Discussion
A grey surface has an emissivity ∈ = 0.35 at a temperature T = 550°K source. If the surface is opaque, calculate its reflectivity for a black body radiation coming from 550°K source.
Answer: D
The reflectivity is 0.65.
For any surface, the incident radiation is absorbed, reflected or transmitted: \(\alpha + \rho + \tau = 1\). An opaque surface transmits nothing, so \(\tau = 0\). For a grey surface, Kirchhoff's law makes the absorptivity equal to the emissivity, whatever the source temperature:
\[ \begin{aligned} \alpha &= \varepsilon = 0.35 \\ \rho &= 1 - \alpha = 1 - 0.35 \\ &= 0.65 \end{aligned} \]
0.35 is the absorptivity, not the reflectivity, which is the trap. The source temperature of 550 K matters only to confirm that source and surface are at the same temperature; for a grey body the answer would not change anyway. 0.85 and 0.75 do not follow from the emissivity given.
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