13.3

Radiation

Energy from the sun is brought to earth by large amounts of visible light waves, as well as by substantial amounts of infrared and ultraviolet waves. These waves are known as electromagnetic waves, a class that also includes the microwaves used for cooking and the radio waves used for AM and FM broadcasts. The sunbather in Figure 13-14 feels hot because her body absorbs energy from the sun’s electromagnetic waves. And anyone who has stood by a roaring fire or put a hand near an incandescent light bulb has experienced a similar effect. Thus, fires and light bulbs also emit electromagnetic waves, and when the energy of such waves is absorbed, it can have the same effect as heat.

Figure 13-14 Suntans are produced by ultraviolet rays. (Ron Chapple/Thinkstock/Alamy Images)

The process of transferring energy via electromagnetic waves is called radiation, and, unlike convection or conduction, it does not require a material medium. Electromagnetic waves from the sun, for example, travel through the void of space during their journey to earth.

| Radiation
| Radiation is the process in which energy is transferred by means of electromagnetic waves.

All bodies continuously radiate energy in the form of electromagnetic waves. Even an ice cube radiates energy, although so little of it is in the form of visible light that an ice cube cannot be seen in the dark. Likewise, the human body emits insufficient visible light to be seen in the dark. However, as Figures 12-6 and 12-7 illustrate, the infrared waves radiating from the body can be detected in the dark by electronic cameras. Generally, an object does not emit much visible light until the temperature of the object exceeds about 1000 K. Then a characteristic red glow appears, like that of a heating coil on an electric stove. When its temperature reaches about 1700 K, an object begins to glow white-hot, like the tungsten filament in an incandescent light bulb.

The physics of summer clothing.

In the 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. The two blocks in sunlight in Figure 13-15, for example, are identical, except that one has a rough surface coated with lampblack (a fine black soot), while the other has a highly polished silver surface. As the thermometers indicate, the temperature of the black block rises at a much faster rate than that of the silvery block. This is because lampblack absorbs about 97% of the incident radiant energy, while the silvery surface absorbs only about 10%. The remaining part of the incident energy is reflected in each case. We see the lampblack as black in color because it reflects so little of the light falling on it, while the silvery surface looks like a mirror because it reflects so much 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 all the electromagnetic waves falling on it.

Figure 13-15 The temperature of the block coated with lampblack rises faster than the temperature of the block coated with silver because the black surface absorbs radiant energy from the sun at the greater rate.

The physics of a white sifaka lemur warming up.

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. The block coated with lampblack absorbs and emits the same amount of radiant energy, and the silvery block does too. In either case, if absorption were greater than emission, the block would experience a net gain in energy. As a result, the temperature of the block would rise and not be constant. Similarly, if emission were greater than absorption, the temperature would fall. Since absorption and emission are balanced, a material that is a good absorber, like lampblack, is also a good emitter, and a material that is a poor absorber, like polished silver, is also a poor emitter. A perfect blackbody, being a perfect absorber, is also a perfect emitter.
The fact that a black surface is both a good absorber and a good emitter is the reason people are uncomfortable wearing dark clothes during the summer. Dark clothes absorb a large fraction of the sun’s radiation and then reemit it in all directions. About one-half of the emitted radiation is directed inward toward the body and creates the sensation of warmth. Light-colored clothes, in contrast, are cooler to wear, since they absorb and reemit relatively little of the incident radiation.
The use of light colors for comfort also occurs in nature. Most lemurs, for instance, are nocturnal and have dark fur like the one shown in Figure 13-16a. Since they are active at night, the dark fur poses no disadvantage in absorbing excessive sunlight. Figure 13-16b shows a species of lemur called the white sifaka, which lives in semiarid regions where there is little shade. The white color of the fur may help in thermoregulation, by reflecting sunlight during the hot part of the day. However, during the cool mornings, reflection of sunlight would be a hindrance in warming up. It is interesting to note that these lemurs have black skin and only sparse fur on their bellies, and that to warm up in the morning, they turn their dark bellies toward the sun. The dark color enhances the absorption of sunlight.

