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

linda300
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

jartsa
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.

linda300
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.

jartsa
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.

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