For a totally opaque surface, emissivity plus reflectivity is approximately equal to what?

For a totally opaque surface, emissivity plus reflectivity is approximately equal to what?

• A. Emissivity Equals Reflectivity
• B. Emissivity Plus Reflectivity Equals 1
• C. Emissivity Plus Reflectivity Equals 0
• D. Emissivity Plus Reflectivity Equals 2

Answer: (B)

For a totally opaque surface, all incident energy is either absorbed or reflected, since there is no transmission. Emissivity measures how well the surface emits (and, by Kirchhoff’s law, how well it absorbs) radiation. Because absorption (or absorptivity) equals emissivity for a surface in thermal equilibrium, and the remaining portion of the incident energy is reflected, the sum of absorptivity (which equals emissivity) and reflectivity must be 1. Therefore emissivity plus reflectivity is approximately 1. Choosing a sum of 0 would imply no absorption or emission, which isn’t possible for a real surface under thermal illumination, and a sum of 2 would imply more than all the incident energy is accounted for, which violates energy conservation.

For a totally opaque surface, all incident energy is either absorbed or reflected, since there is no transmission. Emissivity measures how well the surface emits (and, by Kirchhoff’s law, how well it absorbs) radiation. Because absorption (or absorptivity) equals emissivity for a surface in thermal equilibrium, and the remaining portion of the incident energy is reflected, the sum of absorptivity (which equals emissivity) and reflectivity must be 1. Therefore emissivity plus reflectivity is approximately 1.

Choosing a sum of 0 would imply no absorption or emission, which isn’t possible for a real surface under thermal illumination, and a sum of 2 would imply more than all the incident energy is accounted for, which violates energy conservation.