Figure 13-16 (a) Most lemurs, like this one, are nocturnal and have dark fur. (Wolfgang Kaehler/Corbis-Bettmann) (b) The species of lemur called the white sifaka, however, is active during the day and has white fur. (Nigel Dennis/Wildlife Pictures/Peter Arnold Inc.)

The amount of radiant energy Q emitted by a perfect blackbody is proportional to the radiation time interval . The longer the time, the greater is the amount of energy radiated. Experiment shows that Q is also proportional to the surface area . An object with a large surface area radiates more energy than one with a small surface area, other things being equal. Finally, experiment reveals that Q is proportional to the fourth power of the Kelvin temperature , so the emitted energy increases markedly with increasing temperature. If, for example, the Kelvin temperature of an object doubles, the object emits 24 or 16 times more energy. Combining these factors into a single proportionality, we see that . This proportionality is converted into an equation by inserting a proportionality constant , known as the Stefan–Boltzmann constant. It has been found experimentally that :

The relationship above holds only for a perfect emitter. Most objects are not perfect emitters, however. Suppose that an object radiates only about 80% of the visible light energy that a perfect emitter would radiate, so . The factor such as the 0.80 in this equation is called the emissivity e and is a dimensionless number between zero and one. The emissivity is the ratio of the energy an object actually radiates to the energy the object would radiate if it were a perfect emitter. For visible light, the value of e for the human body, for instance, varies between about 0.65 and 0.80, the smaller values pertaining to lighter skin colors. For infrared radiation, e is nearly one for all skin colors. For a perfect blackbody emitter, e = 1. Including the factor e on the right side of the expression leads to the Stefan–Boltzmann law of radiation.

| The Stefan–Boltzmann Law of Radiation
| The radiant energy Q, emitted in a time t by an object that has a Kelvin temperature T, a surface area A, and an emissivity e, is given by

| (13.2)

where is the Stefan–Boltzmann constant and has a value of .

In Equation 13.2, the Stefan–Boltzmann constant is a universal constant in the sense that its value is the same for all bodies, regardless of the nature of their surfaces. The emissivity e, however, depends on the condition of the surface.
Example 6 shows how the Stefan–Boltzmann law can be used to determine the size of a star.

| Example 6 | | | A Supergiant Star

The supergiant star Betelgeuse has a surface temperature of about 2900 K (about one-half that of our sun) and emits a radiant power (in joules per second, or watts) of approximately (about 10 000 times as great as that of our sun). Assuming that Betelgeuse is a perfect emitter (emissivity e = 1) and spherical, find its radius.

Problem solving insight First solve an equation for the unknown in terms of the known variables. Then substitute numbers for the known variables, as this example shows.

Reasoning According to the Stefan–Boltzmann law, the power emitted is . A star with a relatively small temperature T can have a relatively large radiant power Q/t only if the area A is large. As we will see, Betelgeuse has a very large surface area, so its radius is enormous.
Solution Solving the Stefan–Boltzmann law for the area, we find

But the surface area of a sphere is , so . Therefore,we have

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

The next example explains how to apply the Stefan–Boltzmann law when an object, such as a wood stove, simultaneously emits and absorbs radiant energy.

| Example 7 | | | An Unused Wood-Burning Stove

An unused wood-burning stove has a constant temperature of 18 °C (291 K), which is also the temperature of the room in which the stove stands. The stove has an emissivity of 0.900 and a surface area of 3.50 m2. What is the net radiant power generated by the stove?

Problem solving insight In the Stefan–Boltzmann law of radiation, the temperature T must be expressed in kelvins, not in degrees Celsius or degrees Fahrenheit.

Reasoning Power is the change in energy per unit time (Equation 6.10b), or Q/t, which, according to the Stefan-Boltzmann law, is (Equation 13.2). But in this problem we need to find the net power produced by the stove. The net power is the power the stove emits minus the power the stove absorbs. The power the stove absorbs comes from the walls, ceiling, and floor of the room, all of which emit radiation.
Solution Remembering that temperature must be expressed in kelvins when using the Stefan–Boltzmann law, we find that

| (13.2)

The fact that the unheated stove emits 1280 W of power and yet maintains a constant temperature means that the stove also absorbs 1280 W of radiant power from its surroundings. Thus, the net power generated by the unheated stove is zero:

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 8 | | | A Heated Wood-Burning Stove

The wood-burning stove in Example 7 (emissivity = 0.900 and surface area = 3.50 m2) is being used to heat a room. The fire keeps the stove surface at a constant 198 °C (471 K) and the room at a constant 29 °C (302 K). Determine the net radiant power generated by the stove.

