Calculation of the temperature of the Sun's core? Why is this value so low? From the Stefan-Boltzmann Law ($j^*=\sigma T^4$), the luminosity of the Sun is $L_{sun}=j^*A=[\sigma T^4][4\pi R_{sun}^4]=3.85\times 10^{26}$ W.
If I assume the Sun to be in a steady state then the energy exiting the core (I will assume $R_{core}\sim 0.25R_{sun}$) should be equal to the energy the sun radiates away into space at the surface.
If I set the the power that is exiting the Sun's core (heading into the radiative zone) equal to the Sun's luminosity and apply the $j^*=\sigma T^4$ and $L=j^*A$, I find the temperature at the boundary between the core and radiative zone is 11558 K (much lower than the 7,000,000 K value quoted online).
I realise a more accurate picture would be to solve the equations of hydrostatic equilibrium, but I would think that if all the Sun's energy is created in the core, and the Sun operates at a steady state, than energy conservation would require that the energy exiting the core should equal the luminosity of the sun.
Is there something wrong in my reasoning about energy conservation, or is/are there ways that the sun emit energy formed in the core other than by blackbody radiation? Am I wrong to assume the Sun's core emits as a black body (if this is the case, I don't think a non-zero emissivity would matter much since I'm off by almost 3 orders of magnitude)?
 A: 
If I assume the Sun to be in a steady state then the energy exiting
  the core (I will assume Rcore∼0.25Rsun) should be equal to the energy
  the sun radiates away into space at the surface.

That's where your problem is: The energy flux existing the surface of the sun is only $~10^{-12}$ of the flux exiting the core. This is because the interior of the sun is not a naked Stefan-Boltzmann-surface, but radiation bounces back and forth through diffusion, and is thus mostly trapped. The tiny fraction that escapes in the end is what we see as surface luminosity.
If you want the core photon fluxes, I'm afraid you'll need a structure model for that. Or go full hydrostatic (assuming all fluxes are in balance) and use a polytrope. Once you have the structure, you can play the Stefan-Boltzmann game again and compare the fluxes exiting the core vs. exiting the surface.
A: Your arguments are all nearly 100% spot on, but you are ignoring the fact that the outer shell radiates nearly equal amount of energy back towards the core.  A bit more reasoning will convince you that if there were no temperature gradient, no energy could possibly escape at all. If you are a physicist with ambition to become as good as Fermi, or Feynman, try to make a quick estimate and then calculate more precisely later. HINT: to first approximation it will depend on the ratio of the optical thickness of the sun to the radius.
