91Ó°ÊÓ

Table of microstates, multiplicities, and entropies:

Both the solids, solid A and solid B, are sharing $100units$of energy.

Hence,

$q_A+q_B=100$

The temperature of both the solids is to be determined when $q_A=1$and $q_A=60$.

role="math" localid="1646894597879" $\mathrm{Î\mu }=0.1eV=0.1×1.6×{10}^{-19}=1.6×{10}^{-20}J$

Temperature can be defined in terms of entropy and energy of systems as:

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

Where,

$∂U$can be replaced as role="math" localid="1646893107696" $\mathrm{Î\mu }dq$

Hence, the simplified equation can be written as,

role="math" localid="1646893344744" $T=\left(\frac{\mathrm{Î\mu }dq}{dS}\right)..........\left(2\right)$

For $q_A=1$, slope between $q_A=0$and $q_A=2$are used.

Similarly for solid B, $q_B$will be $100-q_A$.

Hence,

$q_A=0,q_B=100-0=100$, and

$q_A=2,q_B=100-2=98$

Now, from the table, the corresponding values of $dS$for the values of $q_A=0,q_A=2$are $0,10.7k$respectively.

By substituting these values in equation 2, temperature of solid A can be calculated as:

role="math" localid="1646895028376" $$\begin{gathered} T_A=\left(\frac{\mathrm{Î\mu }(2-0)}{S_A\left(2\right)-S_A\left(0\right)}\right) \\ {T}_A=\left(\frac{\mathrm{Î\mu }(2-0)}{10.7k-0}\right) \\ {T}_A=0.187\frac{\mathrm{Î\mu }}{k} \end{gathered}$$

We know that,

$k=1.38×{10}^{-23}J/K$

By substituting the values, we get,

$$\begin{gathered} T_A=\frac{0.187×1.6×{10}^{-20}}{1.38×{10}^{-23}} \\ {T}_A=216.8K \end{gathered}$$

Similarly, temperature of solid B can be calculated as:

$$\begin{gathered} T_B=\frac{\mathrm{Î\mu }(100-98)}{S_B\left(100\right)-S_B\left(98\right)} \\ {T}_B=\left(\frac{\mathrm{Î\mu }(100-98)}{(187.53-185.33)k}\right) \\ {T}_B=0.91\frac{\mathrm{Î\mu }}{k} \end{gathered}$$

By substituting the values, we get,

$$\begin{gathered} T_B=\frac{0.91×1.6×{10}^{-20}}{1.38×{10}^{-23}} \\ {T}_B=1055.07K \end{gathered}$$

Now for $q_A=60$, the slope between $q_A=59$and $q_A=61$are used.

Similarly for solid B, $q_B$will be $100-q_A$.

In this case,

$q_A=59,q_B=100-59=41$, and

$q_A=61,q_B=100-61=39$

Temperature for solid A can be calculated as:

$T_A=\left(\frac{\mathrm{Î\mu }(61-59)}{S_A\left(61\right)-S_A\left(59\right)}\right)$

Now, from the table, the corresponding values of $S_A\left(61\right),S_A\left(59\right)$are $160.9,157.4$respectivley.

By substituting the values in the above equation, we get,

$$\begin{gathered} T_A=\left(\frac{\mathrm{Î\mu }(61-59)}{(160.9-157.4)k}\right) \\ {T}_A==0.57\frac{\mathrm{Î\mu }}{k} \end{gathered}$$

Now,

By substituting the values of $\mathrm{Î\mu },k$in the above equation, we get,

$$\begin{gathered} T_A=0.57×\frac{1.6×{10}^{-20}}{1.38×{10}^{-23}} \\ {T}_A=660.87\mathrm{K} \end{gathered}$$

Similarly, temperature of solid B can be calculated as:

$$\begin{gathered} T_B=\frac{\mathrm{Î\mu }(41-39)}{S_B\left(41\right)-S_B\left(39\right)} \\ {T}_B=\left(\frac{\mathrm{Î\mu }(41-39)}{(107-103.5)k}\right) \\ {T}_B=0.57\frac{\mathrm{Î\mu }}{k} \end{gathered}$$

By substituting the values, we get,

role="math" localid="1646916541515" $$\begin{gathered} T_B=0.57×\frac{1.6×{10}^{-20}}{1.38×{10}^{-23}} \\ {T}_B=660.87\mathrm{K} \end{gathered}$$

Hence, the temperature in both the cases for solids A and B is calculated as:

For$q_A=1,$

$$\begin{gathered} T_A=0.187\mathrm{Î\mu }/k=216.8K \\ {T}_B=0.91\mathrm{Î\mu }/k=1055.07K \end{gathered}$$

For$q_A=60,$

$$\begin{gathered} T_A=0.57\mathrm{Î\mu }/k=660.87K \\ {T}_B=0.57\mathrm{Î\mu }/k=660.87K \end{gathered}$$