1206 - Concepts in Physics

November 09th 2009

A special day for Germany - 20 years after ...

NotesRemember Assignment #5 is due on Wednesday

(no later than NOON)

The midterms can be picked up at my office: Today: 11:30 am to 1:00 pm or 4:00 pm to 5:00pm

Tomorrow: 12:30 pm - 2:30 pmWednesday: 9:00 am to 10:15 am or after 11:30am(We do it this way, because I would like to say a sentence of two to many of you and the

exams are also personal, so you can decide who will see it)

Also problems have been de-rated, so that only 45 points (not 60) are required

ConvectionConvection is the process in which heat is carried from place to place by the bulk

movement of a fluid. The fluid flow itself is called convection current.

When heat is transferred to or from a substance, the internal energy of the substance can change. This change in internal energy is accompanied by a change in temperature (or

change in phase)

During a volcanic eruption, smoke at the top rises thousands of meters because of

convection.

Transfer of heat is very familiar to us - home furnaces distribute heat on cold days and air conditioners remove it on

hot days. Our bodies constantly transfer heat in one direction or another.

Virtually all our energy originates in the sun and is transferred to us over a

distance of 150 million km in form of heat. Convection is one form of heat

transfer.

Convection current - warm air rises, cold air sinks

Refrigerator

Hot water baseboard heating units are frequently used in homes, where they are mounted on the wall next to the floor. In contrast, the cooling coil in a refrigerator is mounted near the top. The locations for these heating and cooling devices are different, yet each location

is designed to maximize the production of convection currents. Explain how.

Example:

An important goal for a heating system is to distribute heat throughout the room. In case of the refrigerator, the goal is to remove the heat from all of the space within. Convection will help to achieve these goals. The ari above the baseboard unit is heated. Buoyant forces from the surrounding cooler air push the warm air upward. Cooler air near the ceiling is

displaced downward and then warmed by the baseboard heating unit. Within the refrigerator, the air in contact with the top-mounted coil is cooled, its volume decreases,

and its density increases. The surrounding warmer and less dense air cannot provide sufficient buoyant force to support the colder air, so it sinks downward. In the process

warmer air near the bottom is displaced upward and is then cooled by the coil.

YOUR TURN: What would happen in the refrigerator, if the coil would be placed at the bottom?

YOUR TURN: What would happen in the refrigerator, if the coil would be placed at the bottom?

The cooling coil at the bottom would cool the air down there, which would stay there, since it can’t sink any further. The air is stagnant as a result with very little convection to carry the heat from other parts of the refrigerator to the coil to be removed. The top

part would not get cooled.

ConductionAnyone who has ever used all-metal skillets (or similar things) for cooking knows, that the metal handle becomes hot. Somehow heat is transferred from the burner to the handles. Clearly heat is not transferred by bulk movement of the metal or the surrounding air, so convection can be ruled out. Instead, heat is transferred

directly through the metal and this process is called conduction.

CONDUCTIONConduction is the process whereby heat is transferred directly through a material,

with any bulk motion of the material playing no role in the transfer.

One mechanism for conduction occurs when the atoms or molecules in a hotter part of the material vibrate or move with greater energy than those in a cooler part. By means of collision, the more energetic molecules pass on some of their

energy to their less energetic neighbors. (Imagine gas)In metals a similar mechanism for conduction of heat occurs. Metals are different

from most substances in having a pool of electrons that are more or less free, which means they can move easily throughout the metal. These free electrons can transport energy and allow metals to transfer heat very well. These free electrons

are also responsible for the excellent electrical conductivity that metals have.

Materials that conduct heat well are called thermal conductors, and those that conduct heat poorly are known as thermal insulators. Most metals are excellent

thermal conductors, while wood, glass and most plastics are common thermal insulators.

Simple picture of heat conduction. Heat flows from the warmer to the cooler end.