The physics of a wood-burning stove.

Reasoning Power is the change in energy per unit time, and according to the Stefan-Boltzmann law, it is (Equation 13.2). As in Example 7, however, we seek the net power, which is the power the stove emits minus the power the stove absorbs from its environment. Since the stove has a higher temperature than its environment, the stove emits more power than it absorbs. Thus, we will find that the net radiant power emitted by the stove is no longer zero, as it is in Example 7. In fact, it is the net power that the stove radiates that has warmed the room to its temperature of 29 °C.
Knowns and Unknowns The following data are available:

Description | Symbol | Value | Comment
Emissivity of stove | e | 0.900 |
Surface area of stove | A | 3.50 m2 |
Temperature of stove surface | T | 198 °C (471 K) | Temperature in Kelvins must be used.
Temperature of room | T0 | 29 °C (302 K) | Temperature in Kelvins must be used.
Unknown Variable |  |  |
Net power generated by stove | Pnet | ? |

Modeling the Problem

| Step 1 net power
| The net power Pnet generated by the stove is the power Pemitted that the stove emits minus the power Pabsorbed that the stove absorbs from its environment, as expressed in Equation 1 at the right. In Steps 2 and 3 we evaluate the emitted and absorbed powers.

| (1)

| Step 2 Emitted Power
| According to the Stefan-Boltzmann law (Equation 13.2), the heated stove (temperature T) emits a power that is

This expression can be substituted into Equation 1, as indicated at the right. In Step 3, we discuss the absorbed power.

| Step 3 Absorbed Power
| The radiant power that the stove absorbs from the room is identical to the power that the stove would emit at the constant room temperature of 29 °C (302 K). The reasoning here is exactly like that in Example 7. With T0 representing the temperature (in Kelvins) of the room, the Stefan-Boltzmann law indicates that

We can also substitute this result into Equation 1, as shown in the right column.

Solution The results of each step can be combined algebraically to show that
Thus, the net radiant power the stove produces from the fuel it burns is

Related Homework: Problems 21, 22, 29, 30

Example 8 illustrates that when an object has a higher temperature than its surroundings, the object emits a net radiant power . The net power is the power the object emits minus the power it absorbs. Applying the Stefan–Boltzmann law as in Example 8 leads to the following expression for Pnet when the temperature of the object is T and the temperature of the environment is T0:

| (13.3)

Need more practice?

Interactive LearningWare13.2
One day, the temperature of a radiator has to be 68 °C to keep the surrounding walls of a room at 24 °C. The next day, however, the temperature of the radiator needs to be only 49 °C to keep the walls at 24 °C because it is warmer outside. Assuming that the room is heated only by radiation, determine the ratio of the net power radiated by the unit on the colder day to that radiated on the warmer day.
Related Homework: Problem 30

Check Your Understanding 2

Two identical cubes have the same temperature. One of them, however, is cut in two and the pieces are separated (see the drawing). What is true about the radiant energy emitted in a given time? (The answer is given at the end of the book.)

a. | The cube cut into two pieces emits twice as much radiant energy as does the uncut cube; .
b. | The cube cut into two pieces emits more radiant energy than does the uncut cube according to .
c. | The cube cut into two pieces emits the same amount of radiant energy as does the uncut cube; .
d. | The cube cut into two pieces emits one-half the radiant energy emitted by the uncut cube; .
e. | The cube cut into two pieces emits less radiant energy than does the uncut cube according to .

Background: The radiant energy emitted in a certain time by an object depends on its surface area, emissivity, and temperature. Only one of these variables changes when the object is cut into two pieces.
For similar questions (including calculational counterparts), consult Self-Assessment Test 13.1. This test is described next.

Test your understanding of the central ideas presented in Sections 13.1, 13.2 and cutnell3158c13-sec1-0003:

· | Convection
· | Conduction
· | Radiation

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