Question

A ceramic cube $3.0cm$ on each side radiates heat at $630W$. At what wavelength, in $mm$, does its emission spectrum peak? Assume$e=1$.

Short answer

$2.421\mu m$

Step-by-step solution

Givenceramic cube = 3cmHeat Radiation (P) = 630 WHere we have to apply Stefan law :-

$$\begin{gathered} P=e\sigma A{T}^4 \\ {A}_{net}=6x3x3 \\ {A}_{net}=54c{m}^2 \\ T={\left(\raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$e\sigma A$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\lambda \left(nm\right)=\raisebox{1ex}{$2.9x{10}^6(nm.K)$}\!\left/ \!\raisebox{-1ex}{$T$}\right. \\ \lambda \left(nm\right)=\raisebox{1ex}{$2.9x{10}^6(nm.K)$}\!\left/ \!\raisebox{-1ex}{${\left({\displaystyle \raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$e\sigma A$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}$}\right. \\ \lambda \left(nm\right)=\raisebox{1ex}{$2.9x{10}^6(nm.K)$}\!\left/ \!\raisebox{-1ex}{${\left({\displaystyle \raisebox{1ex}{$630$}\!\left/ \!\raisebox{-1ex}{$1.5.67.{10}^{-8}.54x{10}^{-4}$}\right.}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}$}\right. \\ \lambda =2421nm=2.421\mu m \end{gathered}$$

Emission spectrum peak at $\lambda =2.421\mu m$of wavelength.