Question

Consider a $\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{g}$sample of ${\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$at ${\mathbf{125}}^o\mathbf{C}$. What phase or phases are present when $\mathbf{215}\mathbf{kJ}$ of energy is removed from this sample? (See Example 93)

Short answer

The sample presents at liquid and solid phases when $215\mathrm{kJ}$ of energy is removed from the sample.

Step-by-step solution

Determine heat released in cooling of steam:

Converting steam at ${125}^{\circ }\mathrm{C}$ to steam at ${100}^{\circ }\mathrm{C}$ .

The heat released (q₁)

$\begin{array}{c}{\mathrm{q}}_1=75.0\mathrm{g}\times 2.0{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}{(125-100)}^{\circ }\mathrm{C}\\ =3.75\mathrm{kJ}\end{array}$

As theheat released calculated is less than the $215\mathrm{kJ}$. So, we need to calculate the next heat released.

Converting steam at ${100}^{\circ }\mathrm{C}$to liquid at ${100}^{\circ }\mathrm{C}$

Heat released (q₂):

$\begin{array}{c}{\mathrm{q}}_2=\frac{75\mathrm{g}}{18.02\mathrm{g}/\mathrm{mol}}\times 40.7\mathrm{kJ}/\mathrm{mol}\\ =169\mathrm{kJ}\end{array}$

As the sum of $({\mathrm{q}}_1+{\mathrm{q}}_2)$ is less than the $215\mathrm{kJ}$, we need to calculate the next heat released.

Determine heat released in cooling of water:

Converting water at ${100}^{\circ }\mathrm{C}$ to water at $0^{\circ }\mathrm{C}$.

Heat released:

$\begin{array}{c}{\mathrm{q}}_3=75.0\mathrm{g}\times 4.2{\mathrm{Jg}}^{-1}{}^{\circ }\mathrm{C}^{-1}{(100-0)}^{\circ }\mathrm{C}\\ =31.5\mathrm{kJ}\end{array}$

As the sum of $({\mathrm{q}}_1+{\mathrm{q}}_2+{\mathrm{q}}_3)$ is less than the $215\mathrm{kJ}$, we need to calculate the next heat released.

Heat released

$\begin{array}{c}{\mathrm{q}}_4=\frac{75\mathrm{g}}{18.02\mathrm{g}/\mathrm{mol}}\times 6.02\mathrm{kJ}/\mathrm{mol}\\ =25.1\mathrm{kJ}\end{array}$

Total heat evolved

$\begin{array}{l}{\mathrm{q}}_{\mathrm{T}}={\mathrm{q}}_1+{\mathrm{q}}_2+{\mathrm{q}}_3+{\mathrm{q}}_4\\ =169+3.75+31.5+25.1\\ =229\mathrm{kJ}\end{array}$

${\mathrm{q}}_{\mathrm{T}}$ is greater than the given energy $215\mathrm{kJ}$, which shows the water is not fully transformed into ice. So, two phases exist that are liquid and solid.