Analysis: if there is no heat loss, then heat loss by water at \(25^{\circ} C\) to a final temperature \(\theta_{f}=\) heat gained by ice at \(0^{\circ} C\) plus water from \({ }^{\circ} C\) to a final temperature \(\theta_{f}\) \(\Longrightarrow(m c \Delta \theta)_{w}=\left(m l_{f}\right)_{\text {ice }}+(m c \Delta \theta)_{w}\)
\(0.825 \times 4200 \times(25-\theta)\)
\(=0.075 \times 333000+0.075 \times 4200 \times\left(\theta_{f}-0\right)\)
\(=86625-3465 \theta_{f}=24975+315 \theta_{f}\)
\(=315 \theta_{f}+3465 \theta_{f}=86625-24975\)
\(=3780 \theta_{f}=61650\)
\(\theta_{f}=\frac{61650}{3780}=16.3^{\circ} C\)