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Figure a shows a graph of EM radiation intensity versus lambda in nm. There are three separate curves on the graph. These are labeled 6000 K or white hot, 4000 K and 3000 K or red hot. All rise, peak and taper down, with the first having the highest intensity and the last one, the lowest. The first curve peaks near the range of visible light, the second and third ones, just after. Figure b shows a flame. It is blue near the bottom, whitish in the middle and orangish at the top.
(a) A graph of the spectrum of electromagnetic waves emitted from an ideal radiator at three different temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the spectrum shifts down in wavelength toward the visible and ultraviolet parts of the spectrum. The shaded portion denotes the visible part of the spectrum. It is apparent that the shift toward the ultraviolet with temperature makes the visible appearance shift from red to white to blue as temperature increases. (b) Note the variations in color corresponding to variations in flame temperature.

The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white is the least effective. On a clear summer day, black asphalt in a parking lot is hotter than adjacent gray sidewalk, because black absorbs better than gray ( [link] ). The reverse is also true—black radiates better than gray. Thus, on a clear summer night, the asphalt is colder than the gray sidewalk, because black radiates the energy more rapidly than gray. A perfectly black object would be an ideal radiator and an ideal absorber , as it would capture all the radiation that falls on it. In contrast, a perfectly white object or a perfect mirror would reflect all radiation, and a perfectly transparent object would transmit it all ( [link] ). Such objects would not emit any radiation. Mathematically, the color is represented by the emissivity     e . A “blackbody” radiator would have an e = 1 , whereas a perfect reflector or transmitter would have e = 0 . For real examples, tungsten light bulb filaments have an e of about 0.5, and carbon black (a material used in printer toner) has an emissivity of about 0.95.

Figure on the left shows ice melting on a light coloured pavement. Figure on the right shows ice melting on a darker pavement. Here the ice has melted more.
The darker pavement is hotter than the lighter pavement (much more of the ice on the right has melted), although both have been in the sunlight for the same time. The thermal conductivities of the pavements are the same.
Figure shows four boxes. The first two are black and the other two are silver coated. The first box is labeled absorb. Incident radiant energy is absorbed and a small part of it is reflected. The second box is labeled radiate. Most energy is emitted and a small part of it is retained. The third box is labeled absorb. Most incident radiant anergy is reflected. A small part is absorbed. The last box is labeled radiate. Most incident energy is retained. A small part of it is emitted.
A black object is a good absorber and a good radiator, whereas a white, clear, or silver object is a poor absorber and a poor radiator.

To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal equilibrium. We know from experience that they will stay in equilibrium (the result of a principle that will be discussed at length in Second Law of Thermodynamics ). For the black object’s temperature to stay constant, it must emit as much radiation as it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the silver object must radiate as little as it absorbs. Thus, one property, emissivity, controls both radiation and absorption.

Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you knock apart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area.

The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation    :

P = σ A e T 4 ,

where σ = 5.67 × 10 −8 J/s · m 2 · K 4 is the Stefan-Boltzmann constant, a combination of fundamental constants of nature; A is the surface area of the object; and T is its temperature in kelvins.

Practice Key Terms 9

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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