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

1 Answer
Meave60
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

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