J = 1/4p  ò 4p  I  dw   = 1/4p  (1.4 x  10 7 Jm-2 s-1) 4p 
So: J » 1.4 x  10 7 Jm-2 s-1
2) The net flux (H) passing through the Sun’s surface is estimated using:  sTo 4/ p = 2H.
Therefore: H = sTo 4/2 p 
Where: sTo 4 =  (5.67 x 10-8 W m-2 K-4)(4800K)4 
sTo 4 =   3.0 x 10 7 W m-2
H = (3.0 x 10 7 W m-2) /2 p = 4.8 x 106 W m-2
3) Consider the solar half-sphere and the energy going into it each second.  We know on average photons are absorbed after traveling a distance with optical depth t =1 in the propagation direction. Averaged over all directions this corresponds to a vertical optical depth of t = 2/3. Based on this find:
a) The energy going into the half sphere each second.
Solution:  
    This is just:   s(Teff)4   =  p F = D 2p S =D 2p I
b) The change in (a) over each absorption and re-emission over vertical optical depth.
Solution:
Over a vertical optical depth one has t = 2/3, then:
D 2p S/D t = p F/ (2/3) = 3p F/ 2
or:
DS/D t   =   3p F/ 2 (2p) = 3F/4
c) The total absorption and total emission and the relationship between them, defined over all space.
Solution:
Over all space, the total emission = 4p S and the total absorption = 4p J, and by radiative equilibrium: 
4p S = 4p J  so that S = J.
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