Answer:
The hotter star radiates 625 times more energy per second from each square meter of its surface
Explanation:
Temperature of the hotter star is 30000 K
temperature of the cooler star = 6000 K
From Stefan-Boltzmann radiation laws, for a non black body
P = εσA[tex]T^{4}[/tex]
where
P is the energy per second or power of radiation
ε is the emissivity of the body
σ is the Stefan-Boltzmann constant of proportionality
A is the area of the sun
T is the temperature of the sun
The sun can be approximated as a black body, and the equation reduces to
P = σA[tex]T^{4}[/tex]
For the hotter body,
P = σA([tex]30000^{4}[/tex]) = 8.1 x 10^17σA J/s
For the cooler body,
P = σA([tex]6000^{4}[/tex]) = 1.296 x 10^15σA J/s
comparing the two stars energy
==> (8.1 x 10^17)/(1.296 x 10^15) = 625
This means that the hotter star radiates 625 times more energy per second from each square meter of its surface
