A capacitor C is fully charged with voltage $V_0$.  After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}.$ The energy loss in the process after the charge is distributed between the two capacitors is

When two capacitors with capacitance $C_1\text{\hspace{0.33em}}and{C}_2$ at potential $V_1\text{\hspace{0.33em}}and{V}_2$ connected to each other by wire, charge begin to flow from higher to lower potential till they acquire common potential. Here, some loss of energy takes place which is given by 

Heat loss, $H=\frac{C_1{C}_2}{2\left(C_1+C_2\right)}{\left(V_1-V_3\right)}^2$

In the equation, put $V_2=0,V_1=V_0, C_1=C,C_2=\frac{C}{2}$

Loss of heat $=\frac{C\times \frac{C}{2}}{2\left(C+\frac{C}{2}\right)}{\left(V_0-0\right)}^2=\frac{C}{6}{V}_0^2$

$H=\frac{1}{6}C{V}_0^2$