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Found in: Page 137

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# Imagine that your dog has eaten the portion of Table 4.1 that gives entropy data; only the enthalpy data remains. Explain how you could reconstruct the missing portion of the table. Use your method to explicitly check a few of the entries for consistency. How much of Table 4.2 could you reconstruct if it were missing? Explain.

We can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.

See the step by step solution

Given table 4.1

And Table 4.2

## Step2: Explanation

First express the change in entropy for steam at zero temperature
$\Delta {S}_{1}=\frac{Q}{T}......\left(1\right)$Where, Q is the heat change (absorbed or lost) and T is the temperature.

We know that the enthalpy is described as the energy that is required to produce a substance at fixed pressure.Now write an expression of the change in entropy for water at given temperature$\Delta {S}_{2}=\frac{\Delta H}{T}......\left(2\right)$Where, $\Delta H$ - change in enthalpy

Substitute Q = 2501 kJ/kg and T =273 K in equation (1), we get
$\Delta {S}_{1}=\frac{\left(2501\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{\left(273\mathrm{K}\right)}\phantom{\rule{0ex}{0ex}}=9.156\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}$This is the same value as in Table

Now, substitute $\Delta H=42\mathrm{kJ}·{\mathrm{kg}}^{-1}$ and T=278 K in equation (2), we get

$\Delta {S}_{2}=\frac{\left(42\mathrm{kJ}.{\mathrm{kg}}^{-1}\right)}{\left(278\mathrm{K}\right)}\phantom{\rule{0ex}{0ex}}=0.151\mathrm{kJ}·{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1}$

This value is the same as in given table

Now write expression for the change in entropy for steam at given temperature
$\Delta S=\frac{\Delta H}{T}-nR\frac{\Delta P}{P}......\left(3\right)$Where, n is number of moles, R is gas constant, P is average pressure and $\Delta P$ is change in pressure.

Substitute $\Delta H=19\mathrm{kJ}·{\mathrm{kg}}^{-1}$ , T=278K, n= 55.55, R= 8.314 J/mole K,$\Delta P$ = 0.006 and P=0.009 bar in expression (3), we get

$\Delta S=\frac{\left(19\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{\left(278\mathrm{K}\right)}-\left(55.55\right)\left(8.314\mathrm{J}/\mathrm{mole}·\mathrm{K}\right)\frac{\left(0.006\mathrm{bar}\right)}{\left(0.009\mathrm{bar}\right)}\phantom{\rule{0ex}{0ex}}=-0.239\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}$

The entropy for steam at finite temperature is:

$\Delta {S}_{1}+\Delta S=\left(9.156\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right)+\left(-0.239\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right)\phantom{\rule{0ex}{0ex}}=8.917\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}$

This value is very close to the value in the table.

So we can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.