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A supernova's Ni/Fe core remnant with mass exceeding about three solar masses would result in a black hole. Although the core's temperature was in the ~1E10K range, the black hole temperature based on predicted Hawking radiation would be very near absolute zero, well below CMB temperature. It seems implausible that a very high temperature supernova remnant can instantly evolve to a near-zero K temperature black hole. How is this consistent with conservation of energy?

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As others have pointed out, there's not a problem with conservation of energy. The black hole still has plenty of energy, even though it's at a low temperature, and the ejecta carry off quite a bit of energy. The total accounting can work out just fine. I think the bigger puzzle here is not with the first law of thermodynamics (energy conservation) but rather with the second law. When a system at a high temperature suddenly turns into a system at a low temperature, it seems like the entropy would drop. It doesn't, but understanding why is quite a puzzle. – Ted Bunn Apr 26 '11 at 15:55
Ted: Are you saying there's no explaination for the second law issue, or that its complicated? – Michael Luciuk Apr 26 '11 at 16:54
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The naïve answer is that self-gravitating systems tend to have a negative specific heat--as they have energy added to them, they cool. This is pretty atypical of most other thermoydyamic systems, which tend to heat as they get more energy. This winds up being true even for a N-particle system obeying Newtonian mechanics. – Jerry Schirmer Apr 26 '11 at 19:29
I was saying that it's complicated, not that it's unexplained. – Ted Bunn Apr 26 '11 at 20:57

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When the matter of the Ni/Fe core collapses, it releases huge amounts of energy in the form of radiation as well as high energy jets (ejected from the rotational poles of the black hole), so a lot of the mass + energy the black hole is trying to engulf (and keep for itself) is actually escaping before being trapped in the black hole.

Also, do not be confused by the fact that the celestial body goes from ca 1E10K range to near 0K in it's black body radiation. Sure, in that respect it becomes less energetic (by far) but it will retain a lot of mass (which is a lot of energy, right there) and it will also be rotating (I won't speculate how much that energy, the one stored in the celestial body's angular momentum changes when it goes from Ni/Fe core to black hole though).

Then of course, if Hawkings radiation really do exist and enough time elapses, all the energy trapped in the black hole will eventually seep out of it as it evaporates, but yes, it would take absurd amount of time (even by cosmological standards) for that to happen.

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IllviJa: Good points. However, the idea that BH temperatures can fall below CMB temps is difficult to believe. – Michael Luciuk Apr 26 '11 at 14:51
IllviJa: Isn't angular momentum concerved in the transition from core remnant to BH? – Michael Luciuk Apr 26 '11 at 23:28
Well, part of the rotating core remnant gets ejected as part of the collapse to a black hole, so I'm uncertain about what happens with the angular momentum as part of that. Another thing worth mentioning is the fact that the supernova remnant collapses into a black hole at the same time as the star as a whole is blown apart as a supernova, so the process where the major part of the star is ejected is interacting with the process that causes part of the star to collapse into a black hole. That makes it more complicated to determine what energy is transferred where during the nova. – IllvilJa Apr 27 '11 at 19:27
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I had to think about this a while before I came up with what I think is a good intuitive picture. Imagine a neutron star emitting radiation with whatever its characteristic energy distribution is. Now, slowly dial up the gravitational constant and watch what happens.

We know that, when it gets high enough, the neutron star will collapse into a black hole. But, what happens in the intermediate time? Well, as G gets larger, the neutron star's radiation will be red-shifted to lower and lower energies as more of the energy is needed to escape the increasing gravitational well. Whatever the lower limit of the detector is, more and more radiation will fall below it, gradually making the star disappear. When the star becomes a black hole, then, we see no radiation coming from it at all (putting aside external sources) except for Hawking radiation, which is of extremely low energy.

I can't say how well this picture translates to actual black hole creation, but it gives an intuitive picture of how the core temperature appears to drop, from an outside perspective, without an instantaneous transition.

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