40% of incident radiant energy on the surface of a thermally transparent body is reflected back. If the transmissivity of the body be 0.15, then the emissivity of surface is

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Quizplus · Accepted Answer

The emissivity of the surface can be found using the relation that the sum of reflectivity, transmissivity, and emissivity equals 1. Given that 40% (or 0.40) of the incident energy is reflected and the transmissivity is 0.15, we can calculate the emissivity as 1 - 0.40 - 0.15 = 0.45.

40% of Incident Radiant Energy on the Surface of a Thermally