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Answers without the blur. Sign up and see all textbooks for free! Q. 3.1

Expert-verified Found in: Page 89 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Use Table 3.1 to compute the temperatures of solid A and solid B when ${q}_{A}=1$. Then compute both temperatures when ${q}_{A}=60$. Express your answers in terms of $\epsilon /k$, and then in kelvins assuming that $\epsilon =0.1eV$.

For ${q}_{A}=1,$

${T}_{A}=0.187\epsilon /k=216.8K\phantom{\rule{0ex}{0ex}}{T}_{B}=0.91\epsilon /k=1055.07K$

For ${q}_{A}=60,$

localid="1646916582956" ${T}_{A}=0.57\epsilon /k=660.87K\phantom{\rule{0ex}{0ex}}{T}_{B}=0.57\epsilon /k=660.87K$

See the step by step solution

## Step 1: Given

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" $\epsilon =0.1eV=0.1×1.6×{10}^{-19}=1.6×{10}^{-20}J$

## Step 2: Calculation of temperatures when qA=1

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

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

Where,

$\partial U$ can be replaced as role="math" localid="1646893107696" $\epsilon dq$

Hence, the simplified equation can be written as,

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

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" ${T}_{A}=\left(\frac{\epsilon \left(2-0\right)}{{S}_{A}\left(2\right)-{S}_{A}\left(0\right)}\right)\phantom{\rule{0ex}{0ex}}{T}_{A}=\left(\frac{\epsilon \left(2-0\right)}{10.7k-0}\right)\phantom{\rule{0ex}{0ex}}{T}_{A}=0.187\frac{\epsilon }{k}$

We know that,

$k=1.38×{10}^{-23}J/K$

By substituting the values, we get,

${T}_{A}=\frac{0.187×1.6×{10}^{-20}}{1.38×{10}^{-23}}\phantom{\rule{0ex}{0ex}}{T}_{A}=216.8K$

Similarly, temperature of solid B can be calculated as:

${T}_{B}=\frac{\epsilon \left(100-98\right)}{{S}_{B}\left(100\right)-{S}_{B}\left(98\right)}\phantom{\rule{0ex}{0ex}}{T}_{B}=\left(\frac{\epsilon \left(100-98\right)}{\left(187.53-185.33\right)k}\right)\phantom{\rule{0ex}{0ex}}{T}_{B}=0.91\frac{\epsilon }{k}$

By substituting the values, we get,

${T}_{B}=\frac{0.91×1.6×{10}^{-20}}{1.38×{10}^{-23}}\phantom{\rule{0ex}{0ex}}{T}_{B}=1055.07K$

## Step 2: Calculation of temperatures when qA=60

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:

${T}_{A}=\left(\frac{\epsilon \left(61-59\right)}{{S}_{A}\left(61\right)-{S}_{A}\left(59\right)}\right)$

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,

${T}_{A}=\left(\frac{\epsilon \left(61-59\right)}{\left(160.9-157.4\right)k}\right)\phantom{\rule{0ex}{0ex}}{T}_{A}==0.57\frac{\epsilon }{k}$

Now,

By substituting the values of $\epsilon ,k$ in the above equation, we get,

${T}_{A}=0.57×\frac{1.6×{10}^{-20}}{1.38×{10}^{-23}}\phantom{\rule{0ex}{0ex}}{T}_{A}=660.87\mathrm{K}$

Similarly, temperature of solid B can be calculated as:

${T}_{B}=\frac{\epsilon \left(41-39\right)}{{S}_{B}\left(41\right)-{S}_{B}\left(39\right)}\phantom{\rule{0ex}{0ex}}{T}_{B}=\left(\frac{\epsilon \left(41-39\right)}{\left(107-103.5\right)k}\right)\phantom{\rule{0ex}{0ex}}{T}_{B}=0.57\frac{\epsilon }{k}$

By substituting the values, we get,

role="math" localid="1646916541515" ${T}_{B}=0.57×\frac{1.6×{10}^{-20}}{1.38×{10}^{-23}}\phantom{\rule{0ex}{0ex}}{T}_{B}=660.87\mathrm{K}$

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

For ${q}_{A}=1,$

${T}_{A}=0.187\epsilon /k=216.8K\phantom{\rule{0ex}{0ex}}{T}_{B}=0.91\epsilon /k=1055.07K$

For ${q}_{A}=60,$

${T}_{A}=0.57\epsilon /k=660.87K\phantom{\rule{0ex}{0ex}}{T}_{B}=0.57\epsilon /k=660.87K$ ### Want to see more solutions like these? 