The heat of neutralization of strong base and strong acid is $-57.0\mathrm{KJ}$. The heat released when $0.5$ mole of ${\mathrm{HNO}}_3$ solution is added to $0.20$ moles of $\mathrm{NaOH}$ solution is

$${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\to {\mathrm{H}}_2\mathrm{O}+\mathrm{Heat}$$

One mole of ${\mathrm{H}}^{+}$and one mole of ${\mathrm{OH}}^{-}$give $57\mathrm{ }\mathrm{k}\mathrm{J}$. When $0.5$ moles of ${\mathrm{H}}^{+}$(from ${\mathrm{HNO}}_3$ ) reacts with $0.2$ mole of ${\mathrm{OH}}^{-}$(from $\mathrm{NaOH}$ ), $0.2$ mole of ${\mathrm{H}}^{+}$is neutralized by $0.2$ mole of ${\mathrm{OH}}^{-}$and $0.3$ mole of ${\mathrm{H}}^{+}$remains unreacted.

$$\begin{gathered} {\mathrm{H}}^{+}+\underset{0.2\text{ mols }}{{\mathrm{OH}}^{-}}\to {\mathrm{H}}_2\mathrm{O} \\ \text{ Heat evolved will be }=\frac{57\times 0.2}{1}=11.4 \mathrm{k}\mathrm{J} \end{gathered}$$.