19.8.3 Radiation

The sun radiates energy to earth through the vacuum of space. Similarly, you feel the warmth from the glowing red coals in a fireplace.

All objects emit energy in the form of radiation, electromagnetic waves generated by oscillating electric charges in the atoms that form the object. These waves transfer energy from the object that emits the radiation to the object that absorbs it. Electromagnetic waves carry energy from the sun; this energy is absorbed when sunlight falls on your skin, warming you by increasing your thermal energy. Your skin also emits electromagnetic radiation, helping to keep your body cool by decreasing your thermal energy. Radiation is a significant part of the energy balance that keeps your body at the proper temperature.

A satellite image of radiation emitted by ocean waters. — This satellite image shows radiation emitted by the ocean waters off the east coast of the United States. You can clearly see the warm waters of the Gulf Stream, a large-scale convection that transfers heat to northern latitudes.

NOTE The word “radiation” comes from “radiate,” meaning “to beam.” Radiation can refer to x rays or to the radioactive decay of nuclei, but it also can refer simply to light and other forms of electromagnetic waves that “beam” from an object. Here we are using this second meaning of the term.

You are familiar with radiation from objects hot enough to glow “red hot” or, at a high enough temperature, “white hot.” The sun is simply a very hot ball of glowing gas, and the white light from an incandescent lightbulb is radiation emitted by a thin wire filament heated to a very high temperature by an electric current. Objects at lower temperatures also radiate, but at infrared wavelengths. You can’t see this radiation (although you can sometimes feel it), but infrared-sensitive detectors can measure it and are used to make thermal images.

The energy radiated by an object depends strongly on temperature. If a small amount of heat energy dQ is radiated during a small time interval dt by an object with surface area $A$ and absolute temperature T, the rate of heat transfer is found to be

\frac{d Q}{d t} = e σ A T^{4}

(19.45)

Because the rate of energy transfer is power $(1 J/s = 1 W), d Q / d t$ is often called the radiated power. Notice the very strong fourth-power dependence on temperature. Doubling the absolute temperature of an object increases the radiated power by a factor of 16!

The parameter $e$ in Equation 19.45 is the emissivity of the surface, a measure of how effectively it radiates. The value of $e$ ranges from 0 to 1. $σ$ is a constant, known as the Stefan-Boltzmann constant, with the value

σ = 5.67 \times 10^{- 8} {W/m}^{2} K^{4}

NOTE Just as in the ideal-gas law, the temperature in Equation 19.45 must be in kelvins.

Objects not only emit radiation, they also absorb radiation emitted by their surroundings. Suppose an object at temperature $T$ is surrounded by an environment at temperature $T_{0}$ . The net rate at which the object radiates heat energy—that is, radiation emitted minus radiation absorbed—is

\frac{d Q_{net}}{d t} = e σ A (T^{4} - T_{0}^{4})

(19.46)

This makes sense. An object should have no net radiation if it’s in thermal equilibrium $(T = T_{0})$ with its surroundings.

Notice that the emissivity $e$ appears for absorption as well as emission; good emitters are also good absorbers. A perfect absorber $(e = 1)$ , one absorbing all light and radiation impinging on it but reflecting none, would appear completely black. Thus a perfect absorber is sometimes called a black body. But a perfect absorber would also be a perfect emitter, so thermal radiation from an ideal emitter is called black-body radiation. It seems strange that black objects are perfect emitters, but think of black charcoal glowing bright red in a fire. At room temperature, it “glows” equally bright with infrared.

EXAMPLE 19.11 Taking the sun’s temperature

The radius of the sun is $6.96 \times 10^{8} m$ . At the distance of the earth, $1.50 \times 10^{11} m,$ the intensity of solar radiation (measured by satellites above the atmosphere) is $1370 {W/m}^{2}$ . What is the temperature of the sun’s surface?

MODEL Assume the sun to be an ideal radiator with $e = 1$ .

