\begin{problem}
In Homework 3, you showed that the sun emits $3.87\E{26}\U{W}$ of
power.  \Part{a} Use Stefan's law to calculate the surface temperature
at the photosphere ($r = 6.96\E{8}\U{m}$).  \Part{b} Estimate the
power needed to produce the same spectrum with an incandescent light
bulb (i.e. to heat the bulb to the same temperature).  Model the light
bulb's tungsten filament as a cylinder $58.0\U{cm}$ long and
$45\U{$\mu$m}$ in diameter with emissivity of $0.45$.
% emissivity from http://www.pyrometry.com/farassociates_tungstenfilaments.pdf
% model the bulb as a cylinder 65\U{mm} bulb diameter.  (48 kW at 2000 K)
\Part{c} Tungsten melts at $3695\U{K}$, so we cannot actually operate the
filament at the same temperature as the sun.  What temperature is the
filament from \Part{b} if we only radiate at $60\U{W}$?
% filament geometry and tungsten melting temperature from
% http://en.wikipedia.org/wiki/Incandescent_light_bulb

Because of their long length, tungsten filaments are usually coiled
twice.  See \url{http://en.wikipedia.org/wiki/Electrical_filament} for
some nice pictures.

This problem is similar to P28.2 and P28.55.
\end{problem} % based on problem 28.2

\begin{solution}
\Part{a}
\begin{align}
  P &= \sigma A e T^4 \\ 
  T &= \p({\frac{P}{\sigma A e}})^{1/4}
     = \p({\frac{P}{\sigma 4 \pi r^2 e}})^{1/4}
     = \p({\frac{3.87\E{26}\U{W}}{5.6696\E{-8}\U{W/m$^2$K$^4$} \cdot 4 \pi (6.96\E{8}\U{m})^2 \cdot 1}})^{1/4}
     = \ans{5790\U{K}} \;,
\end{align}
where we treat the sun as a black body ($e = 1$).

\Part{b}
Let the subscript w denote the tungsten filament and s denote the sun.
\begin{align}
  \frac{P_\text{w}}{P_\text{s}}
     &= \frac{\sigma A_\text{w} e_\text{w} T^4}{\sigma A_\text{s} e_\text{s} T^4}
      = \frac{L_\text{w} 2 \pi r_\text{w} e_\text{w}}{4 \pi r_\text{s}^2} \\
  P_\text{w}
     &= \frac{L_\text{w} r_\text{w} e_\text{w}}{2 r_\text{s}^2} P_\text{s}
      = \frac{0.58\U{m} \cdot \frac{45\E{-6}\U{m}}{2} \cdot 0.45}{2 \cdot (6.96\E{8}\U{m})^2} \cdot 3.87\E{26}
      = \ans{2.35\U{kW}}
\end{align}

\Part{c}
\begin{equation}
  T = \p({\frac{P}{\sigma A e}})^{1/4}
    = \p({\frac{P}{\sigma L 2 \pi r e}})^{1/4}
    = \p({\frac{60\U{W}}{5.6696\E{-8}\U{W/m$^2$K$^4$} \cdot 0.58\U{m}
                         \cdot 2 \pi \cdot \frac{45\E{-6}\U{m}}{2} \cdot 0.45}})^{1/4}
    = \ans{2310\U{K}}
\end{equation}

\end{solution}
