Question

The earth receives solar energy at the rate of $2cal-c{m}^{-2}$ per minute. Assuming the radiation to be black body in character, estimate the surface temperature of the sun. Stefan's $constant=5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}$ and angular diameter of the sun $=32$ minute of arc.

Answer 1

Let $R_s$ be the radius of the sun and $R_0$ its distance from the earth. The angular diameter of the sun is $\theta$.
$\tan \left(\frac{\theta }{2}\right)=\frac{R_s}{R_0}$. Since $R_s<<R_0$,
$\frac{\theta }{2}=\frac{R_s}{R_0}$
But $\theta ={32}^{\prime }=\frac{32}{60}\times \frac{3.142}{180}=9.3\times {10}^{-3}$
$\Rightarrow \theta /2=4.655\times {10}^{-3}$
The energy falling per second per unit area on the earth is
$Q={\left(\frac{R_s}{R_0}\right)}^2\sigma T^4$
Given, $Q=2cal-c{m}^2$ per minute $=\frac{2\times 4.2\times {10}^4}{60}J/m^2{s}^1=1400J{s}^{-1}{m}^{-2}=1400W{m}^{-2}$
Given $\sigma =5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}$. we have
$1400=(4.655\times {10}^{-3}{)}^2\times 5.67\times {10}^{-8}\times T^4\Rightarrow T=5810K$.