Asked by Anonymous

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?
Answered by Steve
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
Answered by Anonymous
I'm so sorry
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