Drop a thermometer in melting ice and watch: heat pours in, and the reading parks at 0 °C until the last crystal is gone. On a heating curve that is the plateau — and it is where $Q = mc\,\Delta T$ goes powerless, because $\Delta T = 0$ there. The energy is still being spent; it is buying a change of phase, prying the molecules apart instead of speeding them up.

The plateau has its own machine: $Q = mL$ (read aloud: Q equals m L): the heat $Q$ in joules, the mass $m$ crossing in kilograms, and $L$, the latent heat — the crossing fee per kilogram, in J/kg, with no $\Delta T$ anywhere in it. Water charges two different fares, and the subscript says which crossing: $L_f$ is fusion, the melting-and-freezing fee, $L_f = 334\ \mathrm{kJ/kg}$; $L_v$ is vaporization, the boiling-and-condensing fee, $L_v = 2260\ \mathrm{kJ/kg}$ — nearly seven times steeper. Match the L to the crossing, and never let $mc\,\Delta T$ anywhere near a plateau.