Chapter 5, Problem 109RQ

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

Explanation of Solution

Given:

The density of the water $\left(\rho \right)$ is $1\text{kg}/\text{L}$.

The volume of the water $\left(V\right)$ is $0.3\text{L}$.

The initial temperature of water $\left(T_1{}_{\text{liquid}}\right)$ is $20°\text{C}$.

The initial temperature of ice $\left(T_1{}_{\text{solid}}\right)$ is $0°\text{C}$.

The final temperature of ice $\left(T_2{}_{\text{liquid}}\right)$ is $5\text{°C}$.

The mass of the water $\left(m_{\text{water}}\right)$ is $0.3\text{kg}$

The heat of fusion of ice $\left(h_{if}\right)$ is $333.7\text{kJ}/\text{kg}$.

Calculation:

Calculate the mass of the water.

  $m_{\text{w}}=\rho V$

  $\begin{array}{c}{m}_{\text{w}}=\left(1\text{kg}/\text{L}\right)\left(0.3\text{L}\right)\\ =0.3\text{kg}\end{array}$

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C$\left(c_{\text{solid}}\right)$  and room temperature $\left(c_{\text{liquid}}\right)$ as $2.11\text{kJ}/\text{kg}°\text{C}$ and $4.18\text{kJ}/\text{kg}°\text{C}$.

Write the expression for the energy balance equation.

  $E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$        (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (II) and write energy balance relation of cold water.

  $Q_{\text{in}}-W_{\text{out}}=\Delta U$        (II)

Here, the heat to be transfer into the system is $Q_{\text{in}}$, the work to be done by the system is $W_{\text{out}}$, and the change in the internal energy is $\Delta U$.

Substitute 0 for $Q_{\text{in}}$ and $0$ for $W_{\text{out}}$ in Equation (II) to obtain the mass of the ice.

  $\begin{array}{l}0=\Delta U\\ 0=\Delta U_{\text{ice}}+\Delta U_{\text{water}}\\ 0={\left[mc{\left(0°\text{C}-T_1\right)}_{\text{solid}}+m{h}_{if}+mc{\left(T_2-0°\text{C}\right)}_{\text{liquid}}\right]}_{\text{ice}}+{\left[mc\left(T_2-T_1\right)\right]}_{\text{water}}\end{array}$        (III)

  $\begin{array}{l}\left[\begin{array}{l}{\left[\begin{array}{l}m\left(2.11\text{kJ}/\text{kg}°\text{C}\right){\left(0°\text{C}-0°\text{C}\right)}_{\text{solid}}+m\left(333.7\text{kJ}/\text{kg}\right)\\ +m\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right){\left(\left(5\text{°C}\right)-0°\text{C}\right)}_{\text{liquid}}\end{array}\right]}_{\text{ice}}\\ +{\left[\left(0.3\text{kJ}/\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5\text{°C}-20°\text{C}\right)\right]}_{\text{water}}\end{array}\right]=0\\ m{\left[0+\left(333.7\text{kJ}/\text{kg}\right)+\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5\text{°C}\right)\right]}_{\text{ice}}+{\left[\begin{array}{l}\left(0.3\text{kJ}/\text{kg}\right)\\ \left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\\ \left(-15°\text{C}\right)\end{array}\right]}_{\text{water}}=0\\ m=0.0546\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\\ m=54.6\text{g}\end{array}$

Thus, the amount of ice that needs to be added into water at 0 C of temperature is $\underset{\_}{54.6\text{g}}$.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature and the amount of water that needs to cool down the cold water at 0 C of temperature.

Explanation of Solution

Substitute 0 for $Q_{\text{in}}$ and $0$ for $W_{\text{out}}$ in Equation (II) and write energy balance relation for cooling down the cold water.

  $\begin{array}{l}0=\Delta U\\ 0=\Delta U_{\text{coldwater}}+\Delta U_{\text{water}}\\ 0={\left[mc\left(T_2-T_1\right)\right]}_{\text{coldwater}}+{\left[mc\left(T_2-T_1\right)\right]}_{\text{water}}\end{array}$        (IV)

For initial temperature of ice as -20 C instead of 0 C.

Substitute $T_{\text{1,cold water}}=-20°\text{C}$, $c_{\text{cold water}}=2.11\text{kJ}/\text{kg}°\text{C}$, $h_{if}=333.7\text{kJ}/\text{kg}$,  for $c_{\text{liquid}}$, $5\text{°C}$ $T_2{}_{\text{,liquid}}=4.18\text{kJ}/\text{kg}\cdot \text{°C}$, $m_{\text{water}}=0.3\text{kg}$, and $T_{\text{1,water}}=20°\text{C}$ in Equation (III).

  $\begin{array}{l}\left[\begin{array}{l}{\left[\begin{array}{l}m\left(2.11\text{kJ}/\text{kg}°\text{C}\right){\left(0°\text{C}-\left(-20°\text{C}\right)\right)}_{\text{solid}}+m\left(333.7\text{kJ}/\text{kg}\right)\\ +m\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right){\left(\left(5\text{°C}\right)-0°\text{C}\right)}_{\text{liquid}}\end{array}\right]}_{\text{ice}}\\ +{\left[\left(0.3\text{kJ}/\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5\text{°C}-20°\text{C}\right)\right]}_{\text{water}}\end{array}\right]=0\\ \left[\begin{array}{l}m{\left[\left(2.11\text{kJ}/\text{kg}°\text{C}\right)\left(20°\text{C}\right)+\left(333.7\text{kJ}/\text{kg}\right)+\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5\text{°C}\right)\right]}_{\text{ice}}\\ +{\left[\left(0.3\text{kJ}/\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(-15°\text{C}\right)\right]}_{\text{water}}\end{array}\right]=0\\ m=0.0487\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\\ m=48.7\text{g}\end{array}$

Thus, the amount of ice that needs to be added into water at -20 C of temperature is $\underset{\_}{48.7\text{g}}$.

Substitute $c=4.18\text{kJ}/\text{kg}\cdot \text{°C}$, $T_{2\text{,coldwater}}=5°\text{C}$,  $T_{1,\text{coldwater}}=0°\text{C}$,  $m_{\text{water}}=0.3\text{kg}$,  for $T_{2\text{,water}}=5°\text{C}$, and  $T_{1,\text{water}}=20°\text{C}$ in Equation (IV).

  $\begin{array}{l}\left[m_{\text{coldwater}}\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5-0\right)°\text{C}\right]+\left(0.3\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5-20\right)°\text{C}=\text{0}\\ \left[m_{\text{coldwater}}\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(5°\text{C}\right)\right]+\left(0.3\text{kg}\right)\left(4.18\text{kJ}/\text{kg}\cdot \text{°C}\right)\left(-15°\text{C}\right)=\text{0}\\ m=0.9\text{kg}\times \left(\frac{1000\text{g}}{1\text{kg}}\right)\\ m=900\text{g}\end{array}$

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is $\underset{\_}{900\text{g}}$.