1. Given data

• First capacitor: $C_1 = 60\,\text{pF} = 60 \times 10^{-12}\,\text{F}$
• Initial voltage: $V = 20\,\text{V}$
• Second capacitor: $C_2 = 60\,\text{pF}$, initially uncharged

2. Initial energy stored in the first capacitor

The electrostatic energy in a capacitor is

$U = \frac{1}{2}CV^2$

So initially,

$U_i = \frac{1}{2}(60 \times 10^{-12})(20)^2$

$U_i = \frac{1}{2}(60 \times 10^{-12})(400)$

$U_i = 12000 \times 10^{-12}\,\text{J} = 12 \times 10^{-9}\,\text{J}$

$U_i = 12\,\text{nJ}$

3. After connecting to another identical uncharged capacitor in parallel

Since the capacitors are identical and connected in parallel, the charge redistributes equally.

Initial charge on first capacitor:

$Q = C_1V = (60 \times 10^{-12})(20) = 1200 \times 10^{-12}\,\text{C}$

Total capacitance after connection:

$C_{\text{eq}} = C_1 + C_2 = 120\,\text{pF}$

Final common voltage:

$V_f = \frac{Q}{C_{\text{eq}}} = \frac{1200 \times 10^{-12}}{120 \times 10^{-12}} = 10\,\text{V}$

4. Final energy stored in the two-capacitor system

$U_f = \frac{1}{2}C_{\text{eq}}V_f^2$

$U_f = \frac{1}{2}(120 \times 10^{-12})(10)^2$

$U_f = \frac{1}{2}(120 \times 10^{-12})(100)$

$U_f = 6000 \times 10^{-12}\,\text{J} = 6 \times 10^{-9}\,\text{J}$

$U_f = 6\,\text{nJ}$

5. Energy lost

$\Delta U = U_i - U_f = 12 - 6 = 6\,\text{nJ}$

6. Final answer

$\boxed{6}$