If a star’s surface temperature is 30,000 K, how much power does a square meter of its surface radiate?

1 Answer
May 23, 2015

To calculate the emitted power per square meter we need to use Stefan-Boltzmann’s Law,
that is,

E=sigmaT^4,

where E = emitted power per square meter of surface

T = temperature in Kelvins

sigma = Stefan-Boltzmann's constant: (5.670373xx10^(-8) "watt") /(1 "m"^2xxK^4)

Given/Known:
T="30000 K"
area = "1 m"^2"
sigma=(5.670373xx10^(-8) "watt")/(1"m"^2xx"K"^4")
Unknown:
E

Solution:

E=sigmaT^4=(5.670373xx10^(-8) "watt") /(1"m"^2xxcancel"K"^4)xx30000cancel"K"^4 = 5xx10^10 "watt/m"^2"
(answer has 1 sig fig due to 1 sig fig in 30000K)

Resources:
http://astroweb.case.edu/ssm/100f09/HW4_soln.pdf
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law