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

Answer:

5.327

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

                        $\frac{W_{2}}{W_{1}} = \frac{79^{4}}{52^{4}}$

                        $\frac{W_{2}}{W_{1}} = 5.327$