Question

Suppose 1.00 mol water at 25°C is flash-evaporated by allowing it to fall into an iron crucible maintained at 150°C. Calculate$\mathbf{∆}\mathbf{S}$for the water, $\mathbf{∆}\mathbf{S}$for the iron crucible, and${\mathbf{\Delta S}}_{\mathbf{tot}}\mathbf{}$ , if ${\mathbf{c}}_{\mathbf{p}}\left({\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right)\right)\mathbf{=}\mathbf{75}\mathbf{.}\mathbf{4}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$and${\mathbf{c}}_{\mathbf{p}}\left({\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{g}\right)\right)\mathbf{=}\mathbf{36}\mathbf{.}\mathbf{0}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$ . Take ${\mathbf{\Delta H}}_{\mathbf{vap}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{68}{\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}}$for water at its boiling point of 100°C.

Short answer

The entropy change for water is 130.516 JK^-1.

The entropy change for crucible is -113.75 JK^-1.

The total entropy change is 16.76 JK^-1.

Step-by-step solution

Latent Heat.

Latent heat is the heat quantity consumed to alter a substance's state concerning a system with a fixed temperature. A matter’s internal energy is also described by latent heat.

The amount of heat obtained from 25° C to 100° C

${\mathrm{q}}_1={\mathrm{mxc}}_{\mathrm{p}}\mathrm{x}∆\mathrm{T}$
Here,

*q₁is the amount of heat.

*m is the moles of a substance, i.e., 1.00.

*c_p is the heat capacity of liquid water, i.e., 75.4 JK^-1mol^-1

*$∆\mathrm{T}$is the temperature difference, i.e.,

$({\mathrm{T}}_1=25°\mathrm{C}+273\mathrm{K}=298\mathrm{K})({\mathrm{T}}_2=100°\mathrm{C}+273\mathrm{K}=373\mathrm{K})$

On substituting the given values, we get
$$\begin{gathered} q_1=m\times c_p\times \Delta T \\ =1.00mol\times 75.4 J{K}^{-1}mo{l}^{-1}\times \left(373-298 \right)K \\ =5655J \\ =5.655kJ \end{gathered}$$

The amount of heat obtained by crucible from 100°C to 150° C .

$q_2=m\times c_p\times \Delta T$

Here,

*q₂ is the amount of heat obtained by the crucible.

*m is the moles, i.e., 1.00 mol.

*c_p is heat capacity of gas, i.e., 36.0 JK^-1 mol^-1

* is the temperature difference, i.e., $\left(T_2-T_1=423 -373=50\right)K$

On substituting the given values;
$$\begin{gathered} q_2=m\times c_p\times \Delta T \\ =1.00mol\times 36.0 J{K}^{-1}mo{l}^{-1}\times 50 K \\ =1800 J \\ =1.8kJ \end{gathered}$$

The amount of heat needed to vaporize 1.00 mol of water.

$q_3=m\times \Delta H_{vap}$

Here,

*q₃ is amount of heat obtained to vaporize.

*m is the mole, i.e., 1.00 mol.

*$\Delta H_{vap}$is the enthalpy of vaporization of water, i.e.,$40.68 kJmo{l}^{-1}$

On substituting the given values;

$$\begin{gathered} q_3=m\times \Delta H_{vap} \\ =1.00mol\times 40.68 kJmo{l}^{-1} \\ =40.68kJ \end{gathered}$$

The total amount of heat absorbed by water.

$$\begin{gathered} \backslash \Delta H=q_1+q_2+q_3 \\ =5.655kJ+40.68 kJ+1.8kJ \\ =48.135 kJ \end{gathered}$$
Thus, the total heat absorbed is 48.135 kJ.

The total entropy.

The entropy of water;
$\Delta S_{water}=n{c}_p\ln \frac{T_{boiling}}{T_{initial}}\left(l\right)+n\frac{\Delta H_{vap}}{T}n{c}_p\ln \frac{T_{final}}{T_{boiling}}\left(g\right)$
On substituting the given values;
$$\begin{gathered} \Delta S_{water}=n{c}_p\ln \frac{T_{boiling}}{T_{initial}}\left(l\right)+n\frac{\Delta H_{vap}}{T}n{c}_p\ln \frac{T_{final}}{T_{boiling}}\left(g\right) \\ =\left(1.00mol\times 75.4J{K}^{-1}mo{l}^{-1}\times \ln \left(\frac{373}{298}\right)K+1.00mol\left(\frac{40680Jmo{l}^{-1}}{373K}\right) \\ +1.00mol\times 36.0J{K}^{-1}mo{l}^{-1}\times \ln \left(\frac{423}{373}\right)K \\ \right) \\ =130.516J{K}^{-1} \end{gathered}$$
Thus, the entropy change for water is 130.516 JK^-1 .
The entropy of crucible;
$$\begin{gathered} \Delta S_{crucible}=\frac{-\Delta H_{crucible}}{T_{crucible}} \\ =\frac{-48.135kJ}{423K} \\ =\frac{-48135J}{423K} \\ =-113.75J{K}^{-1} \end{gathered}$$
Thus, the entropy change for crucible is -113.75 JK^-1 .
The total entropy change;
$$\begin{gathered} \Delta S_{total}=\Delta S_{water}+\Delta S_{crucible} \\ =130.516J{K}^{-1}-113.75J{K}^{-1} \\ =16.76J{K}^{-1} \end{gathered}$$
Therefore, the total entropy change is 16.76 JK^-1