One star has a temperature of 30,000 K and another star has a temperature of 6,000 K. Compared to the cooler star, how much more energy per second will the hotter star radiate from each square meter of its surface?

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

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

For the hotter body,

P = σA() = 8.1 x 10^17σA  J/s

For the cooler body,

P = σA() = 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

v=ir = 3x15 =45 v