Radiative properties of a “white body” Evaluate the statement: For a white body, absorptivity α = 0, reflectivity ρ = 1, and transmissivity τ = 0.

Introduction / Context:Radiative heat transfer uses three basic surface interaction parameters: absorptivity (fraction absorbed), reflectivity (fraction reflected), and transmissivity (fraction transmitted). Idealized surface types like black, white, and grey bodies help students reason about limits.

Given Data / Assumptions:

• A “white body” is an idealization that perfectly reflects all incident radiation.
• Opaque assumption is applied, consistent with τ = 0 for many engineering surfaces.
• Energy conservation requires α + ρ + τ = 1 for any surface at a given wavelength/direction.

Concept / Approach:By definition, a white body reflects all incident radiation across the spectrum of interest, implying ρ = 1. With no transmission (τ = 0 for an opaque white body), the balance α + ρ + τ = 1 forces α = 0. Thus the statement α = 0, ρ = 1, τ = 0 is correct for the idealized, opaque white body model used in thermal engineering.

Step-by-Step Solution:

Verification / Alternative check:Real materials are “nearly white” over limited wavelength bands (e.g., polished aluminum in infrared or titanium dioxide paints in visible). Their ρ is high, α low, τ near zero if opaque. The ideal white body is the limiting case.

Why Other Options Are Wrong:

“Incorrect” conflicts with the ideal definition.Confining correctness to visible wavelengths or incidence angles introduces dependence not specified in the ideal.Diffuse versus specular reflection changes angular distribution, not the integrated value of ρ = 1 for an ideal white body.

Common Pitfalls:Equating “white” only with human visual perception; in heat transfer, the relevant spectrum is often infrared and the definition is spectral-band dependent. Here, the idealized definition holds for the band considered.

Final Answer:

Correct