Use the result of Problem 2.42 to calculate the temperature of a black hole, in terms of its mass $M$. (The energy is $M{c}^2$. ) Evaluate the resulting expression for a one-solar-mass black hole. Also sketch the entropy as a function of energy, and discuss the implications of the shape of the graph.

The expression for the entropy of a black hole is given as:

$S=\frac{8{\pi }^{2}G{M}^2}{hc}k..........\left(1\right)$

Where,

$G$is the gravitational constant, $M$is mass, $h$is Planck's constant, $c$is the speed of light, and $k$is the Boltzmann's constant

The energy of the black hole is given by Einstein's relation as:

$U=M{c}^2..........\left(2\right)$

Mathematically, temperature can be defined as:

$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)..........\left(3\right)$

Where,

$\partial S$is the change in entropy and $\partial U$is the change in the internal energy of the body.

Equation (1) can be modified as:

role="math" localid="1646995322714" $$\begin{gathered} S=\frac{8{\pi }^{2}G{M}^2}{hc}k\times \frac{c^4}{c^4} \\ S=\frac{8{\pi }^{2}G{\left(M{c}^2\right)}^2}{h{c}^5}k \end{gathered}$$

By replacing $M{c}^2$as $U$, we get,

role="math" localid="1646997428892" $S=\frac{8{\pi }^{2}G{U}^2}{h{c}^5}k..........\left(4\right)$

Now, by substituting this value of $S$in equation (3), we get,

$$\begin{gathered} \frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)=\frac{\partial }{\partial U}\left[\frac{8{\pi }^{2}G{U}^2}{h{c}^5}k\right] \\ \frac{1}{T}=\frac{16{\pi }^{2}GU}{h{c}^5}k \\ T=\frac{h{c}^5}{16{\pi }^{2}GUk} \end{gathered}$$

By resusbstituting the value of $U$in the above equation, we get the desired result in terms of mass,

$T=\frac{h{c}^3}{16{\pi }^{2}GM{c}^{2}k}$

For a solar mass black hole, $M=2\times {10}^{30}\mathrm{kg}$.

Also, by substituting $6.67\times {10}^{-11}{\mathrm{m}}^3{\mathrm{kg}}^{-1}{\mathrm{s}}^{-2}$for $G$, $6.62\times {10}^{-34}\mathrm{J}.\mathrm{s}$for $h$, $1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$for $k$and $3\times {10}^8{\mathrm{ms}}^{-1}$for $c$in the above equation, we get,

$$\begin{gathered} T=\frac{\left(6.62\times {10}^{-34}\right){\left(3\times {10}^8\right)}^3}{16{\pi }^2\left(6.67\times {10}^{-11}\right)\left(2\times {10}^{30}\right)\left(1.38\times {10}^{-23}\right)} \\ T=6.15\times {10}^{-8}\mathrm{K} \end{gathered}$$

Consider the equation (4),

$S=\frac{8{\pi }^{2}G{U}^2}{h{c}^5}k$

Here,

$G,K,h,c$are all constants

Hence, it can be modified as:

$S\propto U^2$

Therefore, the graph of entropy as a function of energy can be sketched as follows:

It can be observed that the graph is a concave up graph. Objects exhibiting such behavior would have a negative heat capacity.

The required expression is $T=\frac{h{c}^3}{16{\pi }^{2}GM{c}^{2}k}$and the temperature can be calculated to be $6.15\times {10}^{-8}K$. Also the graph of entropy as a function of energy which is a concave up graph, can be sketched as follows: