How to calculate the Sun’s energy output? What formula can be used to find the total energy being produced from the Sun in Joules per second?
If I recall correctly, the energy is around $2.3012 {\cdot} {10}^{27}\,\frac{\mathrm{J}}{\mathrm{s}}$.
 A: The Stefan-Boltzmann law states that the luminosity of an ideal blackbody is
$$ L = 4\pi\sigma r_*^2 T^4, $$
where in $r_*$ is the stellar radius, $T$ is the effective surface temperature, and $\sigma = 5.7 \times 10^{-5}$ $\rm erg\,cm^{-2} s^{-1} K^{-4}$ in CGS units. However, we cannot directly measure the luminosity of a star. Instead, we measure its flux
$$ f_{\rm total} = \frac{L}{4\pi R^2}, $$
where $R$ is the distance between the star and us, the observer.

See the above blackbody spectrum of the sun (credit to Ocean Optics Web Book). Note the units of its flux (the y-axis): watts per meter-squared per nanometer. This is a measurement of flux per unit wavelength $f_\lambda$. Thus, to get the total flux $f_{\rm total}$, as given in the equation above, we have to compute the total area under this spectrum, i.e.
$$ f_{\rm total} = \int_{0}^{\infty} f_\lambda(\lambda) d\lambda. $$
Typically this is done using a numerical analysis tool, such as Python/SciPy or IDL. As mentioned previously, this value allows observers to know the luminosity of the star. Therefore, it can also given them the radius or surface temperature if they happen to the other. The luminosity also has relationships mass depending on the phase of stellar evolution ($L \propto M^{3.5}$ on average).
A: It is pretty straightforward.  Assume the sun is a black body (that doesn't mean black colored; it just means that it radiates according to the black body law).  The total energy loss per second is the surface temperature raised to the fourth power, times the surface area, times a constant.  Look up the Stefan-Boltzmann law in Wikipedia; it provides a value for the constant.  We assume the sun loses energy as fast as it produces energy.  Just for fun, look up the sun's mass and then compare the rate of heat loss (= energy production) per unit mass for the sun and for a human being (by using the Stefan-Boltzman law on a typical human body).  You will find that we humans produce significantly more power per gram than the sun does.
A: Here is an alternative approach which relies on being able to measure two things, but doesn't rely on the Sun being a black-body or knowing its surface temperature or anything like that.


*

*the top-of-atmosphere flux at the radius of the Earth's orbit, which is about $F = 1360\,\mathrm{W/m^2}$;

*the radius of the Earth's orbit, which is about $R = 1.5\times 10^{11}\,\mathrm{m}$.


Both of these quantities are rather well-known: $F$ is important, for instance, to understanding the climate, and is measured by satellites, $R$ is important to very many things and I'm not sure how it's measured, but it is well-known.
So, now you need to simply calculate the total power passing through a sphere whose radius is the same of the Earth's orbit, and this is
$$
\begin{align}
 P &= 4 \pi R^2 F\\
   &\approx 3.8\times 10^{26}\,\mathrm{W}\\
   &= 3.8\times 10^{26}\,\mathrm{J/s}
\end{align}$$
Since energy is conserved this number is the amount of power passing through any surface surrounding the Sun, including the surface of the Sun itself: it is the power radiated by the Sun.
The number here is correct to within one percent, I think: using better figures for $F$ and $R$, and perhaps dealing with the eccentricity of the Earth's orbit would give something even better.
