A star radiates at the rate of 1.5 X 10^24 W. It's radius is 10.0 X 10^8 m. Assuming that the star is a blackbody radiator, (e=1), what is the surface temperature?
Recall that
P = AεσT^4
So, plug in your numbers and you have
T = ∜(P/4πr^2σ) =∜((1.5*10^24)/(4π(10.0*10^8)^2*(5.670373*10^-8)) = ?°K
I'm so sorry
To determine the surface temperature of a star, we can use the Stefan-Boltzmann law, which relates the total power radiated by a blackbody to its surface area and temperature.
The Stefan-Boltzmann law is given by the equation:
P = σ * A * T^4
Where:
P is the total power radiated (in watts)
σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W m^-2 K^-4)
A is the surface area of the star (in square meters)
T is the temperature of the star (in Kelvins)
In this case, we are given the power radiated by the star (P = 1.5 x 10^24 W) and the radius of the star (r = 10.0 x 10^8 m). The surface area of a sphere is given by the equation:
A = 4πr^2
Substituting the values into the equation, we have:
P = σ * (4πr^2) * T^4
Solving for T, we get:
T = (P / (σ * (4πr^2)))^(1/4)
Now we can plug in the values and calculate:
T = (1.5 x 10^24 W / (5.67 x 10^-8 W m^-2 K^-4 * (4π * (10.0 x 10^8 m)^2)))^(1/4)
T ≈ 5,856 K
Therefore, the surface temperature of the star is approximately 5,856 Kelvins.