Answer:
3.82746e+26 watts
Explanation:
There are two ways to solve this problem. One way is to use the equation
L = 4πσR²T⁴
where
L = the sun's bolometric (all-spectrum) luminous power
σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant
R = 6.957e+8 meters = the sun's radius
T = 5771.8 K = the sun's effective temperature
You find that
L = 3.82746e+26 watts
The other way to solve the problem is to use the Planck integral for radiant flux.
L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv
where
h = 6.62607015e-34 J sec
c = 299792458 m sec⁻¹
k = 1.380649e-23 J K⁻¹
v₁ = 0 = frequency band lower bound, in Hz
v₂ = ∞ = frequency band upper bound, in Hz
You find, once again, that
L = 3.82746e+26 watts
The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use
v₁ = 4.1e+14 = frequency band lower bound, in Hz
v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz
restricting ourselves to the visible spectrum. We find that
L (visible) = 1.56799e+26 watts
So the fraction of the sun's luminosity that is in the visible spectrum is
L (visible) / L = 0.4096686
