Twenty grams of ice at $0^{\circ} \mathrm{C}$ are dropped into a beaker containing 120 grams of water initially at $70^{\circ} \mathrm{C}$. Find the final temperature of the mixture neglecting the heat capacity of the beaker. Heat of fusion of ice is $80$ $\mathrm{cal} / \mathrm{g}$.
We assume the temperature of equilibrium to be $T$ after mixing. Thus

$$
M_{1} L_{\text {fusion }}+M_{1} C_{p, \text { water }} T=M_{2} C_{p, \text { water }}\left(T_{0}-T\right) .
$$

We substitute $M_{1}=20 \mathrm{~g}, M_{2}=120 \mathrm{~g}, T_{0}=70^{\circ} \mathrm{C}, L_{\text {fusion }}=80 \mathrm{cal} / \mathrm{g}$ and $C_{p, \text { water }}=1 \mathrm{cal} / \mathrm{g}$, and obtain the final temperature $T=48.57^{\circ} \mathrm{C}$.