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| color and emissivity |
dex |
11/2/2005 |
does anyone explain on how emissivity affect the colour.what color is hot "red or white
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| Re:color and emissivity |
Raphael Danjoux |
11/3/2005 |
Color (visible ) and emissivity (infrared) are not directly correlated.
In the visible range, color is the result of reflexion of light.
Emissivity represents the ability of a material to emit heat, compared to what a theoretical body called a blackbody does. For thermographic equipment it is an integrated value over a reduced spectral range.
The confusion often comes from the fact that a dark component (a car for example) absorbing more heat from the sun, than a light one, its temperature may be higher after a while. But this is not emissivity ! |
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| Re:color and emissivity |
JKostrzewa |
11/7/2005 |
A slight clarification: In the visible range, color is USUALLY the result of reflection of light from another source. However, emission in the visible range is certainly possible when the source is very hot. Color is then a function of temperature. The hotter the emitting source, the shorter the peak wavelength of emission:
peak wavelength = 2898K / T |
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| Re:color and emissivity |
Laland |
11/7/2005 |
| Does that mean that color has little or no effect on the temperature? |
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| Re:color and emissivity |
brian |
11/8/2005 |
Yes. If we observe two different colored balls in a dark room they will appear to have the same temperature. But if we illuminate them, the darker colored ball will absorb more heat and will appear hotter (as in the example above with the car in sunshine).
If, on the other hand, we observe a red hot and a white hot iron in a dark room, the white hot iron will appear hotter - as it is hotter. |
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| Re:color and emissivity |
JKostrzewa |
11/9/2005 |
| When a photon is incident on a surface, it either transmits through, reflects off, or is absorbed. When photons are absorbed, the temperature of the surface increases, and the energy is then re-emitted (though not necessarily at the same wavelength as the original irradiance). An object that appears black in white light looks that way because it reflects very little of the white light. Instead, it absorbs most of the energy from the light, which causes the object to heat up and re-radiate (typically in the infrared portion of the spectrum unless the object becomes hot enough to radiate in the visible portion of the spectum). An object that appears red in white light looks that way because it is reflecting the red portion of the white light (and asborbing the rest). Therefore, it would not heat as much as an identical black object exposed to the identical white-light source since it is absorbing less radiation. So the color of an opaque bject in white light does provide information about its emissivity in the visible portion of the spectrum. However, it does not provide any information about its emissivity in other portions of the spectrum. Snow, for example, clearly has low emissivity in the visible portion of the spectrum (i.e. it reflects white light), but it is highly emissive in the LWIR portion. |
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