Latent Heat
When a solid melts or a liquid boils, energy
must be added but the temperature remains
constant! (This can be explained by
considering that it takes energy to break the
bonds holding the material together.)
The amount of energy it takes to melt or boil a
certain amount of material is called a latent
heat.
Latent Heat
For water, the latent heat of fusion (heat
needed to melt ice to water) is 79.7 cal/gm.
For water, the latent heat of vaporization
(heat needed to boil water) is 540 cal/gm.
For alcohol, the latent heat of vaporization is
less at 204 cal/gm.
Note: These values are huge compared to the heat capacity of
water which is 1 cal/gmoC.
Latent Heat - Example
Example: how much energy does it take to
vaporize 1 liter of water if the water is
initially at a temperature of 98oF ?
Latent Heat - Example cont.
First we need to find the energy to raise the
temperature of the water up to boiling:
this involves the heat capacity
(which for water is 1 cal/gm*oC)
(density of water is 1 gm/cc, 1 liter = 1000 cc):
C = Q/(m*T) , with T = (212-98)oF *5oC//9oF =63oC
Q = (1 cal/gm*oC)*(1 gm/1cc)*1000 cc/l*63oC
= 63,333 cal/l * (4.186 J/cal) = 265,000 J/l .
Latent Heat - Example cont.
Now we add in the latent heat:
(for water, this is 540 cal/gm)
Q = L*m = (540 cal/gm)*(1 gm/cc)*(1000 cc/l)
= 540,000 cal/l * (4.186 J/cal) = 2,260,000 J/l
Total energy required is:
265,000 J/l + 2,260,000 J/l = 2,525,000 J/l .
Note: the latent heat component is much larger than the heat
capacity component.
Latent Heat - Example #2
Question: how much water at 98oF would be
needed to keep cool for 4 hours by evaporation
if the outside temperature is 100 oF (essentially
same as body’s) and a power output of 200
Watts (doing some work) is desired?
Latent Heat - Example cont.
Since the body generates 200 Watts, or 200
Joules a second, the body must evaporate
water to carry this energy away.
Q =(200 J/sec)*(4 hr)*(3600 sec/hr)= 2,880,000 J.
From the previous considerations, evaporating
1 liter of water carries away 2,525,000 J/l.
Thus we need 2.88MJ / (2.525MJ/liter)
= 1.14 liters of water.
Interesting note: if cool water at 41oF was drunk, the amount
would only change by .005 liters.
Latent Heat - Example cont.
Would more or less alcohol be needed to keep
cool for the same energy output?
(The heat capacity of alcohol is 2.4 J/gm*oC; the
density of [ethanol] .791 gm/cc; the boiling point
is 78oC; latent heat of vaporization is 854 J/gm).
From this you should be able to decide
whether water or alcohol is better for heat
regulation.
Heat Transfer
There are four ways of moving heat:
•
Evaporation (using latent heat - we’ve already
looked at this)
•
Convection (moving heat with a material)
•
Conduction (moving heat through a material)
•
Radiation
We’ll develop equations for conduction and
radiation and talk about convection.
Heat Transfer: Convection
Heat Transfer by Convection is when you heat
some material and then move that material
containing the heat.
The amount of heat energy moved depends on
the heat in the material (heat capacity times
amount of material times the temperature difference)
and how much material you move per time.
The blood and hot air furnaces use this method.
Heat Transfer: Conduction
Heat will flow through a solid material from
the hot end to the cold end. What is
flowing? No matter is flowing!
We can think of energy as flowing in this
case! We measure the flow of energy as
power: 1 Watt = 1 Joule/sec .
Heat Transfer: Conduction
Power = Q/t = k*A*T/L
where k is a constant that depends on the
material, called the thermal conductivity;
where A is the cross sectional area;
L
where L is the distance from the hot end to
the cold end;
A
and T is the temperature difference
hot
k
between the hot and cold ends.
Thi
cold
Tlow
Conduction - R values
Units of thermal conductivity: from
Power = Q/t = k*A*T/L
k has units of W*m/(m2K) or J/(sec*m*K), and
k depends only on the material.
Often a material is given an R value, where R
includes both the material and the thickness of
the material: R = L/k, and R has the units of
m2*sec*K/J (or ft2*oF*hr/BTU), where
1 ft2*oF*hr/BTU = 0.176 m2*sec*K/J )
Conduction - R values
P = Q/t = k*A*T/L = A*T / R
where we define R = L / k .
where if we have several different materials
and thicknesses, we can simply add the
individual R’s to get the total R:
Rtotal = Ri .
