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

At the wavelength of $2.421\mu m$, ceramic cube has its emission spectrum peak.

Step-by-step solution

Step-1 Given information

We have given that cube has a side $l=3.0cm$and radiates heat $P=630W$and $e=1$and

We have to find out the wavelength.

Step 2- Formula used

Using Stefan's Law,

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

where P is heat radiated ,A is the area, T is the temperature and $\sigma$is Stefan-Boltzmann constant

Find net A (net area of surface of the cube)

$A=6\times 3\times 3c{m}^2$

$A=54c{m}^2$

Step 3-Simplify

Using Wien's law,

which states that$\lambda \left(nm\right)=\frac{2.9\times {10}^6(nm.K)}{T}$let this be equation 1

and $T={\left(\frac{P}{\sigma eA}\right)}^{\frac{1}{4}}$

Putting the value of T in equation 1,

$\lambda \left(nm\right)=\frac{2.9\times {10}^6}{{\left({\displaystyle \frac{P}{\sigma eA}}\right)}^{{\displaystyle \frac{1}{4}}}}$

On solving We get

$\lambda =2421nm$

Changing units,

we get,

$\lambda =2.421\mu m$

The wavelength of ceramic cube is$2.421\mu m$