The earth receives solar radiation at a rate of `8.2Jcm^(-2)min^(-1)` . 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 `0.55^(@)C` and the stefan constant `sigma=5.67xx10^(-s)Wm^(-2)K^(-4)` .

Explanation

Let the diameter of the sun be D and its distance from the earth be R. From the questions
`(D)/( R)~~ 0.53xx(pi)/(180)=9.25 xx10^(-3)" "…(i)`

The radiation emitted by the surface of the sun per unit time is
`4pi ((D)/(2))^2 sigma T^4 = pi D^2 sigma T^4`
At distance R, this radiation falls on an area `4pi R^2` in unit time. the radiation received at the earth's surface per unit time per unit area is, therefore,
`(pi D^2 sigma T^4)/(3piR^2) = (sigma T^4 )/(4) ((D)/(R))^2`
Thus, `sigma T^4)/( 4) ((D)/(R ))^2 = 8.2 J//cm^2 -` minute
or, `(1)/(4) xx(5.67xx10^(-8) (W)/( m^2 - K^4))T^4 xx(9.25 xx10^(-3))^2`
`=(8.2)/(10^(-4)xx60 m^2)`
or, `T= 5794 K ~~ 5800 K`.