Conduction - Conductivity
approximate thermal conductivity (k) values
for some materials:
– metals:
k=1
– glass:
k = 2 x 10-3
– wood, brick, fiberglass:
k = 1 x 10-4
– air:
k = 5 x 10-5
cal/(sec*cm*oC)
cal/(sec*cm*oC)
cal/(sec*cm*oC)
cal/(sec*cm*oC)
Conduction - Example
Let’s calculate the R value of brick 4 inches in
thickness: L = 4 in * (.0254 m/in) = .10 m
k = 1.5 x 10-4 cal/(sec*cm*oC)
* (4.186 J/cal)*(100 cm/m) = .063 J/(sec*m*oC)
R = L/k = .10 m / .063 J/(sec*m*oC)
= 1.6 m2*K/Watt
* (1 ft2*oF*hr/BTU) /( .176 m2*K/Watt)
= 9 (1 ft2*oF*hr/BTU)
Conduction - Example
What is the heat loss through the brick walls
(assume no other insulation) of a house that
is 50 ft x 30 ft (floor area: 1500 ft2) x 8 ft
when the temperature inside is 72oF and the
temperature outside is 20oF ?
Conduction - Example
P = Q/t = k*A*T/L = A*T / R
A4 walls = (50ft*8ft) + (30ft*8ft) + (50ft*8ft) + (30ft*8ft)
= 1280 ft2 * (1 m2/10.76 ft2) = 120 m2 .
T = (72-20)oF * (5K/9oF) = 29 K
Rvalue = 1.6 m2*K/Watt
P = 120 m2 * 29 K / (1.6 m2*K/Watt)
= 2,175 Watts = 2.175 kW.
Conduction - Example
Is the heat lost through the ceiling more or less than that through
the walls?
Since the area of the ceiling is 50 ft x 30 ft = 1500 ft2, and the area
of the outside walls is only 1280 ft2, if we assume that the R
values are the same and the ΔT is the same, the heat lost
through the ceilings will be 17% more.
Note that if the heat is provided by electricity, and the cost of that
electricity is 9 cents/kW-hr, then the cost per month to replace
the heat lost by conduction through the walls and ceilings would
be:
P = 2175 W + 2550W = 4725 W = 4.725 kW;
# hrs/month = 30 days/month * 24 hrs/day = 720 hrs/ month
Cost = [$.09/kW*hr] * [4.725 kW * 720 hrs/month] = $306/month.
Note that this does not include heat lost by convection or
radiation.
Convection comparison
For the same size house and situation as in the
previous example: 1500 sq ft with 8 ft
ceilings, inside temperature of 72oF and
outside temperature of 20oF, how much heat
is lost per time via convection (that is air
flow) through the doors, ceilings, and
cracks if we assume one air exchange every
two hours?
Convection example (same house)
Q = Cn*n*ΔT where Cn = (7/2)*R for air
ΔT = 29K (for inside air at 72oF and outside air at 20oF)
From ideal gas law: PV = nRT, we get
n = (P*V)/(R*T) where P = 1 atm = 1.01 x 105 Nt/m2
V = 1500 ft2 * 8 ft = 12,000 ft3 = 340 m3
T = midpoint of temperature range = 46oF = 281 K
n = (1.01x105 Nt/m2)(340 m3)/[(8.3 J/moleK)*281K] =
14,700 moles
Q = (7/2)*(8.3 J/mole*K) * 14,700 moles * 29 K
= 12.4 x 106 J = 3.44 kW-hrs.
P = Q/time = 3.44 kW-hrs / (2 hrs/exchange) = 1.72 kW.
Radiation: Reflected Light versus
Emitted Light
Some objects, like the moon, can be seen by the light they
reflect (from the sun). Other objects can be seen by the
light that they create, like the sun. How do we tell the
difference, and how do we analyze the light created
(emitted)? First, consider this:
A BLACK object absorbs all the light incident on it.
A WHITE object reflects all the light incident on it, usually
in a diffuse way rather than in a specular (mirror-like) way.
black
white - diffuse
white - specular
Blackbody Radiation:
The light from a “blackbody”, then, is light
that comes solely from the object itself
rather than being reflected from some other
source.
A good way of making a blackbody is to force reflected light
to make lots of reflections: inside a bottle with a small
opening.
