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

Concept: When total incident radiant energy flux (Irradiance, \(G\)) strikes a semi-transparent body surface, it splits into three components based on conservation of energy: a fraction is reflected, a fraction is absorbed, and the remaining fraction is transmitted through the material layer. Dividing the energy balance equation by the total incident energy defines the fundamental radiation surface properties: \[ \alpha + \rho + \tau = 1 \] Where:

• = Absorptivity (fraction of incident energy absorbed).

• = Reflectivity (fraction of incident energy reflected).

• = Transmissivity (fraction of incident energy transmitted).
Additionally, Kirchhoff's Law of Thermal Radiation states that for any body in thermodynamic equilibrium with its surroundings, its monochromatic or total surface absorptivity equals its total emissivity (\(\epsilon\)): \[ \epsilon = \alpha \] Where \(\epsilon\) represents the surface emissivity, defined as the ratio of energy radiated by the surface to that radiated by an ideal blackbody at the same temperature.

Step 1: Identifying given parameters from the problem text.
The problem provides the following surface property values:

• Reflectivity: \(\rho = 35\% = 0.35\)

• Transmissivity: \(\tau = 0.3\)

Step 2: Calculating surface absorptivity (\(\alpha\)).
We isolate absorptivity in our property conservation equation: \[ \alpha = 1 - \rho - \tau \] Substitute the given values into the formula: \[ \alpha = 1 - 0.35 - 0.3 \] Performing the subtraction step-by-step: \[ 1 - 0.35 = 0.65 \] \[ 0.65 - 0.3 = 0.35 \] Thus, the absorptivity of the surface is \(\alpha = 0.35\).

Step 3: Finding the emissivity (\(\epsilon\)) using Kirchhoff's Law.
Applying Kirchhoff's law: \[ \epsilon = \alpha = 0.35 \] The calculated emissivity of the surface is 0.35, which matches Option (1).