Drop a thermometer into a glass of melting ice and watch closely: heat pours in from the warm room, and the reading parks at 0 °C until the last sliver of ice is gone. The energy is still being spent — it is buying a change of phase, prying the water molecules loose from the ice, instead of warming anything. On a heating curve this shows up as a flat shelf, and $Q = mc\,\Delta T$ is powerless there, because $\Delta T = 0$.

The plateau has its own formula: $Q = mL$ (read aloud: Q equals m L). $L$ is the latent heat, the crossing fare per kilogram, and notice there is no $\Delta T$ in it at all. Water posts two fares: melting (or freezing) costs $334\ \mathrm{kJ/kg}$, and boiling (or condensing) costs $2260\ \mathrm{kJ/kg}$ — nearly seven times steeper. Same crossing, either direction: melting charges the fare in, freezing pays it back out. Match the fare to the crossing, and never send $mc\,\Delta T$ to do a plateau's job.