Blackbody Radiation:
If very hot objects glow (such as the filaments of
light bulbs and electric burners), do all warm
objects glow?
Do we glow? (Are we warm? Are you HOT?)
Blackbody Radiation:
What are the parameters associated with the
making of light from warm objects?
Blackbody Radiation:
What are the parameters associated with the
making of light from warm objects?
– Temperature of the object.
– Surface area of the object.
– Color of the object ? (If black objects
absorb better than white objects, will black
objects emit better than white objects?)
Blackbody Radiation:
Color Experiment
Consider the following way of making your
stove hot and your freezer cold:
Color Experiment
Put a white object in an insulated and evacuated box
with a black object. The black object will absorb
the radiation from the white object and become
hot, while the white object will reflect the
radiation from the black object and become cool.
Put the white object in the freezer, and the black
object in the stove. Purple arrows come from black
object, orange arrows come from white object.
Color Experiment
Does this violate Conservation of Energy?
Color Experiment
Does this violate Conservation of Energy?
NO
Does this violate any other law?
Color Experiment
Does this violate Conservation of Energy? NO
Does this violate any other law? For instance, the
Second Law of Thermodynamics (entropy tends
to increase) ? YES
(We’ll consider this law shortly.)
This means that a good absorber is also a good
emitter, and a poor absorber is a poor emitter.
Use the symbol  to indicate the
blackness (=1) or whiteness (=0) of an object.
Blackbody Radiation:
What are the parameters associated with the
making of light from warm objects?
– Temperature of the object, T.
– Surface area of the object, A.
– Color of the object, 
Blackbody Radiation:
Is the  for us close to 0 or 1?
(i.e., are we white or black?)
We emit light in the IR, not the visible.
So what is our  for the IR?
Blackbody Radiation:
So what is our  for the IR?
Have you ever been near a fire on a cold
night?
Have you noticed that your front can get hot
at the same time your back can get cold?
Can your hand block this heat from the fire?
Is this due to convection or radiation?
Blackbody radiation:
For humans in the IR, we are all fairly good
absorbers (black). An estimated value for 
for us then is about .97 .
Blackbody Radiation:
Experimental Results
At 310 Kelvin, only get IR
Intensity
(log scale)
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Experimental Results
At much higher temperatures, get visible
look at blue/red ratio to get temperature
Intensity
(log scale)
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Experimental Results
Ptotal = AT4
where  = 5.67 x 10-8 W/m2 *K4
peak = b/T where b = 2.9 x 10-3 m*K
Intensity
(log scale)
UV
blue
yellow
wavelength
red
IR
Blackbody Radiation:
Example
• Given that you eat 2,000 Calories/day,
your power output is around 100 Watts.
• Given that your body temperature is
about 90o F , and
• Given that your surface area is about
1.5 m2,
Blackbody Radiation:
Example
• Given Ptotal = 100 Watts
• Given that Tbody = 90o F
• Given that A = 1.5 m2
WHAT IS THE POWER EMITTED VIA
RADIATION?
Blackbody Radiation:
Example
Pemitted = AT4
–
–
–
–
 = .97
= 5.67 x 10-8 W/m2 *K4
T = 273 + (90-32)*5/9 (in K) = 305 K
A = 1.5 m2
Pemitted = 714 Watts
(compared to 100 Watts generated!)
Blackbody Radiation:
Example
need to consider power absorbed at room T
Pabsorbed = AT4
–
–
–
–
 = .97
= 5.67 x 10-8 W/m2 *K4
T = 273 + (72-32)*5/9 (in K) = 295 K
A = 1.5 m2
Pabsorbed = 625 Watts
(compared to 714 Watts emitted!)
Blackbody Radiation:
Example
Total power lost by radiation =
714 W - 625 W = 89 Watts
(Power generated is about 100 Watts for a
2,000 calorie/day diet.)
Power is also lost by convection (with air)
and by evaporation.
Blackbody Radiation:
Example
At colder temperatures, our emitted power
stays about the same while our absorbed
power gets much lower. This means that
we will get cold unless
– we generate more power, or
– our skin gets colder, or
– we reflect the IR back into our bodies.
Use metal foil for insulation!
Computer Homework
The Computer Homework Program on Heat,
Volume 2, #9, has problems dealing with
ideal gases, heat capacity, and heat flow.
