91Ó°ÊÓ

Given table 4.1

And Table 4.2

First express the change in entropy for steam at zero temperature
$\mathrm{Δ}{S}_1=\frac{Q}{T}......\left(1\right)$
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
$\mathrm{Δ}{S}_2=\frac{\mathrm{Δ}H}{T}......\left(2\right)$
Where, $\mathrm{Δ}H$- change in enthalpy

Substitute Q = 2501 kJ/kg and T =273 K in equation (1), we get
$$\begin{gathered} \mathrm{Δ}{S}_1=\frac{\left(2501\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{(273\mathrm{K})} \\ =9.156\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$
This is the same value as in Table

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

$$\begin{gathered} \mathrm{Δ}{S}_2=\frac{\left(42\mathrm{kJ}.{\mathrm{kg}}^{-1}\right)}{(278\mathrm{K})} \\ =0.151\mathrm{kJ}·{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1} \end{gathered}$$

This value is the same as in given table

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

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

$$\begin{gathered} \mathrm{Δ}S=\frac{\left(19\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{(278\mathrm{K})}-(55.55)(8.314\mathrm{J}/\mathrm{mole}·\mathrm{K})\frac{(0.006\mathrm{bar})}{(0.009\mathrm{bar})} \\ =-0.239\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$

The entropy for steam at finite temperature is:

$$\begin{gathered} \mathrm{Δ}{S}_1+\mathrm{Δ}S=\left(9.156\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right)+\left(-0.239\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1}\right) \\ =8.917\mathrm{kJ}·{\mathrm{kg}}^{-1}·{\mathrm{K}}^{-1} \end{gathered}$$

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.