Assume the ends are isolated (so that heat loss through them is negligible. The “high temperature side” is in contact with a source of heat, the “low temperature side” is in

contact with cooling device. The amount of heat Q conducted through the bas from the warmer end to the cooler end depends on a number of factors:

Q is proportional to the time t during which conduction takes place. Longer time periods means more heat flows. Q ∝ t

Q is proportional to the temperature difference ΔT between the ends of the bar. A larger difference causes more heat to flow. Q ∝ ΔT (no heat flows when both ends

have the same temperature) Q is proportional to the cross-sectional area A of the bar (Q ∝ A)

Q is inversely proportional to the length L of the bar (Q ∝ 1/L). Greater length of material conduct less heat.

CONDUCTION OF HEAT THROUGH A MATERIAL The heat Q conducted during a time t through a bar of length L and cross-sectional area A is Q = k (A ΔT)t * 1/L, where ΔT is the temperature difference between the ends and k is

the thermal expansion conductivity of the material (SI unit: J/(s m C°))

Different materials have different thermal conductivities, that have been determined experimentally. Typical orders of magnitude for metals are 10-500, while other material are

about 0.01 - 0.1 J/(smC)

Example: regulating body temperatureWhen excessive heat is produced within the body, it must be transferred to the skin and dispersed in order to maintain the temperature within (37 °C). One possible mechanism for transfer is conduction through body fat. Suppose that heat travels through 0.039 m of

fat in reaching the skin, which has a total surface area of 1.7 m2 and a temperature of 34.0 °C. Find the amount of heat that reaches the skin in half an hour (1800 s). The

thermal conductivity of body fat is k = 0.20 J/(s m C)

Q = k (A ΔT)t * 1/L

Q = {0.20 J/(smC)}{1.7 m2}{37 C - 34 C}{1800 s}{1/0.030 m} = 6.1 x 104 J

For comparison, a jogger can generate over ten times this amount of heat in half an hour. Conduction through body fat is not a particularly effective way of removing

excess heat. Heat transfer via blood flow to the skin is more effective - that’s why we turn red when we get hot - it also has the advantage that the body can vary the blood

flow as needed.

Example: layered InsulationOne wall of a house consists of 0.019-m-thick plywood backed by 0.076-m-thick

insulation. The temperature at the inside surface is 25.0 C, while the temperature at the outside surface is 4.0 C, both being constant. The thermal conductivities of the insulation and the plywood are, respectively, 0.030 and 0.080 J/(smC). The area of the wall is 35 m2.

Find the heat conducted through the wall in one hour (a) with the insulation and (b) without the insulation.

The temperature T at the insulation-plywood interface must be determined before the heat conducted through the wall can be obtained. In calculating this temperature, we use the fact

that no heat is accumulating in the wall because the inner and outer temperature are constant. Therefore, the heat conducted through the insulation must equal the heat

conducted through the plywood during the same time. Qins = Qply. Each of the Q values can be expressed as Q = k (A ΔT)t * 1/L leading to an expression for the interface temperature.

Once T is available, we can use it to obtain the heat conducted through the wall.

[k (A ΔT)t * 1/L]ins = [k (A ΔT)t * 1/L]ply A and t are the same, so eliminate algebraically{0.030 J/(smC)}{25 C -T}{1/0.076 m} = {0.080 J/(msC)}{T - 4 C}{1/0.019 m} --> T = 5.8 C

Now use original equation to calculate: Q = 9.5 x 105 J

YOUR TURN to do (b)

YOUR TURN to do (b)

Can use Q = k (A ΔT)t * 1/L

with ΔT = 25 C - 4 C = 21 C

Result 110 x 105 J (plywood only)

This is a factor of 12 between the two.

Some fruit growers sometimes protect their crops by spraying them with water when overnight temperature are expected to drop below freezing. Some fruit crops, like the

blueberry plants can withstand temperatures down to freezing, but below that the plants suffer significant damage. When water is sprayed, it can freeze and and act as

insulating layer. The thermal conductivity to ice is fairly small.

RadiationEnergy from the sun is brought to earth by large amounts of visible light waves, as well as by substantila amounts of infrared and ultraviolet waves. These belong to a class of

waves known as electromagnetic waves (they also include microwaves and radio waves). We will get back to these in more detail later in the term.

The process of transferring energy via electromagnetic waves is called radiation and, unlike convection of conduction, it does not require a material as medium.

Electromagnetic waves from the sun, for example, travel through the void of space.

