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

Question

Quizplus · Accepted Answer

The temperature of the star can be calculated using the Stefan-Boltzmann law: $P = 4\pi R^2 \sigma T^4$, where $P$ is the power (rate of radiation), $R$ is the radius of the star, $\sigma$ is the Stefan-Boltzmann constant, and $T$ is the temperature of the star. Rearranging for $T$, we get $T = \left(\frac{P}{4\pi R^2 \sigma}\right)^{1/4}$. Substituting the given values, $P = 5.32 \times 10^{26} \, \text{W}$, $R = 6.95 \times 10^{8} \, \text{m}$, and $\sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4$, we find $T$ to be approximately $6.27 \times 10^{3} \, \text{K}$.

The Radius of a Star Is 6