A capacitor C₁ = 6.0 µF is charged by a 120 V battery and then taken off it…

Step 1. The charges at the start. $Q_1 = C_1E = 720\ \mu\text{C}$ and $Q_2 = C_2V_2 = 180\ \mu\text{C}$.

Step 2. Joining (a). The top wire joins the $+720\ \mu\text{C}$ plate of $C_1$ to the $-180\ \mu\text{C}$ plate of $C_2$. That conductor holds $540\ \mu\text{C}$, now shared by two capacitors in parallel:

Step 3. Charge along a wire (b). $C_1$ now holds $6.0 \times 60 = 360\ \mu\text{C}$, so $720 - 360 = 360\ \mu\text{C}$ has left its plate along the wire.

Step 4. Heat (c). Before: $\tfrac12 C_1E^2 + \tfrac12 C_2V_2^2 = 48.6\ \text{mJ}$. After: $\tfrac12(C_1 + C_2)V^2 = 16.2\ \text{mJ}$. The heat is $32.4\ \text{mJ}$.

Step 5. The battery again (d, e). Both capacitors end at $120\ \text{V}$ the old way round, so the top conductor ends with $9.0 \times 120 = 1080\ \mu\text{C}$. The battery sent $1080 - 540 = 540\ \mu\text{C}$, doing $W = qE = 64.8\ \text{mJ}$.

Step 6. Heat (f). The stored energy rose from $16.2\ \text{mJ}$ to $64.8\ \text{mJ}$, by $48.6\ \text{mJ}$. So the heat is $64.8 - 48.6 = 16.2\ \text{mJ}$. As a check, $\tfrac12(C_1 + C_2)(\Delta V)^2$ gives the same.