A capacitor of 5.0 is initially charged to 12 V . Following this,...

Initial charge on the capacitor: $q_i = CV = 5.0 \times 10^{-6} \times 12 = 60 \mu\text{C}$

Initial energy of the capacitor: $U_i = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 5.0 \times 10^{-6} \times (12)^2 = 360 \mu\text{J}$

When the capacitor is connected to the battery with reversed polarity, the final charge on the originally positive plate becomes $q_f = -60 \mu\text{C}$.

Charge that flows through the battery: $\Delta q = |q_f - q_i| = |-60 - 60| = 120 \mu\text{C}$

Work done by the battery: $W = \Delta q \times V = 120 \mu\text{C} \times 12 \text{ V} = 1440 \mu\text{J}$

Final energy of the capacitor: $U_f = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 5.0 \times 10^{-6} \times (12)^2 = 360 \mu\text{J}$

Heat developed in the connecting wires: $H = W - (U_f - U_i) = 1440 \mu\text{J} - (360 \mu\text{J} - 360 \mu\text{J}) = 1440 \mu\text{J} = 1.44 \times 10^{-3} \text{ J}$