According to Stephen Hawking's theory of black holes, once a sufficient mass has been lost through evaporation, the escape velocity $3 \times 10^8$ meters per second so light is able to escape from the black hole and due to the massive energy it would be white hot. The rate of evaporation after that would be much greater as the mass is significantly less. But how long would it last? Given it's a perfectly round singularity and has a set mass we should be able to calculate it, but have we?
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2$\begingroup$ you got a -1 from somebody because the term "white hole" has a definite meaning in physics. Have a look. en.wikipedia.org/wiki/White_hole . I edited out the contradiction $\endgroup$– anna vCommented May 20, 2015 at 10:35
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$\begingroup$ The title still has "white hole" in a misleading way. $\endgroup$– MBNCommented May 20, 2015 at 13:45
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1$\begingroup$ @MBN must have been a glitch in my editing. corrected it $\endgroup$– anna vCommented Oct 1, 2019 at 6:23
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1 Answer
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Assuming no energy input, the lifetime of a black hole is related to its mass by:
$$ T = \frac{5120\pi G^2}{\hbar c^4}M^3 $$
There is a nice summary of the derivation of the lifetime on this web site.
I make the condition assuming no energy input because for large black holes the Hawking temperature is less than the temperature of the cosmic microwave background so these black holes are curently growing rather than evaporating.
answered May 20, 2015 at 9:56