Solution 8
Let the mass of the ice be $M$ and the mass of the water poured be $M$. Since the problem states water equal to that of ice in mass is poured, $M_w = M_i = M$.
Volume Analysis:
The total volume initially is composed of the water and the submerged ice.
$$ V_{initial} = V_w + V_i = \frac{M}{\rho_w} + \frac{M}{\rho_i} $$
Since this volume corresponds to height $h$ in the vessel of area $A$:
$$ A \cdot h = M \left( \frac{1}{\rho_w} + \frac{1}{\rho_i} \right) \quad \dots(1) $$
When the system reaches thermal equilibrium, some ice melts. The prompt implies partial melting (final temp $0^\circ\text{C}$). Let mass $\Delta m$ of ice melt. The volume change $\Delta V$ corresponds to the drop in height $\Delta h$: $$ \Delta V = A \cdot \Delta h = \Delta m \left( \frac{1}{\rho_i} – \frac{1}{\rho_w} \right) \quad \dots(2) $$
Heat Analysis:
Heat lost by water (cooling from $\theta$ to $0^\circ\text{C}$) = Heat gained by melting ice.
$$ M s_w (\theta – 0) = \Delta m \cdot L $$
$$ \Delta m = \frac{M s_w \theta}{L} $$
Substitute $\Delta m$ into equation (2) and divide by equation (1) to eliminate $M$ and $A$: $$ \frac{A \Delta h}{A h} = \frac{\frac{M s_w \theta}{L} \left( \frac{\rho_w – \rho_i}{\rho_w \rho_i} \right)}{M \left( \frac{\rho_w + \rho_i}{\rho_w \rho_i} \right)} $$ $$ \frac{\Delta h}{h} = \frac{s_w \theta}{L} \left( \frac{\rho_w – \rho_i}{\rho_w + \rho_i} \right) $$
Solving for $\theta$: $$ \theta = \frac{\Delta h}{h} \frac{L}{s_w} \left( \frac{\rho_w + \rho_i}{\rho_w – \rho_i} \right) $$
Using values: $\Delta h = 0.7$, $h = 19$, $L = 3.3 \times 10^5$, $s_w = 4.2 \times 10^3$, $\rho_w = 1000$, $\rho_i = 900$. $$ \theta = \frac{0.7}{19} \times \frac{3.3 \times 10^5}{4.2 \times 10^3} \times \left( \frac{1900}{100} \right) $$ $$ \theta \approx 55^\circ\text{C} $$