Black hole temperature derived from entropy (heat from black hole?)

  • Thread starter linda300
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  • #1
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Main Question or Discussion Point

Hey,

The entropy of a black hole is S = kB (4∏GM2)/(hbar c)

S=Q/T

T= Q/S

T = Q (hbar c)/ (4∏GM2kB)

The temperature derived from hawking radiation is:

T = c3 hbar/ (8 pi G M kB)

Which implies Q = (1/2)M c2

Is this true?

I have found online that the heat should equal to the mass-energy of the black hole,
Mc2

But it was not explained,

Is it correct that Q = (1/2)M c2?

Thanks
 

Answers and Replies

  • #2
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Entropy of an object is not Q/T. Heat capacity of an object is Q/T. Heat capacity stays about the same when an object is cooled or heated, but entropy changes when an object is cooled or heated.

If we extract a small amount of heat energy from an object, that heat energy has entropy Q/T, where T is the temperature that the object has during the extraction.

If we do many small extractions of energy from an object, and sum the entropy changes, then we get the entropy of the object.
 
  • #3
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Ah thanks!

So then it does make sense to use the mass-energy of the black hole then, by considering one particle escaping from the black hole at a time which have energy mc^2 then summing them together to get Mc^2.

But where would the factor of two come in? The factor of two that is required to get the same result as that produced using hawking radiation.

Thanks for your answer!
 
  • #4
1,399
106
Ah thanks!

So then it does make sense to use the mass-energy of the black hole then, by considering one particle escaping from the black hole at a time which have energy mc^2 then summing them together to get Mc^2.

But where would the factor of two come in? The factor of two that is required to get the same result as that produced using hawking radiation.

Thanks for your answer!

The formula for the entropy of an object would be an useful thing to have.

Temperature T rises linearly when we extract heat from a black hole, and temperature falls NOT linearly when we extract heat from an ordinary object.

So maybe:
for a black hole: S=(1/2)Q/T
for an ordinary object: S= something complicated



ADDITION:
The energy coming out from a massive (lot of heat) black hole is cool heat, containing lot of entropy, according to S=Q/T.
When most heat has evaporated, then the heat coming out is at high temperature, and has low entropy, according to S=Q/T.
In the formula S=(1/2)Q/T
S is the entropy of all the heat that the black hole can produce.
T is the temperature of the coolest heat that the the black hole can produce.
Q is the the heat of the black hole when not any heat has escaped yet.
 
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