Foundations of radiative properties and equilibrium Which statement about radiative properties and thermal equilibrium is correct?

Introduction / Context:
Radiative heat transfer depends on material properties: absorptivity (fraction absorbed), emissivity (effectiveness as an emitter compared to a black body), reflectivity, and transmissivity. Kirchhoff’s law provides a powerful relation between emission and absorption at thermal equilibrium.

Given Data / Assumptions:

• Thermal equilibrium between a surface and its surroundings at a given wavelength and direction.
• Properties may be spectral and directional, but the law applies at matching wavelength and direction.

Concept / Approach:
Kirchhoff’s law states that, at thermal equilibrium, emissivity equals absorptivity for each wavelength and direction: ε_λ,θ = α_λ,θ. This does not mean the surface is a perfect absorber; it means its capability to emit relative to a black surface matches its capability to absorb. A black body is the special case where α = ε = 1. A grey body has spectral independence (ε and α constant with wavelength) but not necessarily unity.

Step-by-Step Solution:

Verification / Alternative check:
Thermodynamic arguments show that if ε ≠ α at equilibrium, net radiation exchange would spontaneously drive a temperature change, violating equilibrium.

Why Other Options Are Wrong:

Grey body absorbing all radiation is the definition of a black body, not grey.Emissivity is not “energy absorbed/incident”; that is absorptivity.Visual black appearance does not guarantee black-body behavior across thermal wavelengths.Opaque bodies have τ = 0, but reflectivity need not be zero; α + ρ = 1 with ρ possibly large.

Common Pitfalls:
Confusing color with radiative property over the thermal spectrum; many shiny metals have low ε but high reflectivity in infrared, despite visual appearance.

Final Answer:

At thermal equilibrium, emissivity equals absorptivity.