RADIATIONRadiation is the process in which energy is transferred by means of electromagnetic

waves.

Note! Studying radiation was another step towards modern physics.

All bodies continuously radiate energy in form of electromagnetic waves. Even an ice cube. The human body emits insufficient visible light to be seen in the dark, but

enough infrared radiation to be picked up by a camera sensitive to these wave length.

Transfer of energy by radiation, the absorption of electromagnetic waves is just as important as their emission. The surface of an object plays a significant role in

determining how much radiant energy the object will absorb or emit. For example let’s assume we have two blocks (one coated in silver, the other painted black) sitting in the garden where the sun shines onto them. The temperature of the black block

rises at a much faster rate then that of the silvery block.

The silvery block acts like a mirror and reflects most of the light, while the black one absorbs most of the light. Since the color black is associated with nearly complete absorption of visible light, the term perfect blackbody or simply

blackbody is used when referring to an object that absorbs light.

All objects emit and absorb electromagnetic waves simultaneously. When a body has the same constant temperature as its surroundings, the amount of radiant energy being absorbed must balance the amount being emitted in a given interval of time.

Therefore both blocks (the silvery one and the black one) emit and absorb the same amount of rdiant energy. If the emission and absorption would be different the block would experience a net gain in energy and as a result the temperature of the

block would rise.

A material that is a good absorber is also a good emitter and a material that is a poor absorber is also a poor emitter.

This is the reason why wearing black cloth in the sun will make you hot faster than wearing light colors

The amount of radiant energy Q emitted by a perfect blackbody is proportional to the radiation time interval t as well as the area A. Q ~ tA

Experimentally, we can show that Q is proportional to the fourth power of the Kelvin temperature Q ~ T4

We can combine these factors in a single proportionality, where the proportionality constant σ is called the Stefan-Boltzmann constant and has been found experimentally

to be σ = 5.67 x 10-8 J/(s m2 K4)

THE STEFAN-BOLTZMANN LAW OF RADIATION:

The radiant energy Q, emitted in a time ti by an object that has a Kelvin temperature T, a surface A and an emissivity of e (emission of visible light - unit less), is given by

Q = eσT4At where σ is the Stefan-Boltzmann constant with a value of σ = 5.67 x 10-8 J/(s m2 K4)

Example: A supergiant starThe supergiant star betelgeuse has a surface temperature of about 2900 K (half of what

our sun has) and emits a radiant power (in J/s, or W) of approximately 4x1030 W (about 10 times greater than our sun). Assuming that Betelgeuse is a perfect emitter (e = 1) and

spherical, find its radius.

According to the Stefan-Boltzmann law, the power emitted is Q/t = eσT4A. A star with a relatively small temperature T can have a relatively large radian power Q/t only if the area

A is large. We will see, that this is the case for Betelgeuse.

Re-arranging for A = (Q/t) / eσT4

surface area of a sphere is A = 4πr2, so r = sqrt{A/4π}

r = sqrt{(Q/t)/(4πeσT4)} = 3x1011 m

For comparison, Mars orbits the sun at a distance of 2.28 x 1011 m. Betelgeuse is certainly a “supergiant”

A few remarks ...

• Still a number of people using pencil (for other things than drawings) - this will result in no points for the final exam

• Making up your own numbering scheme (not using the numbers given) will also result in no points for the final exam

• Please use space, it is not problem to get a second and third booklet - it is very hard to mark two different parts in one line

more remarks• Many of you lost points, because numbers

showed up out of nowhere without explanation (words) or the calculation shown. (total distance in ship problem, total mass in elevator problem) and in many cases these numbers were wrong !!!!

• Show all calculation (no matter how simple) or at least write the words.

more remarks• Yes, there can be questions in exams, that

have not been shown in the lecture - this is called transfer and you will have to show that you can use the learned concepts on something “new”

• Scalars and Vectors are different things and cannot be equal !!! (force not equal to energy)

more remarks

• Many of you have to look at basic geometry and trigonometry (this is a requirement for this class)

• sin or cos or tan of a unit doesn’t make any sense ...

• if you have an expressing cos(angle) and want the angle you have to use the arccos function (or cos-1)