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

May 23, 2015

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

$E = \sigma {T}^{4}$,

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

$T$ = temperature in Kelvins

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

Given/Known:
$T = \text{30000 K}$
area = $\text{1 m"^2}$
$\sigma = \left(5.670373 \times {10}^{- 8} \text{watt")/(1"m"^2xx"K"^4}\right)$
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)