume is V=xyz. Then the relative
change in volume is
And the coefficient of volume expansion is
Thermal Expansion
The coefficient of volume expansion is:
If the material is isotropic (all directions equal) then
What about the relative change in cross sectional areas?
Consider an aluminum toroid (donut), does the radius of
hole increase or decrease as the aluminum is heated up?
Thermal Expansion - example
A glass marble is 1.000cm in diameter. At room temperature
the hole diameter in a steel plate is .997cm. How much must
the temperature of the steel be raised for the marble to fit through?
For steel α = 12x10-6/K
What is the relative change in the surface area of the steel?
Could we have just written down this result?
You betcha!
Summary of Temperature and Heat
Zeroth law of thermodynamics => basis for measuring temperature
Thermal balance at equilibrium – identical temperatures!
Temperature scales, Fahrenheit, Celsius, Kelvin
Heat capacity and specific heat
Heat Conduction and thermal conductivity –
Thermal resistance –
Blackbody radiation –
Wien’s & Stephan-Boltzmann laws
Thermal expansion