Question

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?

Answer 1

The hotter star radiates 625 times more energy per second from each square meter of its surface

Step-by-step 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
`T^(4)`

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
`T^(4)`

For the hotter body,

P = σA(
`30000^(4)`) = 8.1 x 10^17σA J/s

For the cooler body,

P = σA(
`6000^(4)`) = 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