Solution:

To solve this, we will use the concept of common potential and energy conservation. Let's derive it step-by-step:

1. Initial Energy Stored in Capacitor 1
   Let the initial capacitance of the first capacitor be $C$, and the battery's voltage be $V$.
   The energy stored in the capacitor is:

   $$U = \frac{1}{2} C V^2$$

2. When the second capacitor is connected
   After disconnecting the battery, an identical capacitor (with capacitance $C$) is connected across the first one. The total charge remains conserved because the battery is removed. Let the final voltage be $V_f$.

   Total charge initially:

   $$Q_{\text{initial}} = CV$$After connecting the second capacitor, the total capacitance becomes:

   $$C_{\text{total}} = C + C = 2C$$Common potential $V_f$:

   $$V_f = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{CV}{2C} = \frac{V}{2}$$

3. Final Energy Stored in Both Capacitors
   The final energy stored in the system is the sum of the energy in both capacitors:

   $$U_{\text{final}} = \frac{1}{2} C V_f^2 + \frac{1}{2} C V_f^2 = C V_f^2$$Substituting $V_f = \frac{V}{2}$:

   $$U_{\text{final}} = C \left( \frac{V}{2} \right)^2 = C \frac{V^2}{4}$$Simplifying:

   $$U_{\text{final}} = \frac{1}{2} \cdot C V^2 \cdot \frac{1}{2} = \frac{U}{2}$$

Thus, the final energy stored in the system is $\frac{U}{2}$.