Chapter 4.5, Problem 134RP

A 0.3-L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 0°C and (b) −20°C. Also determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L.

To determine

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

Answer to Problem 134RP

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

Explanation of Solution

Write the expression for the mass of the water.

$m_{\text{w}}=\rho V$ (I)

Here, the density of the water is $\rho$ and the volume of the water is $V$.

Write the expression for the energy balance equation.

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

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

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

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 $\Delta U$.

Conclusion:

Substitute 0 for $Q_{\text{in}}$ and $0$ for $W_{\text{out}}$ in Equation (III).

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

Here, the mass of the ice is $m$, the heat of fusion of ice is $h_{if}$, the specific heat of ice at liquid state is $c$, the initial temperature of ice is $T_1$, and the final temperature of ice is $T_2$

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and room temperature as $2.11\text{kJ}/\text{kg}°\text{C}$ and $4.18\text{kJ}/\text{kg}°\text{C}$.

Substitute $1\text{kg}/\text{L}$ for $\rho$ and $0.3\text{L}$ for $V$ in Equation (I).

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

For initial temperature of ice as 0 C.

Substitute $0°\text{C}$ for $T_{\text{1,solid}}$, $2.11\text{kJ}/\text{kg}°\text{C}$ for $c_{\text{solid}}$, $333.7\text{kJ}/\text{kg}$ for $h_{if}$, $4.18\text{kJ}/\text{kg}\cdot \text{°C}$ for $c_{\text{liquid}}$, $5\text{°C}$ for $T_2{}_{\text{,liquid}}$, $0.3\text{kg}$ for $m_{\text{water}}$, and $20°\text{C}$ for $T_{\text{1,water}}$ in Equation (IV).

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

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

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

Answer to Problem 134RP

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

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

Explanation of Solution

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

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

Here, the mass of cold water is $m_{\text{coldwater}}$, the specific heat of cold water is $c_{\text{coldwater}}$, the initial temperature of cold water is $T_{\text{1,coldwater}}$, the final temperature of cold water is $T_{\text{2,coldwater}}$, the mass of water is $m_{\text{water}}$, the specific heat of water is $c_{\text{water}}$, the initial temperature of water is $T_{\text{1,water}}$, and the final temperature of water is $T_{\text{2,water}}$.

Conclusion:

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

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

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

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

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

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

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