Question

our sun emits a radiation spectrum similar to a black body at temperature 5 8 0 0 5800k. what would be the temperature of a star that emits twice as much total power in the form of black-body radiation, as compared to our sun?

Answer 1

The temperature of a star that emits twice as much total power as our Sun, which has a temperature of 5800 K, can be calculated using the Stefan-Boltzmann law. The new star's temperature comes out to be approximately 6895 K.

Step-by-step explanation:

The temperature of a star that emits twice as much total power as the Sun, given that the Sun is at a temperature of 5800 K, can be found using the Stefan-Boltzmann law, which states that the energy flux (Φ) is proportional to the fourth power of the temperature (T).

The Stefan-Boltzmann law is represented by the formula Φ = σT4, where σ is the Stefan-Boltzmann constant (5.67 × 10−8 W/m2K4). To emit twice the power, the new star's temperature must be such that (2)Φ = σT4, which means that T4 = (2)Φ/σ. If we take the fourth root of both sides, we get the new temperature in terms of the Sun's temperature: T = (21/4) × Sun's temperature. Plugging in 5800 K for the Sun's temperature, we find T ≈ 1.189 × 5800 K, which yields a temperature of approximately 6895 K for the new star.