A capacitor with a capacitance of 100 is charged to a potential...

Initial charge on the capacitor, $Q_i = C V_i = 100 \times 10^{-6} \times 24 = 2400 \mu\text{C}$

Initial energy stored in the capacitor, $U_i = \dfrac{1}{2} C V_i^2 = \dfrac{1}{2} \times 100 \times 10^{-6} \times (24)^2 = 28.8 \text{ mJ}$

When connected to the $12 \text{ V}$ battery with the positive plate to the positive terminal, the final charge on the capacitor is $Q_f = C V_f = 100 \times 10^{-6} \times 12 = 1200 \mu\text{C}$

Final energy stored in the capacitor, $U_f = \dfrac{1}{2} C V_f^2 = \dfrac{1}{2} \times 100 \times 10^{-6} \times (12)^2 = 7.2 \text{ mJ}$

Charge flown through the battery is $\Delta Q = Q_f - Q_i = 1200 \mu\text{C} - 2400 \mu\text{C} = -1200 \mu\text{C}$

Work done by the battery, $W = \Delta Q \times V_f = -1200 \times 10^{-6} \times 12 = -14.4 \text{ mJ}$

Using the energy conservation principle, the heat developed is given by:

$H = W - (U_f - U_i)$

$H = -14.4 \text{ mJ} - (7.2 \text{ mJ} - 28.8 \text{ mJ}) = -14.4 \text{ mJ} + 21.6 \text{ mJ} = 7.2 \text{ mJ}$