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

In summary, the conversation discusses the topic of black hole entropy and the relationship between heat, temperature, and entropy. It is mentioned that the entropy of an object is not equal to Q/T, but rather Q/T is the heat capacity. The formula for entropy is also discussed, with different equations for a black hole and an ordinary object. The conversation also touches on the factor of two in the formula and the concept of heat evaporating from a black hole.
  • #1
linda300
61
3
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
 
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  • #2
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
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
linda300 said:
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.
 
Last edited:
  • #5
for your question. Yes, it is correct that the heat (Q) of a black hole is equal to half of its mass-energy (Mc^2). This is a result of the famous equation E=mc^2, which states that energy (E) is equal to the mass (m) multiplied by the speed of light (c) squared. In the case of a black hole, this equation can be applied to the mass-energy of the black hole itself, resulting in Q = (1/2)Mc^2.

This relationship between heat and mass-energy is a fundamental concept in thermodynamics and is known as the first law of thermodynamics. It states that energy cannot be created or destroyed, only transformed from one form to another. In the case of a black hole, the heat (Q) is the energy being radiated away through Hawking radiation, while the mass-energy (Mc^2) is the total energy contained within the black hole.

I hope this explanation helps clarify the relationship between heat and mass-energy in a black hole. Please let me know if you have any further questions.
 

What is a black hole?

A black hole is a region in space where the gravitational pull is so strong that it prevents even light from escaping.

How is the temperature of a black hole derived from entropy?

The temperature of a black hole is derived from its entropy through a mathematical relationship known as the Bekenstein-Hawking formula. This formula relates the entropy of a black hole to its surface area, and from there, the temperature can be calculated.

Why does a black hole have a temperature?

Although black holes are known for absorbing all matter and energy, they also emit a type of radiation called Hawking radiation. This radiation is a result of quantum effects near the event horizon of the black hole and is what gives the black hole a temperature.

Is the temperature of a black hole constant?

No, the temperature of a black hole is not constant. It decreases as the black hole grows in mass, and it also varies depending on the size and type of black hole.

Can we measure the temperature of a black hole?

Yes, the temperature of a black hole can be indirectly measured through observing the Hawking radiation it emits. However, this is a challenging task as the radiation is very faint and difficult to detect with current technology.

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