Question

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

Short answer

The wavelenght will be$\lambda =2.421\mu m$

Step-by-step solution

Given Information

We have been given that:

Side of cube = $3.0cm$

Power =$630W$

Calculation

On applying Stefen's Law:

$P=e\sigma A{T}^4$

Here$P$stands for emmisive power,$e$stands for emissivity,$\sigma$is Stefen's constant,$A$is area and$T$is temperature in Kelvin.

We get the expression for T : localid="1650119591603" role="math" $T={\left(\frac{P}{e\sigma A}\right)}^{\frac{1}{4}}$ - ( i )

Total surface of the cube: localid="1650119604767" $A_{net}=6\times 3\times 3=54c{m}^2$

On Applying Wien's law:

localid="1649748840661" $\lambda \left(nm\right)=\frac{2.9\times {10}^6(nm·K)}{T}$

localid="1650119631751" $\lambda \left(nm\right)=\frac{2.9\times {10}^6(nm·K)}{{{\displaystyle \left(\frac{P}{e\sigma A}\right)}}^{{\displaystyle \frac{1}{4}}}}$ - using ( i )

localid="1650119670148" $\lambda \left(nm\right)=\frac{2.9\times {10}^6(nm·K)}{{{\displaystyle \left(\frac{630}{1\times 5.67\times {10}^{-8}\times 54\times {10}^{-4}}\right)}}^{{\displaystyle \frac{1}{4}}}}$

localid="1649748874919" $$\begin{gathered} \lambda =2421nm \\ =2.421\mu m \end{gathered}$$