WAEC Physics 2013 Paper

Concept:Ice first absorbs heat to warm up to its melting point, then absorbs latent heat to change into water at constant temperature.Explanation:Since the final state is water at $0^\circ\mathrm{C}$, the ice must have been at $-5^\circ\mathrm{C}$ initially and warmed by $5^\circ\mathrm{C}$.Mass of ice, $m = 10\,\mathrm{g}$.Heat needed to warm the ice from $-5^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$:$$Q_1 = mc\Delta\theta = 10 \times 2.1 \times 5 = 105\,\mathrm{J}$$Heat needed to melt the ice at $0^\circ\mathrm{C}$ into water at $0^\circ\mathrm{C}$:$$Q_2 = mL = 10 \times 336 = 3360\,\mathrm{J}$$Total quantity of heat used:$$Q = Q_1 + Q_2 = 105 + 3360 = 3465\,\mathrm{J}$$Answer:The quantity of heat used is $3465\,\mathrm{J}$, which is option C.