91Ó°ÊÓ

The electromagnetic radiation inside a kiln, with a volume of V=I m³ and a temperature of 1500 K.

(a) Suppose we have an electromagnetic radiation inside a kiln, which has a volume of V = 1 m³ and temperature of T = 1500 K, the total energy of this radiation is given by:

$U=\frac{8{\mathrm{Ï€}}^5(kT{)}^4}{15(hc{)}^3}V$

Substitute with the values,

$$\begin{gathered} U=\frac{8{\mathrm{π}}^5{\left(\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(1500\mathrm{K})\right)}^4}{15{\left(\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)\right)}^3}\left(1{\mathrm{m}}^3\right) \\ U=3.815×{10}^{-3}\mathrm{J} \end{gathered}$$

(b) The radiation's spectrum is provided by:

$u\left(\mathrm{Ï\mu }\right)=\frac{8\mathrm{Ï€}}{(hc{)}^3}\frac{{\mathrm{Ï\mu }}^3}{e^{\mathrm{Ï\mu }/kT}-1}$

Using python and the code is:

From the graph we can see that the peak occurs at energy of $\mathrm{Ï\mu }=0.36\mathrm{eV}$

(c) Now we need to discover the component of the spectrum that represents visible light (the fraction of the energy), which we can do by integrating the following equation all across the visible light (equation 785):

$U=\frac{8\mathrm{Ï€}V}{(hc{)}^3}{∫}_{{\mathrm{Ï\mu }}_1}^{{\mathrm{Ï\mu }}_2}\frac{x^3}{e^x-1}dx$

The energy in terms of wavelength is:

$\mathrm{Ï\mu }=\frac{hc}{\mathrm{λ}}$

Therefore,

$$\begin{gathered} {\mathrm{Ï\mu }}_2=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{400×{10}^{-9}\mathrm{m}}=4.9695×{10}^{-19}\mathrm{J}=3.1\mathrm{eV} \\ {\mathrm{Ï\mu }}_1=\frac{\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}{700×{10}^{-9}\mathrm{m}}=2.84×{10}^{-19}\mathrm{J}=1.77\mathrm{eV} \end{gathered}$$

The fraction of the energy in the visible light is given by:

$\frac{U_{\text{vis.}}}{U_{\text{tot.}}}=\frac{{∫}_{1.77\mathrm{eV}}^{3.1\mathrm{eV}}\frac{x^3}{e^x-1}dx}{{∫}_0^{∞}\frac{x^3}{e^x-1}dx}$

The integration in the denominator equals ${\mathrm{Ï€}}^4/15$, therefore

$\frac{U_{\text{vis.}}}{U_{tot}}=\frac{15}{{\mathrm{Ï€}}^4}{∫}_{1.77\mathrm{eV}}^{3.1\mathrm{eV}}\frac{x^3}{e^x-1}d\mathrm{Ï\mu }$

But this is not the real boundaries since the integration over x not $\mathrm{Ï\mu }$, we have:

$x=\frac{\mathrm{Ï\mu }}{kT}$

But temperature T=1500K

localid="1647756536663">x1=ϵ1kT=1.77eV8.62×10-5eV/K(1500K)=13.7x2=ϵ2kT=3.1eVeV8.62×10-5eV/K(1500K)=24

Therefore, the integral will be:

$\frac{U_{\text{vis.}}}{U_{tot.}}=\frac{15}{{\mathrm{Ï€}}^4}{∫}_{13.7}^{24}\frac{x^3}{e^x-1}d\mathrm{Ï\mu }$

To solve this integral python is used, the code is and the ratio is:

$\begin{array}{|l|}\hline \frac{U_{\text{vis.}}}{U_{\text{tot.}}}=0.000557\\ \hline\end{array}$