Radiation fundamentals — emissivity of an ideal black body What is the emissivity value assigned to a perfect black body in thermal radiation analysis?

Introduction / Context:
Emissivity quantifies how effectively a surface emits thermal radiation compared to an ideal emitter at the same temperature. Black bodies define the theoretical upper limit for emission and absorption, forming the basis for many radiative heat transfer models.

Given Data / Assumptions:

• Idealized black body concept per classical radiation theory.
• Diffuse-gray simplifications are standard unless wavelength dependence is discussed separately.

Concept / Approach:
A black body is defined as one that absorbs all incident radiation (absorptivity = 1) and, by Kirchhoff’s law, has emissivity equal to absorptivity for surfaces in thermal equilibrium. Therefore, a perfect black body has emissivity epsilon = 1, the maximum possible value.

Step-by-Step Solution:
Definition: black body → absorptivity = 1.Kirchhoff’s law: emissivity = absorptivity at the same wavelength and temperature.Hence, epsilon_black = 1.

Verification / Alternative check:
Stefan–Boltzmann law for a real surface is q = epsilon * sigma * T^4; for a black body, epsilon = 1, giving the maximum radiant exitance.

Why Other Options Are Wrong:
(a), (b), (c) are sub-unity values for gray bodies, not black bodies. (e) While real surfaces have spectral emissivity, the ideal black body has emissivity 1 at all wavelengths.

Common Pitfalls:
Confusing “black paint” with “black body.” Color and visual blackness are not reliable indicators of emissivity across thermal infrared wavelengths.

Final Answer:
1