Chapter 15, Problem 15.77P

Interpretation Introduction

Interpretation:

The final temperature of ${\text{H}}_{\text{2}}\text{O}$ in the glass after all the ice melts has to be determined.

Concept Introduction:

Molar enthalpy for a process is heat energy needed by one mole of a substance to undergo change in the process. The heat energy that change the temperature of $m$ mass of substance from $T_1$ to $T_2$ is as follows:

  $q=mC\left(T_1-T_2\right)$

Here,

$q$ is heat energy in $\text{kJ}$

$m$ is mass of substance.

$C$ is specific heat capacity of substance.

$T_1,T_2$ is initial and final temperature of substance.

The heat energy absorbed by n mole of substance to undergo change is as follows:

  $q=n\left(\Delta H\right)$

Here,

$n$ is moles of substance.

$\Delta H$ is enthalpy for process in $\text{kJ/mol}$

Answer to Problem 15.77P

The final temperature for mixture of ice and water is $35.0254\text{K}$.

Explanation of Solution

Consider the final temperature to be $T_{\text{f}}$.

Ice undergoes 3 processes when put in a water container. In the first process it cools from $-21\text{°C}$ to $0\text{°C}$ and in second process it melts from solid ice to liquid water and lastly cooling form $0\text{°C}$ to $T_{\text{f}}$.

The density of ${\text{H}}_{\text{2}}\text{O}$ is $1{\text{g/cm}}^{\text{3}}$ so the volume and mass of ${\text{H}}_{\text{2}}\text{O}$ will be equal to each other. So if volume of ${\text{H}}_{\text{2}}\text{O}$ is $250\text{mL}$ then mass of ${\text{H}}_{\text{2}}\text{O}$ will be $250\text{g}$.

The formula to determine moles of ice is as follows:

  $\text{number}\text{of}\text{moles}\text{of}\text{ice}=\frac{\text{mass}\text{of}\text{ice}}{\text{molecular}\text{weight}\text{of}\text{ice}}$        (1)

Substitute $18.01528\text{g/mol}$ for molecular weight of ice and $20\text{g}$ for mass of ice in equation (1).

  $\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}\text{ice}=\frac{20\text{g}}{18.01528\text{g/mol}}\\ =1.11016\text{mol}\end{array}$

The temperature difference ($\Delta T$) between $T_2$ and $T_1$ is calculated as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =0°\text{C}-\left(-21°\text{C}\right)\\ =21°\text{C}\end{array}$

The size difference on kelvin and Celsius scale is equal so $\Delta T$ is equal to $21\text{K}$.

The formula to calculate heat energy required to cool $20\text{g}$ of ice form $-21°\text{C}$ to $0°\text{C}$.is as follows:

  $q_{\text{ice}\text{1}}=n_{\text{ice}}{C}_{\text{ice}}\left(\Delta T\right)$        (2)

Substitute $37.7\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}$ for $C_{\text{ice}}$, $1.11016\text{mol}$ for $n_{\text{ice}}$, $21\text{K}$ for $\Delta T$ in equation (2).

  $\begin{array}{c}{q}_{\text{ice}\text{1}}=\left(1.11016\text{mol}\right)\left(37.7\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\left(21\text{K}\right)\\ =0.8789\text{kJ}\end{array}$

The expression to estimate the heat absorbed to by ice to melt is as follows:

  $q_{\text{ice}\text{melting}}=n_{\text{ice}}\left(\Delta H_{\text{ice melting}}\right)$        (3)

Substitute $6.01\text{kJ}\cdot {\text{mol}}^{-\text{1}}$ for $\Delta H_{\text{ice melting}}$ and $1.11016\text{mol}$ for $n_{\text{ice}}$ in equation (3) to calculate the heat absorbed to by ice to melt.

  $\begin{array}{c}{q}_{\text{ice}\text{melting}}=\left(1.11016\text{mol}\right)\left(6.01\text{kJ}\cdot {\text{mol}}^{-\text{1}}\right)\\ =6.6720\text{kJ}\end{array}$

The temperature difference ($\Delta T$) between $T_2$ and $T_1$ is calculated as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =T_{\text{f}}°\text{C}-0°\text{C}\\ =\left(T_{\text{f}}-0\right)°\text{C}\end{array}$

The size difference on kelvin and Celsius scale is equal so $\Delta T$ is equal to $\left(T_{\text{f}}-0\right)\text{K}$.

