Question

The laws of thermal radiation tell us that a star emits light with a total power per unit area that depends only on the star's surface temperature. the formula is power per unit area=σt4 where σ is a constant with a measured value of σ=5.7×10−8wm2×k4 (as usual, m stands for meters and k stands for kelvins.) you can then find the total power (or luminosity) emitted by the star by multiplying its total surface area by the power per unit area. use these facts to calculate the total power emitted into space by a star with a radius of 8.0×108m and a surface temperature of 6500 k. you will need the formula for the surface area of a sphere, which is surface area (sphere)=4πr2

Answer 1

Answer:$8.18\times10^{26}W$

Total Power emitted in space by the star is given by:

$P=\sigma T^4 A$

where total power emitted per unit area is given by $\sigma T^4$

where T is the temeprature in Kelvins and $\sigma=5.7\times10^{-8}Wm^2K^4$

A is the surface area of spherical star. $A =4\pi r^2$.

It is given that:

$T=6500 K\\ r=8.0\times10^8m$

Insert the values in the formula for total power emitted:

$P=5.7 \times10^{-8}\times 6500^4 \times 4\times 3.14 \times (8.0\times10^8)^2=8.18\times10^{26}W$