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If the definition of a black body is:

"A blackbody is an object that absorbs all of the radiation that it receives (that is, it does not reflect any light, nor does it allow any light to pass through it and out the other side). The energy that the blackbody absorbs heats it up, and then it will emit its own radiation."

Then why are light bulbs black bodies? Don't they generate heat themselves instead of absorbing radiation, then giving radiation out?

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    $\begingroup$ A "black body" is a fancy name for a heated object. An incandescent bulb is a heated object and therefore a black body. And so is a hot charcoal. Or a cold charcoal, because it is still heated to a room temperature. An ideal black body does not reflect light. The filament in a bulb does reflect light. So bulbs are not 100% black bodies. But seriously, how much does the filament reflect compared to what it emits? Virtually nothing. So incandescent bulbs (including halogen) are near prefect black bodies. And this why their light is so much more pleasant. $\endgroup$ – safesphere Aug 6 '17 at 5:43
  • $\begingroup$ @safesphere. Good comment. Should have written it as an answer, IMO. $\endgroup$ – Bob Bee Aug 6 '17 at 5:50
  • $\begingroup$ @Bob Bee: Thanks Bob! I appreciate it. I am new to this forum and only learning its draconian rules ;) So often I am just afraid to ask or answer, cause people would just downvore me out of here completely. For example, no one could answer the very first question I asked, yet did not hesitate to vote it down. $\endgroup$ – safesphere Aug 6 '17 at 6:28
  • $\begingroup$ Thankyou! So does that mean LED's and CFL's aren't black bodies? $\endgroup$ – paradox124 Aug 6 '17 at 6:35
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    $\begingroup$ Correct, LED and CFL bulbs are not black bodies. There are only a few black body type sources of light in everyday life. These are approximately black bodies: incandescent and halogen bulbs, hot metal or charcoal, sun and stars. And these are NOT black bodies: LED, CFL, HID (mercury or sodium) bulbs, moon, or fireworks. $\endgroup$ – safesphere Aug 6 '17 at 15:53
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"The energy that the blackbody absorbs heats it up, and then it will emit its own radiation." --- This part may be a bit misleading. It should read as

The energy that the blackbody abosrbs heats it up, and will be emitted as part of its own radiation.

In fact as long as all radiation is absorbed, i.e. not reflected/passed through, it is a black body. A black body does not need to absorb heat first, then to begin radiating out. A light bulb can be roughly viewed as a black body; as a result if you subject it to radiations with a temperature far higher than its own typical temperature, it would heat up to (possibly more than) that one.

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I think your perplexity arises since the definition your cited is easy to misunderstand if you don't look at it carefully.

Let's try and break it in two parts:

  1. A black body is a body which absorbs all the radiation it receives (or, equivalently, reflects none of it).

  2. A black body is a body that gets heated by the energy it absorbs, and emits thermal radiation accordingly (indeed, black body radiation if it's in thermal equilibrium with the environment).

If you look carefully, it doesn't say that such a body is heated just by the radiation it absorbs. It says energy.

In the specific case, a powered-on light bulb get just a very tiny fraction of the energy that heats it up in the form of radiation from the external evironment. All the rest (almost all) is supplied by the means of an electric current carried by a wire: the electrons strike the tungsten atoms making them to vibrate furiously, and you know heat is just kinetic energy.

This explains why the filament (NOT the whole bulb) is a good approximation of a black body. It's not perfect though, as other users mentioned, since it's not a perfect absorber (perfect absorbers do not exist in nature, as far as we know).

Note that the light bulb filament is no longer a good approximation of a black body once you turn it off.

Another (even better) good approximation of a black body is the surface of the Sun (or of a star whatsoever). It does not certainly emits so much EM energy because it's heated by the radiation absorbed from other sources (that is present nonetheless). Rather it is heated by the fusion occurring hundreds of thousands kilometers below.

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  • $\begingroup$ A simple statement about thermal equilibrium would be good. A blackbody is something that absorbs all radiation incident upon it and is in thermal equilibrium at some specific temperature. The Sun is not a great blackbody because different wavelengths arise from layers at different temperatures. $\endgroup$ – Rob Jeffries Jul 6 '18 at 22:48
  • $\begingroup$ I'd dare not to concur. Indeed, the very definition of black body does not require thermal equilibrium for the black body itself. Thermal equilibrium, however, is important in the definition of black body radiation, and you did good in addressing my post about that (in fact, I edited it to take into account your suggestions). Indeed, the definition of black body radiation is just thermal radiation from a black body which is in t. equilibrium with its surroundings. See my other comment below for addressing your other objection. $\endgroup$ – MadHatter Jul 7 '18 at 14:27
  • $\begingroup$ "The Sun is not a great blackbody because different wavelengths arise from layers at different temperatures." If you refer to the different layers of the sun, note that when the Sun is presented as a good approximation of a black body, what is generally meant is that its photosphere is a good approximation of a black body (note that I wrote 'the surface of the Sun'). Which it indeed is, if you leave apart sunspots. $\endgroup$ – MadHatter Jul 7 '18 at 14:31
  • $\begingroup$ Finally, note that even a perfect radiator does not generally emit on a single frequency for a given temperature. Rather, it has a typical frequency distribution for a given temp (the number of frequencies corresponds to the vibrational modes available to the system, so that for a system composed of a great number of coupled particles, you can consider it continuous) $\endgroup$ – MadHatter Jul 7 '18 at 14:52
  • $\begingroup$ No, the photosphere is not a blackbody, otherwise all we could learn about would be it's temperature and emitting area. Look at any solar spectrum to see that it is trivially not a Planck function. $\endgroup$ – Rob Jeffries Jul 7 '18 at 14:58

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