0
$\begingroup$

Saw this article come by on Facebook the other day from I Fucking Love Science. It has a video of someone demoing what happens when you shine a laser on a piece of the newest version of Vantablack. The old version supposedly absorbed 99.96% of the light that hit it, and this is claimed to absorb even more, so almost 100%.

I got thinking though, light is after all just a form of energy. So if it's absorbing all this energy, from the laser in the video, or even just from the lights being on and the sun being on,

where is all that energy going?

Granted the energy of sunlight or room lights at any given moment isn't too high (otherwise we'd all be fried!) but if it's absorbing basically ALL of it every second that it's in the light, it's got to be building up pretty quickly.

My brother posited the logical sounding idea that it just re-emits it as infrared or something. My friend put forth the much more exciting (though probably less likely - lol) idea that it's falling through some dimensional rift or something.

As an ancillary question, I was wondering about applications though, if it's such a perfect absorber of energy, would it be a contender for some awesome new battery tech, or incredible breakthrough in solar cells?

$\endgroup$
  • 3
    $\begingroup$ you want to look up "black body radiation" (spoiler: your brother is closest to being right) $\endgroup$ – Bort Oct 27 '16 at 14:10
5
$\begingroup$

The light absorbed is converted to heat. But as per the Vantablack web page:

Very high front to back thermal conduction - excellent for Black Body calibration sources

The material has very high heat conductivity (thermal diffusivity) so provided there's a cold sink behind it, it will not actually heat up much because the heat absorbed diffuses through it very quickly.

Granted the energy of sunlight or room lights at any given moment isn't too high (otherwise we'd all be fried!) but if it's absorbing basically ALL of it every second that it's in the light, it's got to be building up pretty quickly.

The light from room lights does indeed contribute to heating the room, as the light energy is also converted to heat on absorption by the room's walls. The reason we don't 'get fried' is that the power of these lights is modest compared to what's needed to keep a room at a comfortable temperature and because rooms constantly lose heat as well.

A perfectly,100 % insulated room would nonetheless heat up and up and up, until we got fried. Some have experimented with very well insulated cavities heated only by a light bulb, to cook small amounts of food in!

As an ancillary question, I was wondering about applications though, if it's such a perfect absorber of energy, would it be a contender for some awesome new battery tech, or incredible breakthrough in solar cells?

Solar cells use materials that convert light energy directly to electrical potential, i.e. photovoltaic cells. Vantablack doesn't do that. But some solar systems are happy to convert light energy to heat, e.g. for domestic heating purposes. A radiator, coated in Vantablack, mounted South facing and with water running through it, would have a high efficiency for capturing the sunlight's energy content and converting it to hot water. But I don't know whether Vantablack's market price would make the initial investment and ROI attractive or not.

$\endgroup$
  • $\begingroup$ Some have experimented with very well insulated cavities heated only by a light bulb, to cook small amounts of food in! My sister had one of those as a kid! $\endgroup$ – Devsman Jun 7 '17 at 17:58
1
$\begingroup$

The energy vantablack absorbs goes to the same place the energy the 99.96% black object absorbs goes - into heating the black painted object up. As @Bort hinted in the comments, blackbody radiation is important because, as Planck's law and the derived Stefan-Boltzmann equation shows, the hotter something is, the more radiation it emits. It just so happens that we cannot see that radiation until a body is really hot, on the order of a thousand kelvins, before we can see it in visible light. Get an object hot enough, like welders do, and you'll need glasses to block the ultraviolet light and avoid burning your retinas.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.