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This question already has an answer here:

I understand that a black hole happens after a star exhaustions elements capable of undergoing fusion. Without the energy produced by fusion the remaining mass of the star collapses on itself and forms a black hole. I conclude black holes must be deficient in hydrogen, helium and elements that undergo fusion in stars. I understand that hydrogen and the like undergoes fusion in stars in part (or completely?) because of the gravitational pressure which occurs in stars.

Suppose a young star replete with hydrogen merged with a black hole. My question: would the result be a larger black hole, or a radiant star with a very massive heart? Could a massive infusion of (fusible?) elements reboot the black hole back into a star?

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marked as duplicate by James K, Hohmannfan, uhoh, called2voyage Mar 21 '17 at 13:34

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  • $\begingroup$ Massive stars don't really finish hydrogen burning. Instead, they undergo shell burning; hydrogen fusion just happens in layers at greater radii. $\endgroup$ – HDE 226868 Mar 19 '17 at 17:06
  • $\begingroup$ Just to add to @JamesK's answer below, which I think is spot on. If a young star "merged" or crashed into a black hole the result would be a very very hot accretion disk and a mini quasar. Think of trying to squeeze a million mile across ball of super-hot hydrogen down a spherical hole about 20 miles across. It would be one crazy spectacle. Once inside the event horizon, nothing escapes, but the process of the hydrogen star being pulled inside, would be a visual display millions of times brighter than the star. $\endgroup$ – userLTK Mar 19 '17 at 17:26
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No.

A black hole is characterised by the formation of an event horizon. Matter which travels past an event horizon cannot return to the space outside the event horizon, to do so would require the matter to travel faster than light, and so would require infinite energy.

There is nothing like a regular star "inside" a black hole. The general relativity model of a black hole contains a singularity. Matter which enters a black hole is certain to reach this singularity in a (usually short) amount of time. Adding hydrogen to a black hole would just make the black hole more massive. There is a theorem in mathematical physics that the only properties that a black hole can have are mass, angular momentum and charge. In other words, a black hole made of collapsed hydrogen would be identical to one made of collapsed iron. There is no matter in a black hole, only mass.

The formation of a black hole normally occurs when a core in a massive star has a certain amount of iron, which saps the core of energy causing collapse. If the resulting object has more than about 3 solar masses, then nothing in the universe can stop it from collapsing all the way to a singularity.

As with much about general relativity, your intuition is a very bad guide to how black holes work. However General relativity is a very good model for how black holes should form and behave

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  • $\begingroup$ I've wondered whether ultimately a convention of black holes exceeding a critical mass could be the beginning of something new. Pure speculation, but entertaining ;-) $\endgroup$ – laune Mar 19 '17 at 16:37
  • $\begingroup$ @laune Unlikely. We would have to have seen an upper limit on black hole masses for that to be true. But we routinely find black holes with masses of several million suns. $\endgroup$ – probably_someone Mar 19 '17 at 18:19
  • $\begingroup$ -1 (if I had the reputation). This answer is based purely on classical physics which we know to be an incomplete description of black holes. A more elegant answer is provided in the duplicate link above: the entropy of a black hole far exceeds that of a star. $\endgroup$ – lemon Mar 19 '17 at 21:01
  • $\begingroup$ The best model we have of a black hole is the classical (non-quantum) model of General Relativity. We know it is incomplete; all models are incomplete. We use the best model for the task. If you have a better model of gravitation I (and the Nobel committee) would like to know. (I do agree with the duplicate) $\endgroup$ – James K Mar 20 '17 at 21:30

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