C

Step 1: Calculate Initial Energy in the First Capacitor
The initial charge ($Q_1$) on the first capacitor can be calculated using the formula:
$$Q_1 = C_1 imes V = 0.1 imes 10 = 1 ext{ C}$$
The initial energy ($U_1$) stored in the first capacitor can be calculated using the formula:
$$U_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2} \times 0.1 \times 10^2 = 5 ext{ J}$$

Step 2: Charge Distribution on the Two Capacitors
When the charged capacitor is connected to an uncharged capacitor (let's call it $C_2$), the total charge remains conserved.
Therefore, the total charge ($Q_{total}$) on both capacitors after connecting them is:
$$Q_{total} = Q_1 = 1 ext{ C}$$
Since both capacitors are of the same capacity ($C_1 = C_2 = 0.1 ext{ F}$), the voltage across both will equalize.
The total capacitance ($C_{total}$) for two capacitors in parallel is:
$$C_{total} = C_1 + C_2 = 0.1 + 0.1 = 0.2 ext{ F}$$
Using this total capacitance, we can find the new voltage ($V_f$) across the combined capacitors:
$$V_f = \frac{Q_{total}}{C_{total}} = \frac{1}{0.2} = 5 ext{ V}$$

Step 3: Calculate the Energy Stored in the Two Capacitors
The energy ($U_{total}$) stored in the two capacitors after the charge redistributes can be calculated using the formula:
$$U_{total} = \frac{1}{2} C_{total} V_f^2 = \frac{1}{2} \times 0.2 \times 5^2 = \frac{1}{2} \times 0.2 imes 25 = 2.5 ext{ J}$$

Step 4: Compare the Energies
The initial energy ($U_1$) was 5 J, and the final energy after connecting the two capacitors ($U_{total}$) is 2.5 J.
Therefore, the energy is reduced after connecting the second capacitor.
The total energy stored in the two capacitors is 2.5 J, which is half of the initial energy stored in the first capacitor.
Thus, the correct answer for the total energy stored in the two capacitors is 2.5 J.