Energy in capacitors - loss?

The energy stored in a capacitor is

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

So when I have a 1F supercap charged to 1V the energy is 0.5 J. When I connect a second supercap, also 1F in parallel the charge will distribute and the voltage will halve. Then

$$ U = \dfrac{1}{2} 2F (0.5V)^2 = 0.25 J $$

What happened to the other 0.25 J?

Answer

You moved energy from one place to another and you can't do that unpunished. If you connected the two capacitors via a resistor the 0.25J went as heat in the resistor. If you just shorted the caps together much of the energy will have radiated in the spark, the rest again is lost as heat in the internal resistances of the capacitors.

further reading
Energy loss in charging a capacitor