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A common grade school experiment is to compare how hot different materials get in direct sunlight. See examples here, here and here for example. The premise is generally two identical objects of different colors are compared with respect to how much heat they absorb, e.g. by heating a glass of water or melting ice.

To what extent does color contribute to the results of these experiments? Anecdotally, it seems like black things heat more than white things, colored things land somewhere in the middle.

I'm looking for a formula or some quantitative data about the impact of color on heat absorption. It seems surprisingly difficult to find explanations or information, despite it being a commonly held view. Trying to keep as close to the same material as possible, how much hotter will a black thing be than a white thing when sitting under the same sunlight?

Specifically I'm looking to understand how much color contributes to heat absorption compared to other considerations. E.g. is a material being black the primary determinant of absorption (black paint and black fabric), or are other facets (shiny vs. matte) bigger contributors.

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    $\begingroup$ The color of a material has little to do with how much heat it absorbs. If you are referring to heat absorption from direct electromagnetic radiation, then the answer depends on the imaginary part of the refractive index at the particular wavelength of interest. $\endgroup$ – Feel My Black Hole Aug 7 at 2:41
  • $\begingroup$ I'm open to improving this question or rephrasing if it's fundamentally not correct. I'd appreciate some constructive criticism more than downvotes and no answer. $\endgroup$ – user2152081 Aug 7 at 2:48
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    $\begingroup$ Those "kid" experiments are highly misleading in their conclusions. Using Maxwell's theory of electromagnetism, we can characterize each material with a refractive index. The refractive index completely determines the optical properties of a material. Using only the refractive index, you can calculate the Fresnel coefficients, which tell you the reflection and transmission at an interface. The imaginary part of the refractive index tells you how quickly the transmitted light (at each particular wavelength) decays in the medium, and is transformed into heat. $\endgroup$ – Feel My Black Hole Aug 7 at 2:59
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    $\begingroup$ @FeelMyBlackHole I think you are saying something correct, but it's misleading in the context of the question. The imaginary part of the index determines absorption, but absorption is part of what determines color. Also, the index does not completely determine the optical properties unless the material is perfectly uniform, with a perfectly flat surface. What is the index of wood? I think your comments, while technically true, don't address the question in the same spirit as it is asked. $\endgroup$ – garyp Aug 7 at 3:30
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    $\begingroup$ @user2152081 You asked an understandable enough question. Those attacking/downvoting you are essentially blaming you for not knowing enough to answer your own question, and they fundamentally misunderstand the act of learning. $\endgroup$ – alexchandel Aug 7 at 6:14
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I assume that you are asking about visible wavelength photons.

The color of a material depends on its emission and reflection of photons.

The wavelength of the photons that are emitted (or reflected) by the surface lattice determine the color of the object.

Basically the wavelength of emitted photon does not depend on the wavelength of the absorbed photons, but it depends on the lattice structure of the material.

You are saying that the object being black is a primary determinant of absorption, and if it is shiny or matte.

Now there is two types of reflection:

  1. specular

  2. diffuse

In the case of shiny objects, like mirrors, it is specular reflection (elastic scattering). This keeps the relative energy and angle of the photons.

In the case of matte objects, it is usually diffuse reflection, or absorption and re-emission.

Now an object being black is because it does not emit any visible wavelength photons.

Though, it could absorb any wavelength photons.

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