SOLVE The total power radiated by the sun is the power per $m^{2}$ multiplied by the surface area of a sphere extending to the earth:

P = \frac{1370 W}{1 m^{2}} \times 4 π (1.50 \times 10^{11} {m)}^{2} = 3.87 \times 10^{26} W

That is, the sun radiates energy at the rate $d Q / d t = 3.87 \times 10^{26} J / s$ . That’s a lot of power! This energy is radiated from the surface of a sphere of radius $R_{S .}$ Using this information in Equation 19.45, we find that the sun’s surface temperature is

\begin{array}{l} T & = {[\frac{d Q / d t}{e σ (4 π R_{S}^{2})}]}^{1 / 4} \\ = {[\frac{3.87 \times 10^{26} W}{(1) (5.67 \times 10^{- 8} {W/m}^{2} K^{4}) 4 π {(6.96 \times 10^{8} m)}^{2}}]}^{1 / 4} \\ = 5790 K \end{array}

REVIEW This temperature is confirmed by measurements of the solar spectrum, a topic we’ll explore in Part VIII.

Thermal radiation plays a prominent role in climate and global warming. The earth as a whole is in thermal equilibrium. Consequently, it must radiate back into space exactly as much energy as it receives from the sun. The incoming radiation from the hot sun is mostly visible light. The earth’s atmosphere is transparent to visible light, so this radiation reaches the surface and is absorbed. The cooler earth radiates infrared radiation, but the atmosphere is not completely transparent to infrared. Some components of the atmosphere, notably water vapor and carbon dioxide, are strong absorbers of infrared radiation. They hinder the emission of radiation and, rather like a blanket, keep the earth’s surface warmer than it would be without these gases in the atmosphere.

The greenhouse effect, as it’s called, is a natural part of the earth’s climate. The earth would be much colder and mostly frozen were it not for naturally occurring carbon dioxide in the atmosphere. But carbon dioxide also results from the burning of fossil fuels, and human activities since the beginning of the industrial revolution have increased the atmospheric concentration of carbon dioxide by over 50%. This human contribution has amplified the greenhouse effect and is the primary cause of global warming.

CHAPTER 19 CHALLENGE EXAMPLE Boiling water

$400 mL$ of water is poured into a covered $8.0 -cm-diameter$ , 150 g glass beaker with a $2.0 -mm-thick$ bottom; then the beaker is placed on a $400 ° C$ hot plate. Once the water reaches the boiling point, how long will it take to boil away all the water?

MODEL The bottom of the beaker is a heat-conducting material transferring heat energy from the $400 ° C$ hot plate to the $100 ° C$ boiling water. The temperature of both the water and the beaker remains constant until the water has boiled away. We’ll assume that heat losses due to convection and radiation are negligible, in which case the heat energy entering the system is used entirely for the phase change of the water. The beaker’s mass isn’t relevant because its temperature isn’t changing.

SOLVE The heat energy required to boil mass $M$ of water is

Q = M L_{v}

where $L_{v} = 2.26 \times 10^{6} J / kg$ is the heat of vaporization. The heat energy transferred through the bottom of the beaker during a time interval $Δ t$ is

Q = k A \frac{Δ T}{L} Δ t

where $k = 0.80 W / m K$ is the thermal conductivity of glass. Because the heat transferred by conduction is used entirely for boiling the water, we can combine these two expressions:

k A \frac{Δ T}{L} Δ t = M L_{v}

and then solve for $Δ t :$

\begin{matrix} Δ t & = & \frac{M L L_{v}}{k A Δ T} = \frac{(0.40 kg) (0.0020 m) (2.26 \times 10^{6} J / kg)}{(0.80 W / m K) (0.0050 m^{2}) (300 K)} \\ = & 1500 s = 25 min \end{matrix}

We used the density of water to find that $M = 400 g = 0.40 kg$ and calculated $A = π r^{2} = 0.0050 m^{2}$ as the area through which heat conduction occurs.

REVIEW $400 mL$ is roughly $2 cups,$ a small hot plate can bring $2 cups$ of water to a boil in 5 min or so, and boiling the water away takes quite a bit longer than bringing it to a boil. $25 min$ is a slight underestimate since we neglected energy losses due to convection and radiation, but it seems reasonable. A stove could boil the water away much faster because the burner temperature (gas flame or red-hot heating coil) is much higher.