Dominance of Infra Red and Heat
Name: Luke N.
I understand that all heated objects emit radiation. The
intensity depends on the emissivity value of the material. The frequency
range of the emitted electromagnetic waves is a function of the temperature.
So why, when we talk about heat radiation, are infrared frequencies
focused on as opposed to any other range of electromagnetic waves?
I have read several articles which seem to imply that we only obtain heat
energy from the Sun's infrared spectrum and not so from the other
frequencies (like those of visible light).
Infrared frequencies are not different from other frequencies in their
ability to carry heat energy. However, near room temperature, the
black body spectrum peaks in the infrared, and most real objects we
encounter behave pretty much like these idealized radiators, so we've
come to associate infrared with heat.
Well, Luke, if that is what they say, they are wrong. The energy intensity of
the solar spectrum is at a maximum in the visible, at the wavelength of
yellow-green light. (This is why the Sun is considered a yellow star.) In
fact, the sensitivity of the human eye to different frequencies of light
roughly parallels the intensity of those frequencies in the solar spectrum.
(No coincidence: we evolved seeing sunlight.)
We get heat from other wavelengths in the solar spectrum because many
substances on earth (leaves, rocks, dirt, animals, etc.) absorb light,
converting the energy to heat. It is not necessary for the infrared
component even to be present. It is not a very large portion of the energy
Richard E. Barrans Jr., Ph.D.
PG Research Foundation, Darien, Illinois
You are correct, objects at any temperature emit electromagnetic radiation
of all frequencies
(wavelengths) that depends upon the temperature (black body radiation) and
the relative amounts are given by Planck's distribution law.
The reason that the infrared portion of the electromagnetic spectrum is
stressed when we talk about heat energy is that is where the maximum in the
Planck distribution occurs for what we nominally call "hot objects".
Of course some hot objects emit radiation in the visible range -- they glow.
Radiation from the sun or other sources is not necessarily a continuum. In
addition, light of various wavelengths can be converted to other wavelengths
by a variety of processes. Some examples are:
microwave (oven) radiation converted to infrared radiation (hot food).
visible radiation converted to infrared radiation (heat) when it is absorbed
by the ground, or even more so by a black asphalt driveway.
ultraviolet radiation converted to visible radiation by fluoresence ("black
light" U.V. lamps)
incandescent lamps that produce both visible light and heat.
fluorescent lights that produce a higher amount of visible light compared to
infrared radiation because of the phosphors.
It is just not correct that the sun provides heat only from its infrared
portion of the solar spectrum.
All of these processes above and others produce "heat radiation" from solar
As you know, heat is just energy.
The sensation of heat from a radiative source requires that the energy be
in the radiation AND that the radiation is absorbed. Almost all infrared
wavelengths are absorbed in the skin and thus creates the sensation of
Visible light, on the other hand, is mostly reflected. Thus, while there
energy available in the visible light portion of the solar spectrum it is not
efficiently converted to heat -- unless a good absorber is present (such
Solar heaters are black to improve the efficiency of absorption of
both the visible and infrared wavelengths.
The answer to your question is already in your question; the intensity
of the emitted radiation is a function of the wavelength and that
function depends on the temperature of the emitter. It is a smooth
curve which extends from very short wavelengths to infinitely long
wavelengths, but most of the energy transmitted is in the region of
To become a little quantitative, the maximum is given by the Wien
LT = 3E6 nm K
Here L is the wavelength in nanometers (1 nanometer = 1 nm = 1E-9 m)
and T is the temperature of the emitter in degrees Kelvin. If you're
not happy with exponential notation, 3E6 = 3,000,000 and
1E-9 = 0.000000001.
To get some feeling for these numbers, 300 K is approximately room
temperature (around 80 F) and the temperature of the sun is around
6,000 K. The sun is the brightest (emits the highest intensity)
around a wavelength of 500 nm (plug in the above equation to get
this), right in the middle of the visible range of wavelengths.
(Obviously our eyes developed to take advantage of the properties of
our sun -- luckily the molecules we are made of could take advantage
of these properties).
Here are some wavelengths to keep in mind: (These numbers denote
regions and are not exact).
< 400 470 570 670 700
Ultraviolet Blue Yellow Red Infrared
Now, if you take the temperature of the earth to be 300K, the above
equation will tell you that the maximum intensity of the radiation
emitted is be at a wavelength of 10,000 nm, in the far infrared.
And that is why the sun's energy mostly gets through to the earth,
since the earth's atmosphere is quite transparent in the visible
region, but the earth's radiation is increasingly trapped because CO2
and other "greenhouse" gases are transparent to visible light, but
tend to reflect light of infrared wavelengths.
Best, Dick Plano...
Hi, Luke !!!
As you know, all the radiant energy is in the form
of electromagnetic waves, and they have a broad range of frequencies, coming from
radio waves, microwaves, infrared, visible, ultraviolet, X-rays and
gamma-rays. When you say : "I understand that all heated objects emit radiation."
I would point out that not only heated objects emit radiation, but
lets say, every object ( with the temperature greater than 0o Kelvin) emits
radiation. All objects emit a mixture of radiant wavelenghts of energy.
If the temperature of the body becomes hotter, some of the radiant energy
is in the visible range of wavelenghts like - for instance - the filament of a light
bulb. The Sun emit energy across the whole range of wavelenghts.
You mention that you have read several articles which seem to imply that
we only obtain heat energy from the Suns INFRARED SPECTRUM and not
from other frequencies. Well, for sure that thermal radiation is defined as
electromagnetic radiation in the wavelenght range of 0.1 to 100 micra, which
contains the visible light ( ca.from 0,7 till 0,4 micron ). Our atmosphere filters
out many harmful types of electromagnetic radiation.
I belive that your question focus attention on how much energy is delivered
by the Sun when you take into account the INFRARED SPECTRUM. Certainly
depending upon the temperature of a body, you have a different wavelenght
allocation. Just to recall, the energy of a vibrating molecule is proportional to
its frequency of vibration. And they came - as Planck said - in discrete lumps.
The Plancks radiation law describes the wavelenght distribution of electromag-
netic emission of a blackbody
Well, it is necessary to count all the frequencies in to correctly evaluate
the delivered energy of the Sun.
At common temperatures: 0 to 100 degrees Celsius, most radiation emitted is
in the infrared range. At much higher temperatures, objects begin to have a
red glow: visible light is emitted. As temperature increases, the color
passes through the visible spectrum: orange, yellow, white(full spectrum),
As for the sun, the earth receives the full spectrum. Most of the radiation
above visible light is reflected back by the ozone layer of our atmosphere.
The ultraviolet light that does get through causes sunburn. Radio waves
don't have enough energy to be noticed: some pass right through the earth.
The two parts of the spectrum that have greatest effect are infrared and
visible. Since visible light is such a narrow range of frequencies,
infrared does provide a great deal of the heat we receive.
Dr. Ken Mellendorf
Illinois Central College
I believe that near the normal body temperature (say
from 60 deg F to 120 deg F), objects tend to give up
waves in the infrared frequency range.
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Update: June 2012