Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

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 $ε / k$ , and then in kelvins assuming that $ε = 0.1 e V$ .

For $q_{A} = 1,$

$T_{A} = 0.187 ε / k = 216.8 K T_{B} = 0.91 ε / k = 1055.07 K$

For $q_{A} = 60,$

localid="1646916582956" $T_{A} = 0.57 ε / k = 660.87 K T_{B} = 0.57 ε / k = 660.87 K$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Table of microstates, multiplicities, and entropies:

Both the solids, solid A and solid B, are sharing $100 u n i t s$ 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" $ε = 0.1 e V = 0.1 \times 1.6 \times 10^{- 19} = 1.6 \times 10^{- 20} J$

Step 2: Calculation of temperatures when qA=1

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

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

Where,

$\partial U$ can be replaced as role="math" localid="1646893107696" $ε d q$

Hence, the simplified equation can be written as,

role="math" localid="1646893344744" $T = (\frac{ε d q}{d S}) . . . . . . . . . . (2)$

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 $d S$ for the values of $q_{A} = 0, q_{A} = 2$ are $0, 10.7 k$ respectively.

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

role="math" localid="1646895028376" $T_{A} = (\frac{ε (2 - 0)}{S_{A} (2) - S_{A} (0)}) T_{A} = (\frac{ε (2 - 0)}{10.7 k - 0}) T_{A} = 0.187 \frac{ε}{k}$

We know that,

$k = 1.38 \times 10^{- 23} J / K$

By substituting the values, we get,

$T_{A} = \frac{0.187 \times 1.6 \times 10^{- 20}}{1.38 \times 10^{- 23}} T_{A} = 216.8 K$

Similarly, temperature of solid B can be calculated as:

$T_{B} = \frac{ε (100 - 98)}{S_{B} (100) - S_{B} (98)} T_{B} = (\frac{ε (100 - 98)}{(187.53 - 185.33) k}) T_{B} = 0.91 \frac{ε}{k}$

By substituting the values, we get,

$T_{B} = \frac{0.91 \times 1.6 \times 10^{- 20}}{1.38 \times 10^{- 23}} T_{B} = 1055.07 K$

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:

$T_{A} = (\frac{ε (61 - 59)}{S_{A} (61) - S_{A} (59)})$

Now, from the table, the corresponding values of $S_{A} (61), S_{A} (59)$ are $160.9, 157.4$ respectivley.

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

$T_{A} = (\frac{ε (61 - 59)}{(160.9 - 157.4) k}) T_{A} = = 0.57 \frac{ε}{k}$

Now,

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

$T_{A} = 0.57 \times \frac{1.6 \times 10^{- 20}}{1.38 \times 10^{- 23}} T_{A} = 660.87 K$

Similarly, temperature of solid B can be calculated as:

$T_{B} = \frac{ε (41 - 39)}{S_{B} (41) - S_{B} (39)} T_{B} = (\frac{ε (41 - 39)}{(107 - 103.5) k}) T_{B} = 0.57 \frac{ε}{k}$

By substituting the values, we get,

role="math" localid="1646916541515" $T_{B} = 0.57 \times \frac{1.6 \times 10^{- 20}}{1.38 \times 10^{- 23}} T_{B} = 660.87 K$

Step 4: Final answer

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

For $q_{A} = 1,$

$T_{A} = 0.187 ε / k = 216.8 K T_{B} = 0.91 ε / k = 1055.07 K$

For $q_{A} = 60,$

$T_{A} = 0.57 ε / k = 660.87 K T_{B} = 0.57 ε / k = 660.87 K$

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An Introduction to Thermal Physics

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Short Answer

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 $ε / k$ , and then in kelvins assuming that $ε = 0.1 e V$ .

Step by Step Solution

Step 1: Given

Step 2: Calculation of temperatures when qA=1

Step 2: Calculation of temperatures when qA=60

Step 4: Final answer

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Use Table 3.1 to compute the temperatures of solid A and solid B when qA=1. Then compute both temperatures when qA=60. Express your answers in terms of ε/k, and then in kelvins assuming that ε=0.1 eV.

Step by Step Solution

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Step 2: Calculation of temperatures when qA=60

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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 $ε / k$ , and then in kelvins assuming that $ε = 0.1 e V$ .