Question

On the specified surface, if \(35 \%\) of incident radiation energy is reflected and the transmissivity of the body is \(0.3\), then the emissivity of the surface is

• A. 0.35
• B. 0.3
• C. 1.0
• D. 0.65

Hint:

Radiation properties summary for any material surface: - $\alpha + \rho + \tau = 1$ - For an opaque body, transmissivity is zero ($\tau = 0 \implies \alpha + \rho = 1$). - By Kirchhoff's Law, always equate Emissivity directly to Absorptivity ($\epsilon = \alpha$) under thermal equilibrium conditions.

Answer 1

The Correct Option is A

Step 1: Write the radiation energy balance.
Every unit of incident energy on a surface must be reflected, absorbed, or transmitted, so: \[ \alpha + \rho + \tau = 1 \] Here we are given the reflected fraction \(\rho = 0.35\) and the transmitted fraction \(\tau = 0.3\).

Step 2: Solve for absorptivity and apply Kirchhoff's law.
\[ \alpha = 1 - \rho - \tau = 1 - 0.35 - 0.3 = 0.35 \] For a surface in thermal equilibrium with its surroundings, Kirchhoff's law tells us emissivity equals absorptivity, \(\epsilon = \alpha\), so: \[ \boxed{\epsilon = 0.35} \] This matches option 1.