The amount of radiant power produced by the sun is approximately 3.9 x 1026 W. Assuming the sun to be a perfect blackbody (e = 1) sphere with a radius of 6.96 x 108 m, find its surface temperature (in kelvins).
here:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
NOW, that is the power PER UNIT AREA
so
3.9 *10^26 / (4 pi r^2) = power per unit area
= constant*T^4
as shown in Stefan/ B link
To find the surface temperature of the sun, we can use the Stefan-Boltzmann Law, which relates the radiant power emitted by a perfect blackbody to its surface temperature.
The Stefan-Boltzmann Law states that the total radiant power (P) emitted by a blackbody is directly proportional to the fourth power of its surface temperature (T) and is given by the equation:
P = σ * A * T^4
where σ is the Stefan-Boltzmann constant (σ ≈ 5.67 x 10^-8 W m^-2 K^-4), A is the surface area of the blackbody, and T is the surface temperature.
In this case, we are given the radiant power of the sun (P = 3.9 x 10^26 W) and its radius (r = 6.96 x 10^8 m). The surface area (A) of the sphere can be calculated using the formula:
A = 4πr^2
Substituting the given values, we can solve for T:
3.9 x 10^26 = σ * (4π * (6.96 x 10^8)^2) * T^4
Now, let's solve for T:
First, calculate the surface area A:
A = 4π * (6.96 x 10^8)^2
A ≈ 6.085 x 10^18 m^2
Substitute the values into the equation:
3.9 x 10^26 = (5.67 x 10^-8) * (6.085 x 10^18) * T^4
Divide both sides by (5.67 x 10^-8) * (6.085 x 10^18):
T^4 ≈ (3.9 x 10^26) / [(5.67 x 10^-8) * (6.085 x 10^18)]
T^4 ≈ 1.1524 x 10^15
Now, take the fourth root of both sides to isolate T:
T ≈ (1.1524 x 10^15)^(1/4)
T ≈ 5778 K (rounded to the nearest kelvin)
Therefore, the surface temperature of the sun is approximately 5778 kelvins.