For the following reaction

${\mathrm{H}}^{+}{}_{\left(\mathrm{aq}\right)}+{\mathrm{OH}}_{\left(\mathrm{aq}\right)}^{-}\to {\mathrm{H}}_2{\mathrm{O}}_{\left(l\right)};\Delta \mathrm{H}=-\mathrm{Q}$

Where $\Delta \mathrm{H}$ represents

The heat of neutralization is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt. The heat (or enthalpy) of neutralization $\left(\Delta \mathrm{H}\right)$ is the heat evolved when an acid and a base react to form a salt plus water. $Q$ in the above equation is $-\Delta \mathrm{H}$ and is expressed in $\mathrm{kJ}/\mathrm{mol}$ of water. Neutralization reactions are generally exothermic and thus $\Delta H$ is negative.

Hence the correct answer is (C) Heat of neutralization.