Radiative properties check: If a surface has absorptivity α = 1, reflectivity ρ = 0, and transmissivity τ = 0, then in heat transfer terminology this surface is classified as which type of body?

Introduction / Context:
In thermal radiation, every surface is described by three fundamental radiative properties: absorptivity (α), reflectivity (ρ), and transmissivity (τ). Understanding the extreme cases helps students classify real materials and approximate them in engineering calculations.

Given Data / Assumptions:

• Absorptivity α = 1 (the surface absorbs all incident radiation).
• Reflectivity ρ = 0 (no portion is reflected).
• Transmissivity τ = 0 (no portion passes through).
• Energy conservation for radiation: α + ρ + τ = 1.

Concept / Approach:
The definition of a black body is a surface that absorbs all incident radiant energy at every wavelength and direction. Therefore its α = 1. Because total must sum to 1, a black body necessarily has ρ = 0 and τ = 0. It is also the ideal emitter: its emissive power at a given temperature is the maximum possible among all bodies.

Step-by-Step Solution:

Verification / Alternative check:
Kirchhoff’s law at thermal equilibrium states emissivity equals absorptivity. If α = 1, emissivity = 1, matching the notion that a black body is also a perfect emitter (reference curve for radiation exchange).

Why Other Options Are Wrong:

Grey body: absorptivity/emissivity are constant with wavelength but less than 1.Opaque body: simply means τ = 0; it can still reflect. Here ρ = 0, so this is a special case, not the general term.White body: ideal reflector with ρ = 1, not consistent with α = 1.Translucent body: implies τ > 0, which contradicts τ = 0.

Common Pitfalls:
Confusing “black” in radiation with visible color; a surface may appear shiny black visually yet not behave as an ideal black body across all wavelengths.

Final Answer:

black body