Define the absorptivity of a surface.

Hint: Thermal radiation is electromagnetic radiation produced by the thermal motion of matter particles. Thermal radiation is emitted by all matter with a temperature greater than absolute zero. Particle motion causes charge-acceleration or dipole oscillation, which results in electromagnetic radiation.

Complete answer:
There is a fundamental relationship (Gustav Kirchhoff's 1859 thermal radiation law) that equates a surface's emissivity with its absorption of incident radiation (the "absorptivity" of a surface). Kirchhoff's Law explains why emissivities cannot be greater than one, because the greatest absorptivity - corresponding to complete absorption of all incident light by a truly black object - is also one.

Even though reflected light is not absorbed, mirror-like, metallic surfaces that reflect light will have low emissivities. Near room temperature, a polished silver surface has an emissivity of about \[0.02\] Because black soot absorbs thermal radiation very well and has an emissivity as high as \[0.97\], it is a good approximation to an ideal black body.

The appearance of a surface to the eye, except bare, polished metals, is not a good guide to emissivities near room temperature. White paint, as a result, absorbs very little visible light.Paint, on the other hand, absorbs light very well and has a high emissivity at an infrared wavelength of \[10\times {{10}^{-6}}\] meters. Similarly, pure water absorbs very little visible light, but it is still a substance.

Thus, a surface's absorptivity is defined as the ratio of thermal radiation absorbed per second to the radiation incident on it.

Note: The emissivity of a material's surface is its ability to emit energy as thermal radiation. Thermal radiation is a type of electromagnetic radiation that includes both visible (light) radiation and infrared radiation, which is invisible to the human eye. Thermal radiation from extremely hot objects (as seen in the photograph) is easily visible to the naked eye.