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?

Question

Quizplus · Accepted Answer

The energy radiated per second per unit area of a star's surface is given by the Stefan-Boltzmann law, E = σT⁴, where σ is the Stefan-Boltzmann constant and T is the temperature in Kelvin. Using this formula, we can calculate the energy radiated by each square meter of surface area for each star.

For the hotter star with T = 30,000 K, 
E = σT⁴ = (5.67 × 10⁻⁸ W/m²K⁴) × (30,000 K)⁴
E = 7.21 × 10¹⁰ W/m²

For the cooler star with T = 6,000 K, 
E = σT⁴ = (5.67 × 10⁻⁸ W/m²K⁴) × (6,000 K)⁴
E = 5.76 × 10⁶ W/m²

The ratio of the energy radiated by the hotter star to that radiated by the cooler star is:
7.21 × 10¹⁰ W/m² ÷ 5.76 × 10⁶ W/m² = 1.25 × 10⁴

This means that the hotter star radiates 625 times more energy per second per unit area than the cooler star. Therefore, the answer is D) 625 times.

One Star Has a Temperature of 30,000 K and Another