I How to calculate the energy emitted through IR waves

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The energy emitted by a body in watts/m2 is calculated using the formula εσT4, where ε represents emissivity, which varies based on material and surface characteristics. For perfect black bodies, ε equals 1, but for bodies emitting only infrared (IR) light, ε can differ and is not always defined, especially for non-thermal light sources. Emissivity is a ratio comparing the spectral radiance of a body to that of a black body at the same temperature, ranging from 0 to 1. Factors such as surface roughness and material type significantly influence emissivity values, which are often determined empirically. Understanding these principles is essential for calculating thermal radiation, particularly in contexts like Earth's heat emission when only IR wavelengths are considered.
amukher
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The energy emitted by a body in watts/m2 is = εσT4. In the case of a perfect black body, ε=1. If the body only emits IR light, what should be the value of ε?
 
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amukher said:
If the body only emits IR light, what should be the value of ε?
If the "body" is a light source not due to thermal radiation, then there is no ε. A CO2 laser will not emit the same power as a remote control, even if they are at the same temperature.
 
amukher said:
The energy emitted by a body in watts/m2 is = εσT4. In the case of a perfect black body, ε=1. If the body only emits IR light, what should be the value of ε?
The emissivity is material dependent and surface roughness dependent. For example polished metal and foil of the same metal will have different emissivities, as shown here.
 
amukher said:
If the body only emits IR light, what should be the value of ε?

such bodies exist which only emits a particular radiation - then its perhaps not a black body ; call it a real body or grey body ;
The emissivity of the surface of a material is its effectiveness in emitting energy as thermal radiation
then one can define emissivity with respect to black body as a ratio and it can be from 0 to 1.
it can be taken as ratio of the following
Emissivity = L(1) of the body/ L(2) for a black body
where
L(1)
the spectral radiance in frequency of that surface;
L(2)
is the spectral radiance in frequency of a black body at the same temperature as that surface;
For details see<https://en.wikipedia.org/wiki/Emissivity#Spectral_hemispherical_emissivity>
 
DrClaude said:
If the "body" is a light source not due to thermal radiation, then there is no ε. A CO2 laser will not emit the same power as a remote control, even if they are at the same temperature.

Let us say that the atmosphere blocks all visible light wavelengths and allows only IR wavelengths to reach the earth. The Earth would then be a source of thermal radiation. To calculate the heat emitted by the earth, I would need the value of ε.
 
amukher said:
Let us say that the atmosphere blocks all visible light wavelengths and allows only IR wavelengths to reach the earth. The Earth would then be a source of thermal radiation. To calculate the heat emitted by the earth, I would need the value of ε.
The emissivity depends on the body shape (e.g how smooth it is) and the material. There's no easy way to predict them, they are basically empirical values
 
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