Three ice cubes each of side $$2\ \text{cm}$$ at $$0^\circ\text{C}$$ are dropped into a glass of $$18^\circ\text{C}$$ water. The volume of the water is $$200\ \text{cm}^3$$. What are the volume and the temperature of the water after the ice cubes have melted?<br>(Density of water $$= 1\ \text{g/cm}^3$$, density of ice $$= 0.9\ \text{g/cm}^3$$, latent heat of ice $$= 80\ \text{cal/g}$$, specific heat of water $$= 1\ \text{cal/g}^\circ\text{C}$$)

The ice has volume $$3 \times 2^3 = 24\ \text{cm}^3$$ and mass $$0.9 \times 24 = 21.6\ \text{g}$$, and melting it needs $$21.6 \times 80 = 1728\ \text{cal}$$, less than the $$200 \times 18 = 3600\ \text{cal}$$ the water can give up in cooling to $$0^\circ\text{C}$$, so all the ice melts. If the final temperature is $$T$$, then $$200(18 - T) = 1728 + 21.6T$$, so $$221.6T = 1872$$ and $$T \approx 8.45^\circ\text{C}$$. The melted ice adds $$21.6\ \text{g}$$, i.e. $$21.6\ \text{cm}^3$$ of water, so the final volume is $$200 + 21.6 = 221.6\ \text{cm}^3$$.