# Temperature of a black hole

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.7.

$\displaystyle S=\frac{8\pi^{2}GM^{2}}{hc}k \ \ \ \ \ (1)$

Taking ${U=Mc^{2}}$ as the energy of a black hole, we can write this as

$\displaystyle S=\frac{8\pi^{2}Gk}{hc^{5}}U^{2} \ \ \ \ \ (2)$

The temperature is therefore

$\displaystyle T=\left(\frac{\partial S}{\partial U}\right)^{-1}=\frac{hc^{5}}{16\pi^{2}GkU}=\frac{hc^{3}}{16\pi^{2}GkM} \ \ \ \ \ (3)$

which agrees with our earlier result from general relativity.

For a solar mass black hole, this gives a value of

 $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\left(6.62\times10^{-34}\right)\left(3\times10^{8}\right)^{3}}{16\pi^{2}\left(6.67\times10^{-11}\right)\left(1.38\times10^{-23}\right)\left(2\times10^{30}\right)}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 6\times10^{-8}\mbox{ K} \ \ \ \ \ (5)$

Not only are black holes dark, they are also very cold.

The entropy-versus-energy curve 2 is a parabola so its slope ${\frac{\partial S}{\partial U}}$ increases as ${U}$ increases. As we’ve seen, this means that a black hole has negative heat capacity, and thus decreases in temperature as more energy (mass) is added. This is also clear from 3, since ${T\propto\frac{1}{M}}$.