If you had an ideal battery and a non-ideal wire, the wire would have some very small but nonzero resistance. Then there would be a very large current, and 6 Coulombs would go by very quickly. The 6 Joules would all be deposited as heat in the wire.
If you had an ideal wire as well, then the energy would become kinetic energy of electrons, at least until they slammed into the positive terminal at the end of the wire. Then their energy would be converted to heat. But such things as ideal batteries and ideal wires don't exist, so this last scenario is something of a physics fantasy.
Batteries and capacitors are not the same. If you connect the two leads of a battery with a low resistance wire, the maximum current the battery can deliver will travel down the wires turning to heat. current squared x resistance will be the power dissipated
The wire will burn up if the battery has enough current capacity.
I'm not going to double check your equalivance factor; but here is the real-world of electricity.
First of all there is no such thing as a zero resistance wire, the end connections alone would make for minor resistance.
Secondly; what you propose is called "a dead short," because either the wire or battery will overheat. One of them will open circuit, that's why a fuse works. I've seen wires get red hot because of this.
You'd see it as heat in the battery.
I have read that the P.E of a coulomb at the end of a circuit is zero. So if a battery has a voltage of 6V, this would mean that each coulomb would have to use up 6 joules of energy in the circuit right? So if we connect the battery with a length of wire (no resistance) , how would this energy be spent? Would it be spent as the kinetic energy of the electrons?
Please correct me if i have made any mistakes in my logic.
