Question 11.7: Partial Melting Goal Understand how to handle an incomplete ...
Partial Melting
Goal Understand how to handle an incomplete phase change.
Problem A 5.00-\mathrm{kg} block of ice at 0^{\circ} \mathrm{C} is added to an insulated container partially filled with 10.0 \mathrm{~kg} of water at 15.0^{\circ} \mathrm{C}. (a) Find the final temperature, neglecting the heat capacity of the container. (b) Find the mass of the ice that was melted.
Strategy Part (a) is tricky, because the ice does not entirely melt in this example. When there is any doubt concerning whether there will be a complete phase change, some preliminary calculations are necessary. First, find the total energy required to melt the ice, Q_{\text {melt }}, and then find Q_{\text {water }}, the maximum energy that can be delivered by the water above 0^{\circ} \mathrm{C}. If the energy delivered by the water is high enough, all the ice melts. If not, there will usually be a final mixture of ice and water at 0^{\circ} \mathrm{C}, unless the ice starts at a temperature far below 0^{\circ} \mathrm{C}, in which case all the liquid water freezes.
(a) Find the equilibrium temperature.
First, compute the amount of energy necessary to completely melt the ice:
\begin{aligned} Q_{\text {melt }} & =m_{\text {ice }} L_{f}=(5.00 \mathrm{~kg})\left(3.33 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right) \\ & =1.67 \times 10^{6} \mathrm{~J} \end{aligned}
Next, calculate the maximum energy that can be lost by the initial mass of liquid water without freezing it:
\begin{aligned} Q_{\text {water }} & =m_{\text {water }} c \Delta T \\ & =(10.0 \mathrm{~kg})\left(4190 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(0^{\circ} \mathrm{C}-15.0^{\circ} \mathrm{C}\right) \\ & =-6.29 \times 10^{5} \mathrm{~J} \end{aligned}
This is less than half the energy necessary to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point:
T=0^{\circ} \mathrm{C}
(b) Compute the mass of ice melted. Set the total available energy equal to the heat of fusion of m grams of ice, m L_{f} :
\begin{aligned} 6.29 \times 10^{5} \mathrm{~J} & =m L_{f}=m\left(3.33 \times 10^{5} \mathrm{~J} / \mathrm{kg}\right) \\ m & =1.89 \mathrm{~kg} \end{aligned}
Remarks If this problem is solved assuming (wrongly) that all the ice melts, a final temperature of T=-16.5^{\circ} \mathrm{C} is obtained. The only way that could happen is if the system were not isolated, contrary to the statement of the problem. In the following exercise, you must also compute the thermal energy needed to warm the ice to its melting point.