Question

In Example 12.3, a process was considered in which 72.4 g iron initially at 100.0°C was added to 100.0 g water initially at 10.0°C, and an equilibrium temperature of 16.5°C was reached. Take ${\mathbf{c}}_{\mathbf{p}}\left(\mathbf{Fe}\right)$to be $\mathbf{25}\mathbf{.}\mathbf{1}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$and ${\mathbf{c}}_{\mathbf{p}}\left(H_{2}O\right)$to be $\mathbf{75}\mathbf{.}\mathbf{3}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$, independent of temperature. Calculate for the iron, for the water, and$∆\mathrm{S}$total in this process.

Short answer

The entropy change for iron is $-8.32J{K}^{-1}$

The entropy change for water is $8.74J{K}^{-1}$

The total entropy change for the process is $+0.42J{K}^{-1}$

Step-by-step solution

The Given data.

*The mass of iron is 72.4 g.

*The mass of water is 100.0 g.

*The initial temperature of iron is$100.0°\mathrm{C}=100.0+273=373\mathrm{K}$.

*The initial temperature of water is$10.0°\mathrm{C}=10.0+273=283\mathrm{K}$.

*The equilibrium temperature is $165°\mathrm{C}=165+273=289\mathrm{K}$.

*The heat capacity of iron is$25.1{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$.

*The heat capacity of water is $75.3{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$.

Heat Capacity

At constant volume (V) and pressure (P), the molar heat capacity is symbolized as C_v, and C_p, respectively.

The molar heat capacity is the estimate of “heat energy” appropriate for elevating the temperature change.

The number of moles of iron and water.

$$\begin{gathered} Moles of iron=\frac{Mass}{Molarmass} \\ =\frac{72.4g}{55.85 g/mol} \\ =1.30mol \end{gathered}$$
$$\begin{gathered} Moles of water=\frac{Mass}{Molarmass} \\ =\frac{100.0g}{18.0 g/mol} \\ =5.56mol \end{gathered}$$

The total entropy change.

The entropy change for iron is calculated as:

$\Delta S=n_{Fe}{\left(c_p\right)}_{Fe}\ln \frac{T_f}{{\left(T_i\right)}_{Fe}}$
On substituting the given values, we get
$$\begin{gathered} \Delta S=n_{Fe}{\left(c_p\right)}_{Fe}\ln \frac{T_f}{{\left(T_i\right)}_{Fe}} \\ =1.30mol\times 25.1J{K}^{-1}mo{l}^{-1}\times \ln \frac{289K}{373K} \\ =32.63J{K}^{-1}\times -0.255 \\ =-8.32J{K}^{-1} \end{gathered}$$

Thus, the entropy change for iron is

Theentropy change for water is calculated as:
$$\begin{gathered} \Delta S_{H_{2}O}=n_{H_{2}O}{\left(c_p\right)}_{H_{2}O}\ln \frac{T_f}{{\left(T_i\right)}_{H_{2}O}} \\ =5.56mol\times 75.3J{K}^{-1}mo{l}^{-1}\times \ln \frac{289K}{283K} \\ =418.6J{K}^{-1}\times 0.0209 \\ =8.74J{K}^{-1} \end{gathered}$$
Thus, the entropy change for water is 8.74 JK^-1
Thetotal entropy change for the process is calculated as:
$$\begin{gathered} \Delta S_{H_{2}O}=\Delta S_{Fe}+\Delta S_{H_{2}O} \\ =-8.32J{K}^{-1}+8.74J{K}^{-1} \\ =+0.42J{K}^{-1} \end{gathered}$$
Therefore, the total entropy change for the process is +0.42 JK^-1.