Question

If one star has a temperature of 5200 K and another star has a temperature of 7900 K, how much more energy per second will the hotter star radiate from each square meter of its surface?

Answer 1

5.327

Step-by-step explanation:

Stefan-Boltzmann law states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.

W = σT⁴

Where,

W is the total radiant heat power emitted from a surface

σ is constant of proportionality, called the Stefan–Boltzmann constant = 5.67 × 10⁻⁸ Wm⁻²K⁻⁴

T is absolute temperature in kelvin

For the first star, T = 5200 K

∴ W₁ = σ(5200)⁴

For the second star, T = 7900 K

∴ W₂ = σ(7900)⁴

The amount of energy radiated by the hotter star W₂, with respect to the other star W₁ is,

W₂ / W₁ = σ(7900)⁴ / σ(5200)⁴

`(W_(2))/(W_(1)) = (79^(4))/(52^(4))`

`(W_(2))/(W_(1)) = 5.327`