How to calculate the energy emitted through IR waves

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Discussion Overview

The discussion revolves around the calculation of energy emitted through infrared (IR) waves, focusing on the concept of emissivity (ε) in relation to different types of bodies, including perfect black bodies and real or grey bodies. Participants explore the implications of emissivity in thermal radiation and its dependence on material properties.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants state that the energy emitted by a body is given by the formula εσT4, where ε is the emissivity and σ is the Stefan-Boltzmann constant.
  • There is a question regarding the value of ε for bodies that only emit IR light, with some suggesting that if the body is not a black body, it may be classified as a real or grey body.
  • One participant notes that emissivity is material and surface roughness dependent, providing examples of how polished metal and foil of the same metal can have different emissivities.
  • Another participant argues that if the body is a light source not due to thermal radiation, then ε may not apply, citing the example of a CO2 laser versus a remote control.
  • Some participants discuss the scenario where the atmosphere blocks visible light, allowing only IR wavelengths to reach the Earth, and express the need for the value of ε to calculate the heat emitted by the Earth.
  • It is mentioned that the emissivity depends on the shape and smoothness of the body, and that predicting emissivity values is complex and often empirical.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of emissivity for non-thermal light sources and the classification of bodies emitting only IR light. There is no consensus on the value of ε for such cases, and the discussion remains unresolved.

Contextual Notes

Participants highlight that emissivity can vary based on material properties and surface characteristics, and that there are complexities in defining and predicting these values.

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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