Question

The earth receives solar radiation at a rate of $8.2J/c{m}^2$-minute. Assuming that the sun radiates like a blackbody, calculate the surface temperature of the sun. The angle subtended by the sun on the earth is ${0.53}^{\circ }$ and the Stefan constant
$\sigma =5.67\times {10}^{-8}W/m^2{K}^4$

• A. 5000 K
• B. 5800 K
• C. 5400 K
• D. 6200 K

Answer 1

The correct option is B 5800 K


Let the diameter of the sun be D and its distance from the earth be R. From the question,
$\frac{D}{R}=0.53\times \frac{\pi }{180}=9.25\times {10}^{-3}\dots \dots \left(1\right)$
The radiation emitted by the surface of the sun per unit time is
$\pi \frac{D}{2}{T}^2\sigma T^4=\pi D^2\sigma T^4$
At distance R, this radiation falls on an area $4\pi R^2$ in unit time. The radiation received at the earth's surface per unit time per unit area is, therefore,
$\frac{\pi D^2\sigma T^4}{4\pi R^2}=\frac{\sigma T^4}{4}\frac{D}{R}{T}^2$
Thus,
$\frac{\sigma T^4}{4}\frac{D}{R}{T}^2=8.2J/c{m}^2-minute$
or,
$\frac{\sigma T^4}{4}\frac{D}{R}{T}^2=8.2J/c{m}^2-minute\frac{1}{4}\times 567\times {10}^{-8}\frac{W}{m^2-K^4}{T}^4\times (9.25\times {10}^{-3}{)}^2=\frac{82}{{10}^{-4}\times 60}\frac{W}{m^2}or,T=5794K\approx 5800K$