${\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{KOH}}_{ }\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2{\mathrm{O}}_{\left(0\right)}$; $∆\mathrm{H}$ for the above reaction is

When ${\mathrm{H}}^{+}$combine with one ion - $13.7\mathrm{Kcal}$ energy is released

$2{\mathrm{H}}^{+}$combine with ion so energy released $=2\times -13.7=-27.4\mathrm{Kcal}$.

Hence option C  is the correct answer.