Question 11.6: Ice Water Goal Solve a problem involving heat transfer and a...
Ice Water
Goal Solve a problem involving heat transfer and a phase change from solid to liquid.
Problem At a party, 6.00 \mathrm{~kg} of ice at -5.00^{\circ} \mathrm{C} is added to a cooler holding 30 liters of water at 20.0^{\circ} \mathrm{C}. What is the temperature of the water when it comes to equilibrium?
Strategy In this problem, it’s best to make a table. With the addition of thermal energy Q_{\text {ice }}, the ice will warm to 0^{\circ} \mathrm{C}; then melt at 0^{\circ} \mathrm{C} with the addition of energy Q_{\text {melt }}. Next, the melted ice will warm to some final temperature T by absorbing energy Q_{\text {ice-water }}, obtained from the energy change of the original liquid water, Q_{\text {water }}. By conservation of energy, these quantities must sum to zero.
Calculate the mass of liquid water:
\begin{aligned} m_{\text {water }} & =\rho_{\text {water }} V \\ & =\left(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)(30.0 \mathrm{~L}) \frac{1.00 \mathrm{~m}^{3}}{1.00 \times 10^{3} \mathrm{~L}} \\ & =30.0 \mathrm{~kg} \end{aligned}
Write the equation of thermal equilibrium:
Q_{\text {ice }}+Q_{\text {melt }}+Q_{\text {ice-water }}+Q_{\text {water }}=0 (1)
Construct a comprehensive table:
\begin{array}{lcccccc} \hline \boldsymbol{Q} & \boldsymbol{m}(\mathbf{k g}) & \boldsymbol{c}\left(\mathbf{J} / \mathbf{k g} \cdot{ }^{\circ} \mathbf{C}\right) & \boldsymbol{L}(\mathbf{J} / \mathbf{k g}) & \boldsymbol{T}_{\boldsymbol{f}}\left({ }^{\circ} \mathbf{C}\right) & \boldsymbol{T}_{\boldsymbol{i}}\left({ }^{\circ} \mathbf{C}\right) & \text{Expression }\\ \hline Q_{\text {ice }} & 6.00 & 2090 & & 0 & -5.00 & m_{\text {ice }} c_{\text {ice }}\left(T_{f}-T_{i}\right) \\ Q_{\text {melt }} & 6.00 & & 3.33 \times 10^{5} & 0 & 0 & m_{\text {ice }} L_{f} \\ Q_{\text {ice-water }} & 6.00 & 4190 & & T & 0 & m_{\text {ice }} c_{\text {wat }}\left(T_{f}-T_{i}\right) \\ Q_{\text {water }} & 30.0 & 4190 & & T & 20.0 & m_{\text {wat }} c_{\text {wat }}\left(T_{f}-T_{i}\right) \\ \hline \end{array}
Substitute all quantities in the second through sixth columns into the last column and sum (which is the evaluation of Equation 1), and solve for T :
\begin{aligned} 6.27 \times 10^{4} \mathrm{~J}+ & 2.00 \times 10^{6} \mathrm{~J} \\ + & \left(2.51 \times 10^{4} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)\left(T-0^{\circ} \mathrm{C}\right) \\ + & \left(1.26 \times 10^{5} \mathrm{~J} /{ }^{\circ} \mathrm{C}\right)\left(T-20.0^{\circ} \mathrm{C}\right)=0 \end{aligned}
T=3.03^{\circ} \mathrm{C}
Remarks Making a table is optional. However, simple substitution errors are extremely common, and the table makes such errors less likely.