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

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

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Answer 1 · 2015-05-23T16:53:41

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

$E = σ T^{4}$ $E=sigmaT^4$ ,

where $E$ $E$ = emitted power per square meter of surface

$T$ $T$ = temperature in Kelvins

$σ$ $sigma$ = Stefan-Boltzmann's constant: $\frac{5.670373 \times 10^{- 8} watt}{1 m^{2} \times K^{4}}$ $(5.670373xx10^(-8) "watt") /(1 "m"^2xxK^4)$

Given/Known:
$T = 30000 K$ $T="30000 K"$
area = ${1 m}^{2}$ $"1 m"^2"$
$σ = \frac{5.670373 \times 10^{- 8} watt}{1 m^{2} \times K^{4}}$ $sigma=(5.670373xx10^(-8) "watt")/(1"m"^2xx"K"^4")$
Unknown:
$E$ $E$

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

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