A capacitor is charged by a battery. The battery is removed and...

Charge on capacitor,
$q = CV$

When it is connected with another uncharged capacitor.

${V}_{c} = \dfrac{{q}_{1} + {q}_{2}}{{C}_{1} + {C}_{2}} = \dfrac{q + 0}{C + C}$
${V}_{c} = \dfrac{V}{2}$
Initial energy,
${U}_{i} = \dfrac{1}{2}C{V}^{2}$
Final energy,
${U}_{f} = \dfrac{1}{2}C{\left(\dfrac{V}{2}\right)}^{2} + \dfrac{1}{2}C{\left(\dfrac{V}{2}\right)}^{2}$

$= \dfrac{C{V}^{2}}{4}$

Loss of energy $= {U}_{i} - {U}_{f}$
$= \dfrac{C{V}^{2}}{4}$,
i.e., decreases by a factor of $2$.