$100{\mathrm{ }\mathrm{c}\mathrm{m}}^3$ of $0.1\mathrm{M}\mathrm{HCl}$ and $100{\mathrm{ }\mathrm{c}\mathrm{m}}^3$ of $0.1\mathrm{M}$ $\mathrm{NaOH}$ solutions are mixed in a calorimeter. If the heat liberated is "Q" $\mathrm{K}$. Cal, the heat of neutralization $(\Delta \mathrm{H}$ ) (in $\mathrm{K}.\mathrm{Cal})$ of ${\mathrm{HCl}}_{\text{(aq) }}$ and ${\mathrm{NaOH}}_{\left(\mathrm{oq}\right)}$ is

During the process, following reaction occur$\mathrm{NaOH}+\mathrm{HCl}⟶\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$.

$\therefore$ When 1 mole of $\mathrm{NaOH}$ is neutralized by 1 mole of $\mathrm{HCl}$ then the heat liberated is $\mathrm{Q}\mathrm{kcal}$.

Or $\mathrm{\Delta H}=\mathrm{Q}$ kcal

$\therefore 10$ moles of $\mathrm{NaOH}$ are neutralized by 10 moles of $\mathrm{HCl}$

Heat liberated $=-10\times 10\times \mathrm{Q}\mathrm{kcal}=-100\mathrm{Q}\mathrm{kcal}$.