The expression tocalculate heat energy required to cool $20\text{g}$ of ice from $0°\text{C}$ to $T_{\text{f}}°\text{C}$.is as follows:

  $q_{\text{ice}2}=n_{\text{ice}}{C}_{\text{ice}}\left(\Delta T\right)$        (4)

Substitute $75.3\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}$ for $C_{\text{ice}}$, $1.11016\text{mol}$ for $n_{\text{ice}}$, $\left(T_{\text{f}}-0\right)\text{K}$ for $\Delta T$ in equation (4).

  $\begin{array}{c}{q}_{\text{ice}2}=\left(1.11016\text{mol}\right)\left(75.3\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\left(\left(T_{\text{f}}-0\right)\text{K}\right)\\ =0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)\end{array}$

The formula to determine moles of water is as follows:

  $\text{number}\text{of}\text{moles}\text{of}\text{water}=\frac{\text{mass}\text{of}\text{water}}{\text{molecular}\text{weight}\text{of}\text{water}}$        (5)

Substitute $18.01528\text{g/mol}$ for molecular weight of ${\text{H}}_{\text{2}}\text{O}$ and $250\text{g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (5).

  $\begin{array}{c}\text{number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}=\frac{250\text{g}}{18.01528\text{g/mol}}\\ =13.8771\text{mol}\end{array}$

The temperature difference ($\Delta T$) between $T_2$ and $T_1$ is calculated as follows:

  $\begin{array}{c}\Delta T=T_2-T_1\\ =25°\text{C}-T_{\text{f}}°\text{C}\\ =\left(25-T_{\text{f}}\right)°\text{C}\end{array}$

The size difference on kelvin and Celsius scale is equal so $\Delta T$ is equal to $\left(25-T_{\text{f}}\right)\text{K}$.

The formula to calculate heat energy required to cool $250\text{g}$ of ${\text{H}}_{\text{2}}\text{O}$ from $T_{\text{f}}°\text{C}$ to $25°\text{C}$.is as follows:

  $q_{\text{water}}=n_{\text{water}}{C}_{\text{water}}\left(\Delta T\right)$        (6)

Substitute $75.3\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}$ for $C$, $13.8771\text{mol}$ for $n$, $\left(25-T_{\text{f}}\right)\text{K}$ for $\Delta T$ in equation (6).

  $\begin{array}{c}{q}_{\text{water}}=\left(13.8771\text{mol}\right)\left(75.3\text{J}\cdot {\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(\frac{{10}^{-3}\text{kJ}}{1\text{J}}\right)\left(\left(25-T_{\text{f}}\right)\text{K}\right)\\ =1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)\end{array}$

As there is no loss of heat to or from the surrounding hence there will be conservation of energy that is heat gained by $20\text{g}$ ice cube is equal to the heat lost by $250\text{g}$ of water which can be shown as follow:

  $q_{\text{ice}\text{1}}+q_{\text{ice}\text{melting}}+q_{\text{ice}2}=-q_{\text{water}}$        (7)

Substitute $0.8789\text{kJ}$ for $q_{\text{ice}\text{1}}$, $6.6720\text{kJ}$ for $q_{\text{ice}\text{melting}}$, $0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)$ for $q_{\text{ice}2}$ and $1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)$ for $q_{\text{water}}$.in equation (7) to calculate the $T_{\text{f}}$.

  $\begin{array}{c}0.8789\text{kJ}+6.6720\text{kJ}+0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)=-1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)\\ 7.5509\text{kJ}+0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)=-1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)\end{array}$

Rearrange above equation to calculate $T_{\text{f}}$.

$\begin{array}{c}7.5509\text{kJ}+0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)=-1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)\\ 7.5509\text{kJ}=-1.0449\text{kJ}\left(\left(25-T_{\text{f}}\right)\text{K}\right)-0.0835\text{kJ}\left(\left(T_{\text{f}}-0\right)\text{K}\right)\\ 7.5509\text{kJ}=-26.1225\text{kJ}\cdot \text{K}+\left(1.0449\text{kJ}\right)\left(T_{\text{f}}\right)-\left(0.0835\text{kJ}\right)\left(T_{\text{f}}\right)\\ 26.1225\text{kJ}\cdot \text{K}+7.5509\text{kJ}=0.9614\text{kJ}\left(T_{\text{f}}\right)\end{array}$

Rearrange above equation further to calculate $T_{\text{f}}$.

  $\begin{array}{c}33.6734\text{kJ}=0.9614\text{kJ}\left(T_{\text{f}}\right)\\ T_{\text{f}}=35.0254\text{K}\end{array}$

The final temperature for mixture of ice and water is $35.0254\text{K}$.