Foundations of radiative properties and equilibrium Which statement about radiative properties and thermal equilibrium is correct?
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AA grey body absorbs all incident radiation (definition of a black body).
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BAt thermal equilibrium, emissivity equals absorptivity.
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CEmissivity is the ratio of energy absorbed to energy incident.
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DA perfect body is any surface that appears black to the eye.
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EReflectivity must be zero for any opaque body at equilibrium.
Answer
Correct Answer: At thermal equilibrium, emissivity equals absorptivity.
Explanation
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:
Define absorptivity α: fraction of incident radiation absorbed.Define emissivity ε: emitted intensity relative to a black body at same T.Apply Kirchhoff’s law at equilibrium: ε = α (spectrally and directionally).Conclude option (b) is correct.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.