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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.)

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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.

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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.

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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 ?

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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)*5/9=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 .

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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 .

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Latent Heat - Example #2

Question: how much water 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?

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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.

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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.

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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.

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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.

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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 .

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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;

where L is the distance from the hot end to the cold end;

and T is the temperature difference between the hot and cold ends.









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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 )

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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 .

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Conduction - Conductivity

approximate thermal conductivity (k) values for some materials:

  • metals: k = 1 cal/(sec*cm*oC)

  • glass: k = 2 x 10-3 cal/(sec*cm*oC)

  • wood, brick, fiberglass:

    k = 1 x 10-4 cal/(sec*cm*oC)

  • air: k = 5 x 10-5 cal/(sec*cm*oC)

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Conduction - Example

Let’s calculate the R valueof 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)

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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 ?

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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.

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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.

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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?

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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 getn = (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 Kn = (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.

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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.

white specular


white - diffuse

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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.

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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?)

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Blackbody Radiation:

What are the parameters associated with the making of light from warm objects?

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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?)

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Blackbody Radiation:Color Experiment

Consider the following way of making your stove hot and your freezer cold:

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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. Black arrows come from black object, orange arrows come from white object.

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Color Experiment

Does this violate Conservation of Energy?

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Color Experiment

Does this violate Conservation of Energy? NO

Does this violate any other law?

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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.

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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, 

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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?

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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?

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Blackbody radiation:

For humans in the IR, we are all fairly good absorbers (black). An estimated value for  for us then is about .97 .

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Blackbody Radiation:Experimental Results

At 310 Kelvin, only get IR


(log scale)

blue yellow red




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Blackbody Radiation:Experimental Results

At much higher temperatures, get visible

look at blue/red ratio to get temperature


(log scale)

blue yellow red




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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


(log scale)

blue yellow red




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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,

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Blackbody Radiation:Example

  • Given Ptotal = 100 Watts

  • Given that Tbody = 90o F

  • Given that A = 1.5 m2



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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!)

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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!)

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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.

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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!

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Computer Homework

The Computer Homework Program on Heat, Volume 2, #9, has problems dealing with ideal gases, heat capacity, and heat flow.

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The First Law of Thermodynamics is a fancy name for the Law of Conservation of Energy applied to thermal systems. It says:

DU = Q - W

where DU 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.

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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.

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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 atprobability.

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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:

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Flipping Four Coins

Four heads: (only one way) HHHH

Three heads: (four ways)


Two heads: (six ways)



One head: (four ways)


Zero heads: (only one way) TTTT

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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 .

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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%.

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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!

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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.