Thermodynamics
The First Law of Thermodynamics is a fancy
name for the Law of Conservation of Energy
applied to thermal systems. It says:
U = Q - W
where U indicates the change in the internal energy
of the system. This internal energy is related to the
temperature and heat capacity of the system; Q is the
heat energy added to the system; and W is the work
done by the system.
Thermodynamics
The first law of thermodynamics, like the
conservation of energy, does not indicate
the direction. It does not explain why,
when cold milk is added to hot coffee, the
cold milk warms up and the hot coffee cools
down. The conservation of energy (first law
of thermodynamics) permits the possibility
that the milk would get even colder while
the coffee gets hotter after they are mixed.
Second Law of Thermodynamics
It is the Second Law of Thermodynamics
that explains why the hot coffee does cool
down and the cold milk warms up when
they are mixed.
To understand the second law, however, we
need to first look a little at probability.
Probability
Consider flipping four coins. How many
heads would you expect to get (assuming they
were honest coins)?
Why do you expect this?
Let’s look at all the possible combinations of
flipping four coins:
Flipping Four Coins
Four heads: (only one way)
Three heads: (four ways)
THHH HTHH HHTH
Two heads: (six ways)
TTHH
THTH
THHT
HTTH
HTHT
HHTT
One head: (four ways)
HTTT
THTT
TTHT
Zero heads: (only one way)
HHHH
HHHT
TTTH
TTTT
Probability
We see that there are more ways of getting
two heads and two tails than any other
combination.
The same argument can be made about the
distribution of energy among many
molecules: the highest probability
corresponds to the most ways of having that
outcome.
Probability
In the case of distributing the thermal energy
between the hot coffee and the cold milk,
there are more ways of distributing the
energy equally among the many coffee and
milk molecules than there are ways of
giving it all to just the coffee or just the
milk molecules.
Statement of 2nd Law
A system will tend to go to its most
probable state.
To measure the ways of having the same state
(like determining the number of ways of having
two heads out of four coins), we use the
concept of entropy.
Another Statement
Entropy is a measure of the probability of
being in a state. Since things tend to go to
their most probable state, we can write the
2nd Law of Thermodynamics as: systems
tend to have their entropy increase.
About the 2nd Law
Note that this is not an absolute law like
Conservation of Energy. Rather it is a
probabilistic Law. However, when dealing
with large numbers (recall that Avagadro’s
Number is 6 x 1023, or almost a trillion trillion),
the probabilities become essentially
certainties.
Heat Engines
Heat engines are devices that convert some of the
heat into useful energies such as electrical.
These engines can only work, however, if there
is a difference in temperature.
We can think by analogy: just as gravity tends to
bring masses together, and we can get work out
of a gravitational separation; so entropy tends to
increase by bringing temperatures together, and
we can get work out of a separation of
temperatures.
Efficiency
Efficiency is a measure of how much you get
out versus how much you put in. For heat
engines, then:
Efficiency =  = Work done / Heat Added
By the first law, then, the work done is simply
the difference in the heat going into the engine
minus the heat coming out of the engine. The
total heat added is the heat going into the
engine.  = (Qhot - Qcold) /Qhot
Carnot Efficienty
By considering the most efficient way of
running a heat engine, we come up with the
Carnot cycle. This is the best we can
theoretically do with a heat engine. For a
Carnot efficiency, we have the formula:
Carnot = (Thot - Tcold) / Thot .
Note that these temperatures must be in
absolute (Kelvin), not oC or oF .
Heat Engines
Our heat engines, whether fired by coal, oil or
nuclear energy, are all limited in their
efficiencies by this Carnot efficiency. In
practice we come very close to this theoretical
maximum.
Note that for the best efficiencies, we need Thot
to be very hot and Tcold to be very cold. Due
to material limitations on both Thot and Tcold,
this efficiency is about 30%.
Heat Engines
All heat engines, then, have to get rid of the
excess heat energy. Most major power
stations do this via water because of the
high heat capacity of water relative to air.
The biggest structure at nuclear power stations is the
cooling tower, and this is often depicted as a
symbol of nuclear power. Yet the same heat dump
is present at coal and oil facilities!
Heat Engines
Your car is a heat engine, but it is only about
15% efficient compared to a major power
plant’s efficiency of 30%. Your car’s
engine cannot operate at the same high
temperature as the power plant!
Your car employs both a water transfer (via
water convection inside the engine) as well
as an air transfer (at the radiator) for the
waste heat.