The heat released during mixing of 50 mL of  $0.1M{H}_{2}S{O}_4$ with 50 mL of $0.2MKOH$   is

Milli gram equivalents of  $H_{2}S{O}_4=0.1\times 2\times 50=10$
Milli gram equivalents of  $KOH=0.2\times 50=10$
1 gram equivalent of acidand base neutralization  $57.3\times {10}^{3}J$
$10\times {10}^{-3}$  gram equivalents  $\to ?$
Of acidiandbase neutralization  $10\times {10}^{-3}\times {10}^3\times 57-3=573J$