Say we have a blackbody sphere in a perfect vacuum with surface area of $1 m^2$ at temperature 5.1205 C = 278.2705 K. Let's say inside the sphere it has its own energy source that is outputting enough energy to keep the sphere at this temperature.

According to the Stefen-Boltzmann law, it will radiate $j^* = \sigma T^4$ = 340.00 W/$m^2$. Since the sphere has $1 m^2$ surface area it's radiating 340 W and thus it's at thermal equilibrium (i.e. its internal energy source must be emitting 340 W as well).

Now say we take a flashlight that outputs 100 W of light energy and we shine it on the sphere. Will this result in the sphere heating up past its 278.2705 K?

The answer seems to be obviously that yes, since more energy is going into the sphere. Now besides its own internal energy it has an extra +100 W, so it will at equilibrium net radiate 440 W across its $m^2$ and reach 296.8K = 23.65 C.

Or will this 100 W of energy be reflected and not absorbed because it's from a lower-energy source or because the energy coming in is lower than the energy it's emitting? This seems odd because, why wouldn't it just get absorbed?

However if it does get absorbed, it seems we can get a funny situation. Let's consider a similar sphere but without its own internal energy source. Place it in a perfectly insulating chamber (no heat escapes), except the flashlight energy can still get through (e.g. a chamber transparent to the flashlight's wavelengths but reflective to the wavelengths it would emit as a blackbody). Now we shine the light on it.

The system is net gaining energy at a rate of 100 W --- since anything the sphere is radiating is reflecting back. Does that mean it will get infinitely hot? If so isn't this odd, since it means we could basically melt steel with a flashlight? (i.e. once the sphere gets hot enough, take it out of the chamber and drop it on some steel and it will melt right through). Would that actually work? If not, where's the disconnect?


2 Answers 2


By definition, all the radiation incident on a blackbody is absorbed, and never reflected. In the first hypothetical scenario you are considering, the blackbody receives radiation from two sources:

  1. The internal energy source of 340 W
  2. The flashlight of 100 W

All of this radiation is absorbed until the blackbody reaches a temperature of 296.8 K, when the rate of absorption equals the rate of emission.

In the second hypothetical scenario, the blackbody receives radiation from the following sources:

  1. The flashlight of 100 W
  2. The radiation reflected from the walls of the chamber

Any emissions from the blackbody will ultimately be reflected back onto it, hence the system gains energy at a net power of 100 W.

Yes, the blackbody's temperature should rise indefinitely (or at least until the system melts down into a liquid). This can be understood intuitively.

As the blackbody's temperature rises, the rate of emission increases. Simultaneously, because of the nature of the enclosure, more and more radiation will be incident on the blackbody, which will be absorbed by it. The rate of emission is constantly chasing the rate of absorption, but they can never be equal. Hence, the system never reaches thermal equilibrium.

  • $\begingroup$ Interesting. But doesn't it violate the second law of thermodynamics -- where heat can only flow from hotter to colder? Why can heat flow from the 'colder' flashlight to the 'hotter' insulated spherical blackbody? $\endgroup$
    – Cloudyman
    Mar 4 at 22:41
  • 1
    $\begingroup$ @Claudiu The second law prohibits flow of heat from a body at a lower temperature to a body at a higher temperature. This is because such a transaction would decrease the entropy of the isolated system. Radiation isn't hot or cold. It doesn't have a temperature. Moreover, the presence of the flashlight implies that the system is not isolated. Hence you cannot apply the second law directly. $\endgroup$
    – Amogh
    Mar 4 at 22:48
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    $\begingroup$ Ah I see. Right, the entire closed system would include whatever is powering the flashlight. And this system as a whole wouldn't violate the 2nd law. Makes sense. The assumption of "this thing that constantly inputs energy" is what leads to "it heats up forever"... $\endgroup$
    – Cloudyman
    Mar 4 at 22:50
  • $\begingroup$ @Claudiu Exactly. $\endgroup$
    – Amogh
    Mar 4 at 23:11

If we catch into a mirror lined box a sample of photons emitted by a filament of an incandescent light bulb at temperature 3000 degrees, then temperature in the box will be 3000 degrees, measured by a thermometer.

It's the same thing as when we catch into a box a sample of ideal gas molecules leaked out of a box into vacuum, and then temperatures in the two boxes are the same.

So old fashioned flashlight can't heat anything over 3000 degrees, while LED flashlight can't heat anything over x degrees, where x is some number about which I have no idea. Surely it's higher than 3000, and probably it's lower than 10000.

  • $\begingroup$ Hmm interesting. But what limits it? The other answer was that if you have energy in, it will keep heating up to any temperature. What does it matter the source of the energy? $\endgroup$
    – Cloudyman
    Mar 6 at 8:40
  • $\begingroup$ @Claudiu Entropy = Q/T. And entropy = information. So large amount of information coming out of a LED light means that the entropy of the light is high, which means that the temperature of the light is low. Photons have different momentums and positions and colors - that's the information. Photon wind, or stream, or radiation is even cooler than the original photon gas, but catching said radiation in a box produces photon gas with the original temperature again. $\endgroup$
    – stuffu
    Mar 6 at 20:45

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