A star radiates at the rate of 1.5 X 10^24 W. It's radius is 10.0 X 10^8 m. Assuming that the star is a blackbody radiator, (e=1), what is the surface temperature?

Answer 1

Recall that
P = AεσT^4
So, plug in your numbers and you have
T = ∜(P/4πr^2σ) =∜((1.5*10^24)/(4π(10.0*10^8)^2*(5.670373*10^-8)) = ?°K

Answer 2

I'm so sorry