What causes objects to appear "black"?

Question

To my understanding, the reason objects appear of a certain color is because it is the color (in visible light) that is absorbed least by the electrons in the material's atoms, and the rest is reflected.

It is also my understanding that those wavelengths are absorbed because of the possible energy levels of the electrons.

Also, the reason why that light is not re-emitted is because while the electrons have a tendency to drop in energy level (to a more stable state) the more immediate way to do so is through heat transfer, and that energy is then lost through heat energy in the bonds between atoms giving off energy as black-body radiation.

Is there anything wrong with this understanding? Are there any details missing that would be good to know?

Why does infrared contribute so much more to heat than other wavelengths such as ultraviolet if it is all transferred to heat anyway?

Answer 1

Is there anything wrong with this understanding?

A few minor points:

• Black-body radiation does not refer to a specific microscopic process, and in particular is not necessarily "the bonds between atoms giving off energy". Black-body radiation is an explicitly macroscopic phenomenon, whose usefulness comes from the fact that it doesn't particularly care about the microscopic structure of the material. For more details, see What are the various physical mechanisms for energy transfer to the photon during blackbody emission?.

• Absorption and reflection are not the only things that can happen when light hits a medium. There is also transmission, for one. And in higher-energy cases, you can also have decomposition/photodegradation, ionization, and pair production, to name a few.

• In addition to being dissipated as heat, electromagnetic radiation can also cause an object to accelerate or rotate, if the properties of the object and of the incident radiation are correct.

Why does infrared contribute so much more to heat than other wavelengths such as ultraviolet if it is all transferred to heat anyway?

Well, firstly because not all absorbed radiation is dissipated as heat. If the radiation is high enough in energy, it will break chemical bonds (like UV radiation) and/or ionize the atoms of the material (like X-rays) rather than simply increase the material's temperature. If the radiation is low enough in energy, it will simply be transmitted without interacting significantly with the material. So, in order for radiation to be primarily dissipated as heat, it needs to be around the same energy as a transition that won't actually break the material apart. Many (but not all*) materials have just such a transition in the infrared range, due to interatomic bonds having a typical binding energy of a few eV.

It also tends to help that, for materials around room temperature, the peak of the blackbody radiation spectrum is in the infrared range. This means that the amount of infrared radiation a body emits is correlated with its temperature, and in general macroscopic materials around room temperature will emit and absorb infrared radiation readily.

Of course, if you happen to know a lot about your material, you can use transitions outside the infrared to transfer heat. For example, water molecules have rotational transitions that are very closely spaced in energy; the energy difference between them corresponds to the energy of a microwave photon. So when water is blasted with microwaves, it heats up, which makes microwave ovens possible.

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*For example, zinc selenide is effectively transparent to infrared radiation, so you'd have a hard time heating it up that way if you tried.

Answer 2

To my understanding, the reason objects appear of a certain color is because it is the color (in visible light) that is absorbed least by the electrons in the material's atoms, and the rest is reflected.

For the most part this is correct. But there are other possibilities such as structural color/interference, fluorescence, phosphorescence, etc.

It is also my understanding that those wavelengths are absorbed because of the possible energy levels of the electrons.

Also correct. But in a bulk material, the energy levels of the electrons are very different than the widely-spaced ones in lone atoms.

Also, the reason why that light is not re-emitted is because while the electrons have a tendency to drop in energy level (to a more stable state) the more immediate way to do so is through heat transfer

It can be that the electrons do drop in energy level. But since the bulk material can move and vibrate, there are many, many energy levels available between the exitation and the ground state. This allows the energy to be distributed throughout the material rather than just be re-released as a single photon.

Once the energy is spread through the material, it is no longer energetic enough to be released as a visible photon.

Why does infrared contribute so much more to heat than other wavelengths such as ultraviolet if it is all transferred to heat anyway?

There is very little UV in our environment, so whether a material does or doesn't absorb UV tends not not matter much. On a sunny day, the absorption or reflection of visible light does matter significantly.

Objects near room temperature don't produce any visible or UV photons. So outside of a sunny environment (or high-temperature regimes), IR dominates the radiative heat transfer.