Consider the circuit shown in the figure. The switch S is open for...

When the switch $S$ is open, the two capacitors are in series. The equivalent capacitance is $C_{eq} = \dfrac{C}{2}$.

The initial charge on the left capacitor is $q_i = C_{eq}\mathcal{E} = \dfrac{C\mathcal{E}}{2}$.

The initial energy stored in the system is $U_i = \dfrac{1}{2} C_{eq} \mathcal{E}^2 = \dfrac{C\mathcal{E}^2}{4}$.

When the switch $S$ is closed, the right capacitor is short-circuited. The left capacitor is now directly connected across the battery.

The final charge on the left capacitor is $q_f = C\mathcal{E}$.

The final energy stored in the system is $U_f = \dfrac{1}{2} C \mathcal{E}^2$.

The charge supplied by the battery is $\Delta q = q_f - q_i = C\mathcal{E} - \dfrac{C\mathcal{E}}{2} = \dfrac{C\mathcal{E}}{2}$.

The work done by the battery is $W = \Delta q \mathcal{E} = \dfrac{C\mathcal{E}^2}{2}$.

The heat developed in the system is given by $H = W - (U_f - U_i)$.

$H = \dfrac{C\mathcal{E}^2}{2} - \left(\dfrac{C\mathcal{E}^2}{2} - \dfrac{C\mathcal{E}^2}{4}\right) = \dfrac{C\mathcal{E}^2}{4}$.