What amount of heat will be generated in the circuit shown after the switch w S is shifted from position 1 to position 2

The charge on the condensers in position 1 are shown.
Here, $\frac{\mathrm{q}}{\mathrm{C}}=\frac{{\mathrm{q}}_0}{{\mathrm{C}}_0}=\frac{\mathrm{q}+{\mathrm{q}}_0}{\mathrm{C}+{\mathrm{C}}_0}$
and $\left(\mathrm{q}+{\mathrm{q}}_0\right)\left(\frac{1}{\mathrm{C}+{\mathrm{C}}_0}+\frac{1}{\mathrm{C}}\right)=\mathrm{\epsilon }$
or $\left(\mathrm{q}+{\mathrm{q}}_0\right)=\frac{\mathrm{C}\left(\mathrm{C}+{\mathrm{C}}_0\right)\mathrm{\epsilon }}{{\mathrm{C}}_0+2\mathrm{C}}$
Hence ,  $\mathrm{q}=\frac{{\mathrm{C}}^2\mathrm{\epsilon }}{{\mathrm{C}}_0+2\mathrm{C}}\text{ and }{\mathrm{q}}_0=\frac{{\mathrm{CC}}_0\mathrm{\epsilon }}{{\mathrm{C}}_0+2\mathrm{C}}$
 

After the switch thrown to position 2, the charges change as shown in (Fig - b). A charge q₀  has flown in the right loop through the two  condensers and a charge 0 q through the cell, Because of the symmetery of the problem there is no change in the energy stored in the condensers. Thus H (Heat product) = Energy delivered by the cell 
$=\mathrm{\Delta Q\epsilon }={\mathrm{q}}_0\mathrm{\epsilon }=\frac{{\mathrm{CC}}_0{\mathrm{\epsilon }}^2}{{\mathrm{C}}_0+2\mathrm{C}}$