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

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$T^{4}$

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