Question

A wave of radiation falls on a body, 35% of the radiation is reflected back. If transmissivity of the body is 0.25, then emissivity is:

1. 0.35
2. 0.45
3. 0.40
4. 0.25

Answer 1

Correct Answer - Option 3 : 0.40

Concept:

If Q is the total radiant energy incident upon the surface of a body, some part of it (Q_A) will be absorbed, some part (Q_R) reflected and some part (Q_Tr) transmitted through the body. By energy balance -

Q_A + Q_R + Q_Tr = Q

\(\frac{Q_A}{Q}\;+\;\frac{Q_R}{Q}\;+\;\frac{Q_{Tr}}{Q}=1\)

\(α\;+ρ\;+\tau=1\)

where α = absorptivity, ρ = reflectivity and \(\tau=transmissivity\).

Kirchhoff's law:

It states that the emissivity (ϵ) of the surface of a body is equal to its absorptivity (α) when the body is in thermal equilibrium with its surroundings.

ϵ = α.

Calculation:

Given:

ρ = 0.35, \(\tau=0.25\)

\(α\;+ρ\;+\tau=1\)

⇒ α + 0.35 + 0.25 = 1

∴ α = 0.4

From Kirchoff's Law:

ϵ = α

∴ ϵ = 0.4