Temperature, brightness temperature, emissivity

[Ardit]

Hello,

I have two different materials, wood and iron in two different environments. I have both the wood and iron temperature, respectively Tw, and Ti. I also know their emissivities, Ew, and Ei.

What I want to know is the temperature of the iron if it was wood (i.e. if its material was wood but standing in a different environment from the real wood)?
Does the solution have to do with the brightness temperature Tb whose relation with the real surface temperature Ts and emissivity E is: Tb=E(1/4) x Ts ?

 

[mfb]

The emissivities will depend on the wavelength, so does the incoming light. In general, you cannot calculate the answer with just those 4 values. However, if both materials are prefect gray bodies (emissivity is independent of wavelength), they will have the same temperature in the same environment.

 

[Ardit]

OK, I do have the wavelength λ value for emissivity E. I also have another emissivity E' captured for another wavelength λ' for both the materials.
So, if we say that the materials are not perfect gray bodies, can I get what I am looking for with the data I have above?

 

[mfb]

Just 2 wavelengths? Not a full spectrum?

For all materials, you can calculate:
- The absorption as integral over emissivity multiplied by the radiation density.
- The emission as integral over the (temperature-dependent) Planck spectrum multiplied with the emissivity
In equilibrium, both numbers are the same, this allows to calculate the temperature.

 

[Ardit]

Well, I have two spectral channels, one ranges from 10.780-11.280 µm and the other 11.770-12.270 µm. For both of them, I have the associated emissivities.
Integral...sounds scary to me. It's been years that I haven't dealt with them.

 

[mfb]

Well, I think you will need additional data (or some assumptions).

Consider a green car in green light at room temperature, for example:
The green car might reflect most of the green light. It is at room temperature, so its emission of visible light is negligible, and thermal emission is mainly infrared. It gets heated a bit, but not much.
What happens to a red car? It can absorb most of the green light. Its emission of visible light is negligible, and thermal emission is mainly infrared - the same as the green car. To counter the increased heating from the light, it has to get hotter.

Both cars can have the same emissivity in the infrared, and even the same emissivity for sunlight, but they will behave completely different in specific light conditions.

For the same reason, black cars get hotter than white cars in sunlight. Absorption in the visible light is completely different, but emission is mainly infrared, where the cars are not "black" and "white" any more.

 

[Ardit]

Nice examples.
So, if we have two cars of the same material, shape and texture under the same conditions but one car is white and the other is black, will they have the same emissivities?

 

[mfb]

Emissivity is not a single number as material property - it is a function of frequency (or temperature, if you like).
The white car has a lower emissivity in the visible spectrum (not necessarily for every frequency, but at least in general) and probably a lower emissivity as 6000°C hot object (if it would be solid at that temperature...).