Question

87) The radius of a star is 6.95 × 108 m, and its rate of radiation has been measured to be 5.32 × 1026 W. Assuming that it is a perfect emitter, what is the temperature of the surface of this star? (σ = 5.67 × 10-8 W/m2 ∙ K4)

A) 6.27 × 103 K
B) 8.25 × 103 K
C) 8.87 × 103 K
D) 3.93 × 107 K
E) 5.78 × 107 K

Answer 1

Using the Stefan-Boltzmann law and the given power radiated and radius of a star, we calculate the surface temperature to be approximately 6.27 × 10^3 K, which is option A.

Step-by-step explanation:

To find the temperature of the surface of the star, we can use the Stefan-Boltzmann law, which states that the power radiated per unit area of a black body is proportional to the fourth power of its temperature.

The formula for this law is P = σAT^4, where P is the power radiated, σ is the Stefan-Boltzmann constant, A is the surface area, and T is the temperature in Kelvin.

In this case, we are given the total power (P) radiated by the star as 5.32 × 10^26 W and the radius (r) of the star as 6.95 × 10^8 m. We can calculate the surface area (A) of the star using the formula for the surface area of a sphere, A = 4πr^2.

Substituting the known values into the Stefan-Boltzmann formula and solving for T, we get:

A = 4π(6.95 × 10^8 m)^2

T = √[4(5.32 × 10^26 W) / (4π(6.95 × 10^8 m)^2 × 5.67 × 10^-8 W/m^2 · K^4)]

After calculating T, we find that the closest value that fits one of the answer choices is 6.27 × 10^3 K, which corresponds to option A.