According to the Bekenstein–Hawking formula in black hole thermodynamics, $$S=\frac{kAc^3}{4G\hbar}$$ Consider a Schwarzschild black hole, we know that $$r_{\rm g}=\frac{2GM}{c^2},\ A=4\pi r_{\rm g}^2$$ Therefore, we can get $$M^2=\frac{\hbar c^3}{4\pi k_{\rm B} G}S$$ From thermodynamics, we learn that ##{\rm d}M=T{\rm d}S##, so we have $$T=\frac{{\rm d} M}{{\rm d} S}=\frac{\hbar c^3}{8\pi k_{\rm B}G}\frac{1}{M}$$I believe this is what you want.James Way said:how the mass and temperature change over time and how to calculate them
Hawking Radiation is a theoretical concept proposed by physicist Stephen Hawking in 1974. It suggests that black holes emit radiation due to quantum effects near the event horizon, resulting in a gradual loss of mass over time.
The temperature of a black hole decreases as it emits Hawking Radiation, since the radiation carries away energy from the black hole. This is known as the Hawking temperature, and it is inversely proportional to the mass of the black hole.
Yes, according to Hawking's theory, black holes with a very small mass will eventually evaporate completely due to Hawking Radiation. However, the time scale for this process is incredibly long, on the order of 10^67 years for a black hole with the mass of the sun.
Hawking Radiation challenges the idea that black holes are completely "black" and do not emit any radiation. It also raises questions about the conservation of energy and information in the universe, as the radiation carries away energy and information from the black hole.
Currently, there is no direct experimental evidence for Hawking Radiation. However, indirect evidence has been observed through observations of black holes and their surroundings, as well as theoretical calculations that align with Hawking's predictions.