A black body hypothetically absorbs all wavelengths of light and simultaneously emits all of them at the same time. Since the visible spectrum combines into white light, is true to say that emitted radiations must also appear white?

  • $\begingroup$ In this kind of question we always forget to account for eyes response. Besides this general comment, the black body is not a flat line and its appearance depends on T as you know. Sun is technically a good approximation of a black body and classifies as yellow star. Tough is still a decent source of white light for our eyes. Warmer stars are more bluish etc. A black body can (mostly) emits out of our visible range, $\endgroup$ – Alchimista Dec 15 '17 at 11:42
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    $\begingroup$ No black bodies do not simultaneously emit all wavelengths absorbed. That describes a perfect reflector, not a perfect absorber. $\endgroup$ – sammy gerbil Dec 15 '17 at 12:21

This is not true.

While an ideal absorber (aka a black body) is also an ideal emitter, this does not mean it emits light at all frequencies equally ("white noise"). The spectrum of the electromagnetic radiation from a black body instead follows Planck's Law. This is anything but white noise. The intensity peaks at a wavelength that is inversely proportional to the object's temperature.

A black body at room temperature emits light primarily in the thermal infrared with almost no emissions in the visible range while a black body at 11000 kelvins (e.g., Rigel) emits light primarily in the visible and near ultraviolet, peaking in blue.

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