In the circuit shown switch S is kept open for long time. Switch is closed at t = 0, find total heat produced after closing the switch

Before closing the switch

$Q=C_{eq}V=\left(\frac{C_1{C}_2}{C_1+C_2}\right)\left(V_1-V_2\right)$

$Q=\frac{6}{5}\times 1\mu C$

Initial energy stored capacitors = $\frac{Q^2}{2C_1}+\frac{Q^2}{2C_2}$

$=\frac{{\left(\frac{6}{5}\right)}^2}{2\times 3}+\frac{{\left(\frac{6}{5}\right)}^2}{2\times 2}=\frac{3}{5}\mu J$

After closing the switch 

$Q_2=10\mu C$

$Q_1=18\mu C$

Final energy stored in capacitor

$U=\frac{1}{2}\times 3{\left(6\right)}^2+\frac{1}{2}\times 2{\left(5\right)}^2$

$U=79\mu J$

Work done by batteries = $6\left(Q_1-Q\right)+5\left(Q+Q_2\right)$

$\begin{array}{l}=156.8\mu \mathrm{J}\\ H=\text{ work done by battery - change in potential energy of capacitors }\\ =156.8-\left(79-\frac{3}{5}\right)\\ =78.4\mu \mathrm{J}\end{array}$