The earth receives solar energy at the rate of 2 cal `Cm^(-2)` per minute. Assuming theradiation tobeblack body in character, estimate the surface temperature of the sun. Given that `sigma =5.67 xx10^(-8) Wm^(-2)K^(-4)` and angular diameter of the sun =32 minute of arc.

Explanation

Let R be the radius of the sun and `R_(0)` its distance from the earth. The angular `tan((theta)/2)=(R_(a))/(R_(0))` Since `R_(a) lt lt R_(0)`
`(theta)/2 =(R_(a))/(R_(0))`
But `theta=32=32/60xx3.142/180=9.3xx10^(-3)`
`implies theta//2=4.655xx10^(-3)`
The energy falling per second per unit area on the earth is
`Q=((R_(s))/(R_(0)))^(2)sigma T^(4)`
Given `Q=2` cal `-cm^(-2)` per minute `=(2xx2.4xx10^(4))/60 J//m^(2)s^(1)=1400Js^(-1)m^(-2)=1400Wm^(-2)`
Given `sigma =5.67x10^(-8)Wm^(-2)K^_(1)` we have
`1400=(4.655xx10^(-3))^(2)xx5.67xx10^(-8)xxT^(4)impliesT=5810K`