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If only 1% of a 100W light bulb's energy is emitted in the form of visible light, whats the other 99%? Some of it is heat, I assume - but to say all of the 99% is heat seems excessive... does the tungsten (or whatever element is used as filament these days) emit some other form of electromagnetic energy as well?
Question Date: 2005-01-05
Answer 1:

Actually, Tungsten is supposed to be a pretty good black body emitter at that temperature (about 2500K). So I integrated the black body curve to see how much was visible divided by the flux that was not. I used 670 nmas the red limit and 370 nm as the violet limit. Here is what I got:

at 2000K (old car headlamps) 0.58% is visible
at 2500K (typical bulb) 2.67% is visible
at 3000 (halogen bulb) 7.2% is visible
and for the sun 5700K, ~65% is visible.

The solar value is off a bit since I did not count ultraviolet light which is minimal in regular bulbs and because hydrogen gas is not a great black body.

So I missed it a bit -- but the results do show why halogen bulbs are so much cheaper to run and that even they aren't so good.

1% was a guess -- I should have been more careful. The invisible energy is nearly all infrared light (at 2500K the peak emission is about 1.1 umin the near infrared.) This energy definitely leaves the envelope as the bulb itself would rapidly melt otherwise. (I confirmed this one experimentally some years ago as a TA in a physics lab section.)

By the way -- there is a nice java applet to show the black body light curve as a function of temperature at:
lecture on line

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