Question

Iron has a heat capacity of$\mathbf{25}\mathbf{.}\mathbf{1}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$, approximately independent of temperature between 0°C and 100°C.

(a) Calculate the enthalpy and entropy change of 1.00 mol iron as it is cooled at atmospheric pressure from 100°C to 0°C.

(b) A piece of iron weighing 55.85 g and at 100°C is placed in a large reservoir of water held at 0°C. It cools irreversibly until its temperature equals that of the water. Assuming the water reservoir is large enough that its temperature remains close to 0°C, calculate the entropy changes for the iron and the water and the total entropy change in this process.

Short answer

The enthalpy and entropy change for iron are -2510J and -7.83 J/K.

The entropy change for iron is -7.83J/K.

The entropy change for water is 9.19 J/K.

The total entropy change for the process is 1.36 J/K.

Step-by-step solution

Entropy

The entropy is one of the parameters of thermodynamics that measure the randomness of a system. It is the system’s thermal energy per unit temperature.

Calculation

a.

__Given data__

*The heat capacity of iron is $25.1{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ .
*The initial temperature for iron is $100°\mathrm{C}(100+273=373\mathrm{K})$.

*The final temperature of iron is $0°\mathrm{C}(0+273=273\mathrm{K})$.

*The number of moles of iron is 1.00 mol.

b.

*The mass of iron is 55.85 g/mol.

\\

The enthalpy change is calculated by using the formula as:
$$\begin{gathered} \Delta H=n_{Fe}{\left(c_p\right)}_{Fe}\Delta T \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=1.00mol\times 25.1J{K}^{\mathit{-}\mathit{1}}mo{l}^{\mathit{-}\mathit{1}}\times \left(T_{\mathrm{2}}\mathrm{-}{T}_{\mathrm{1}}\right) \\ =1.00mol\times 25.1J{K}^{-1}mo{l}^{-1}\times \left(\mathrm{273}\mathrm{-}\mathrm{373}\right)K \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=-2510J \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{} \end{gathered}$$
The entropy change is calculated as:
$$\begin{gathered} ∆\mathrm{S}={\mathrm{n}}_{\mathrm{Fe}}{\left({\mathrm{c}}_{\mathrm{p}}\right)}_{\mathrm{Fe}}\mathrm{In}\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ =1.00\mathrm{mol}\mathrm{x}25.1{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\mathrm{x}\mathrm{In}\frac{273\mathrm{K}}{373\mathrm{K}} \\ =1.00\mathrm{mol}\mathrm{x}25.1{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}\mathrm{x}-0.312 \\ =-7.83\mathrm{J}/\mathrm{K} \end{gathered}$$
Thus, the enthalpy and entropy change for iron are -2510J and$-7.83J/K$.

b.

The number of moles of iron is calculated as:

$$\begin{gathered} Moles of iron=\frac{Mass}{Molarmass} \\ =\frac{55.85g}{55.85 g/mol} \\ =1.00mol \end{gathered}$$

The heat evolved by the iron piece is calculated as:

$$\begin{gathered} q_{Fe}=\Delta H_{Fe} \\ =-2510J \end{gathered}$$

The heat absorbed by the water is calculated as:
localid="1663738114442" $$\begin{gathered} q_{water}=\Delta H_{water} \\ =-\Delta H_{Fe} \\ =-\left(-2510J\right) \\ =2510J \end{gathered}$$

The entropy change for iron is as follows:
$$\begin{gathered} \Delta S=n_{Fe}{\left(c_p\right)}_{Fe}\ln \frac{T_2}{T_1} \\ =1.00mol\times 25.1J{K}^{-1}mo{l}^{-1}\times \ln \frac{273K}{373 K} \\ =1.00mol\times 25.1J{K}^{-1}mo{l}^{-1}\times -0.312 \\ =-7.83J/K \end{gathered}$$

The entropy change for water is calculated as:

$$\begin{gathered} \Delta S_{water}=\frac{\Delta H_{water}}{T} \\ =\frac{2510J}{273K} \\ =9.19J{K}^{-1} \end{gathered}$$

The entropy change for the process is the sum of entropy change of iron and water which is given as:
$$\begin{gathered} \Delta S_{total}=\Delta S_{Fe}+\Delta S_{water} \\ =-7.83J{K}^{-1}+9.19J{K}^{-1} \\ =+1.36J/K \end{gathered}$$
Therefore, the total entropy change for the process is 1.36 J/K.