Question #295810

The surface of the Sun has a temperature of about 5800 K. The radius of the Sun is 6.96 x 108 m. Calculate the total energy radiated by the Sun each second. Assume the emissivity of the Sun is 0.965

1
Expert's answer
2022-02-09T16:15:08-0500

Given:

T=5800KT=\rm 5800\:\rm K

R=6.96108mR=\rm 6.96*10^8\:\rm m

ϵ=0.965\epsilon=0.965

σ=5.67108W/(m2K4)\sigma=5.67*10^{-8}\:\rm W/(m^2\cdot K^4)

t=1st=1\:\rm s


The luminosity (radiation power) of Sun

L=ϵσT4A=ϵσT44πR2L=\epsilon \sigma T^4A=\epsilon \sigma T^4*4\pi R^2

The total energy radiated by the Sun each second

E=Lt=ϵσT44πR2tE=Lt=\epsilon \sigma T^4*4\pi R^2t

E=0.9655.67108580044π(6.96108)21=3.771026JE=0.965*5.67*10^{-8}*5800^4\\ *4\pi*(6.96*10^8)^2*1=3.77*10^{26}\: \rm J


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