Heat absorption, IR - light - UV, differences?

[Bad Monkey]

General query about heat absorption. We all know that something that is black both absorbs and radiates heat better.

My question is how does this property vary across the spectrum, if at all, for different materials (if there is any difference?).

For instance, a black painted car will heat up more quickly than a white car in the daytime.

(At night time I assume it is the case that the black car retains heat less efficiently because it radiates the heat as IR - correct?)

Does this happen because

1. There is an IR component in the sunlight, and the black we see in the visible spectrum is a side effect of a property which in fact relates to IR
2. Or, the visible light imparts energy to the car which is not reflected (hence black)

If #2, is the "black" property constant across the spectrum? Does a high UV index have a similar effect? Or can materials have differing black levels across the spectrum?

To further this I will pose some case examples:

• UV tends to destroy plastics, polyester & nylon, etc. Will a dyed white plastic last longer in sunlight than the black version? (I.e. does the visible black speak to its UV reflection factor)
• If there is a difference across the spectrum, does that mean that we can have something white which also radiates efficiently?

I think that will do for now. Hopefully you'll see what I'm getting at!

 

[Andy Resnick]

The spectral absorptivity in the visible is not correlated (in general) with the absorptivity in other parts of the spectrum. For instance, snow looks black in the infrared.

 

[Bad Monkey]

Isn't this then a cause/effect question?

Snow is white, but also doesn't radiate what small amount of heat it has. I am confused by this because it seems obvious - in the real world, clearly snow or ice must look black in IR because it is cold. Duh. Isn't that a statement relative to the snow's environment? If a chunk of ice around 0 deg C is teleported in deep space, wouldn't it look rather bright in the IR until it cooled down?

Anyway your answer also reiterates that black objects radiate heat quickly, conversely white (snow) does not. Again, is this connected to the color, or an independent property (IR level)?

E.g. we know that snow and ice cover contributes to global cooling in an ice age, because the white cover reflects light, and that the reverse is true with shrinking polar snow caps. However is it also the case that snow/ice reflects IR? Which is actually the important factor? Light / IR / both-but-not-related?

 

[Andy Resnick]

I think you missed the point- snow is black in the IR because the absorptivity/emissivity is very high in the IR.

IR means the spectral region from 3-12 microns. room temperature objects radiate most of their energy is the 8-12 micron band, hot things like engine exhaust peaks around 3-5 microns.

The radiated energy is a product of the blackbody curve at the temperature of the object and the spectral emissivity, integrated over wavelength. If the emissivity is zero somewhere, then the absorptivity is also zero, and for opaque objects, the reflectivity is then 1 at that spectral region. For transparent objects, the sum emissivity + reflectivity + transmisstion = 1.

 

[Bad Monkey]

So why do we have the rule of thumb that black (visible color) objects both absorb and emit heat to a higher degree than white?

Is this false, it being only dependent on the properties in the 3-12 micron band, or just a generality? If the latter, what is the connection?

I also don't understand now what you meant with the snow thing. Doesn't "black" mean its emissivity and reflectivity are low?

 

[Andy Resnick]

The rule of thumb exists because many common dyes that are highly absorptive in the visible are also absorptive to the main energy peak in solar irradiance. That's different from *requiring* or *assuming* materials have no spectral variation in properties from the visible and IR. Water, in particular, has a lot of spectral features. Texturing the surface of materials can also be used to tweak the IR emissivity and leave the visible appearance unchanged. Another exploitation from back when I did this sort of thing is that certain paints based on Soviet sources had a particular spectral blip around 9 microns that U.S.-based paints did not.

An object appears 'black' when the absorptivity is high. A high absorptivity means a high emissivity- thermodynamically, a perfectly black object is identical to a perfectly reflecting cavity with a tiny hole to let some light out.

Does that help?

 

[Bad Monkey]

Yes that does help thanks. Moving on then, is the IR absorptivity characteristic the only factor that matters with regard to heat, or does the black visible color also affect this independently? (visible spectrum absorbed and not reflected, does this energy convert to heat?).

If so, we can have something white which also radiates efficiently? If not, the visible color can be unimportant - even dark?

 

[GT1]

An object appears 'black' when the absorptivity is high. A high absorptivity means a high emissivity- thermodynamically, a perfectly black object is identical to a perfectly reflecting cavity with a tiny hole to let some light out.

Does always Emissivity=~Absorptivity? My teacher told us once that this true only for the long wavelengths (IR etc.)

 

[Andy Resnick]

It's identically true (Kirchoff's law). It's true due to the Second Law of Thermodynamics, otherwise you can have a cold object radiate heat onto a hot object, making the cold object colder and the hot object hotter.