A 12 µF capacitor is charged to 100 V. An 8.0 µF capacitor is charged to 50 V…

Step 1. Like signs, the voltage. $V' = \dfrac{12\ \mu\text{F} \times 100\ \text{V} + 8.0\ \mu\text{F} \times 50\ \text{V}}{12\ \mu\text{F} + 8.0\ \mu\text{F}} = \dfrac{1600\ \mu\text{C}}{20\ \mu\text{F}} = 80\ \text{V}$.

Step 2. Like signs, the energy. Before: $\tfrac{1}{2} \times 12\ \mu\text{F} \times (100\ \text{V})^2 + \tfrac{1}{2} \times 8.0\ \mu\text{F} \times (50\ \text{V})^2 = 60\ \text{mJ} + 10\ \text{mJ} = 70\ \text{mJ}$. After: $\tfrac{1}{2} \times 20\ \mu\text{F} \times (80\ \text{V})^2 = 64\ \text{mJ}$. The loss is $6.0\ \text{mJ}$.

Step 3. Opposite signs, the voltage. $V' = \dfrac{1200\ \mu\text{C} - 400\ \mu\text{C}}{20\ \mu\text{F}} = 40\ \text{V}$.

Step 4. Opposite signs, the energy. After: $\tfrac{1}{2} \times 20\ \mu\text{F} \times (40\ \text{V})^2 = 16\ \text{mJ}$. The loss is $70\ \text{mJ} - 16\ \text{mJ} = 54\ \text{mJ}$.