If a totally opaque object reflects 20% of incident IR radiation, what is its emissivity?

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Multiple Choice

If a totally opaque object reflects 20% of incident IR radiation, what is its emissivity?

Explanation:
For an opaque surface, no light or IR passes through; the incident energy either reflects or is absorbed. Emissivity equals absorptivity (Kirchhoff’s law) at thermal equilibrium, and absorbed energy plus reflected energy must add up to 1. Since 20% is reflected, the absorbed portion is 1 − 0.20 = 0.80. Therefore, emissivity is 0.80. This also fits the idea that absorbent, opaque surfaces can emit radiation effectively (up to a maximum of 1), while high reflectivity would reduce emissivity.

For an opaque surface, no light or IR passes through; the incident energy either reflects or is absorbed. Emissivity equals absorptivity (Kirchhoff’s law) at thermal equilibrium, and absorbed energy plus reflected energy must add up to 1. Since 20% is reflected, the absorbed portion is 1 − 0.20 = 0.80. Therefore, emissivity is 0.80.

This also fits the idea that absorbent, opaque surfaces can emit radiation effectively (up to a maximum of 1), while high reflectivity would reduce emissivity.

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