Question

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

Answer 1

Answer:

0.45

Explanation:

a + p + T = 1

a + 0.40 + 0.15 = 1 (p = 0.40, T = 0.15)

a = 1 - 0.55

a = 0.45