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 surface temperature of sun be `T_s`. Then total energy radiated by sun per second is given as
`E=sigmaT_s^4.(4piR_s^2)`
Energy reveibed by earth per second per square metre is give as
`(dQ)/(dt)=(E)/(4piR_(es)^2)=sigmaT_s^4((R_s)/(R_(es)^(2)))` ..(i)
It is given that angular diameter of sun as observed from earth is 32 min of arc. Thus. we have
`(R_s)/(R_(es))=(theta)/(2)=(1)/(2)xx(32)/(60)=4.655xx10^-3`
and it is given that `(dQ)/(dt)=(2xx4.2xx10^4)/(60) J//s-m^(2)` ....(ii)
Now from Eqs. (i) and (ii) we have
`5.67xx10^-8xxT_s^4xx[4.655xx10^-3]^2=(2xx4.2xx10^4)/(60)`
or `T_s^4=(2xx4.2xx10^4)/(60xx5.67xx10^-8xx(4.655xx10^-3)^2)`
or `T_s^4=1.14xx10^(15)`
or `T_s=5810.67K`