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Let's take an arbitrary black physical object you can encounter in your everyday life, like a black chair, e.g. a chair whose whole surface was painted with black paint.

Does the chair appear black because the surface of the chair absorbs any light that hits it, reflecting none of it back?

I am not sure what the answer is, because I am confused by the myriad answers to what "black really is" I found on the web.

P.S. I understand that the "truest black" can only be found in things like black holes that reflect exactly 0% of the visible light that enters them and absorb exactly 100% of the visible light that enters them. Let's not be pedantic.

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The color of something you see is given by the electromagnetic radiation coming off of it. This means that in order that an object appear black, it must satisfy two properties:

  • It must absorb (essentially) all the (visible) light that hits it, reflecting none of it back.
  • It must be cool enough that its thermal glow is not visible.

Every body emits thermal radiation that depends on its temperature. However, for room temperature bodies, almost all the energy is radiated well into the infrared. On the other hand, we know that at temperatures of thousand of degrees, objects will glow visibly; that's what it means to be "red hot" (or hotter).

What it means to be "true" black is a linguistic question, not a physics question. Physicists tend to refer bodies that absorb all the radiation that hits them as "black," regardless of their temperature; so we can approximate the sun as a "black body" with a temperature of approximately 5700 K. However, that's just a choice of terminology; obviously the sun does not look black.

Sometimes, surfaces can have a black color but also be shiny. This is due to reflecting a relatively small amount of light (independent of color) and absorbing the rest. The way this is usually accomplished is by having a thin coating of partially reflective material over the underlying black pigmented surface that absorbs all the light that makes it through the surface layer.

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    $\begingroup$ So in short, in my particular example, the answer to the question in bold (which I changed due to your answer) is "yes"? $\endgroup$
    – Alex
    Dec 30, 2020 at 0:48
  • $\begingroup$ @Alex Almost. See my added paragraph at the end. $\endgroup$
    – Buzz
    Dec 30, 2020 at 0:51

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