12 V car battery short circuit with a wrench that has resistance of 0.5 ohms, how much current?

A wrench does not have a resistance of 0.5 ohms, it's way lower.

Your basic multimeter cannot measure resistances to better than an ohm or so, the resistance of leads, and the unreliability of contact resistance make it impossible.

The way resistances as low as a wrench are measured is to use a 4-terminal Kelvin method. What you do here is to pass a current through the sample using two terminals, then measure the voltage across the sample using a different pair of terminals. With a wrench, if you used perhaps 1A from end to end, you would see a few mV or so voltage drop.

Let's put some numbers on your wrench. I don't like looking up resistivity, the large factors of 10 cause me concern whether I'm going to get them right on the back of an envelope, so I remember just one fact. A 1m length of 1mm\$^2\$ copper wire is about 17mohm, and then work from there.

Let's assume your wrench is 250mm long, and has a 10mm x 5mm shaft. It's 1/4 of 1m long, and 50mm\$^2\$, so is 1/200th of the resistance of my 1m x 1mm\$^2\$ wire. If it was made of copper, it would have a resistance of 17mohm/200, which is roughly 100μohm. But it's not copper, it's steel, and probably an alloy. After a quick rush around Wikipedia, let's assume it's 50x more resistive than copper, so has a resistance of about 5mohm.

12v dropped across 5mohm would give a current of 2400A. The CCA of the battery is way below that, so the wrench is not limiting the current, the battery is.

Contact resistance is a further complication. In the case of a battery shorted by a wrench, there's likely to be a plasma arc between the contacts, which can have a very low resistance indeed. The small contact area is also worth considering, though as that region is very short, it's often insignificant compared to the length of the conductor.

In practical terms the true resistance of the wrench is close to zero.The battery will deliver the maximum instantaneous current that can be extracted from its cells, which will be way lower than any calculations you make. The net effect is that the wrench will become essentially a fuse: it will burn through at its narrowest point. I have seen it happen to a crescent spanner, and it is spectacular, as it blew the head clean off. Fortunately, the person who did it was not hurt, but it was very dangerous and he was very lucky. It may well also explode the battery, particularly if the spanner is big enough to sustain the current for a little longer.

DO NOT RISK DOING THIS, IT MAY WELL KILL YOU OR AT LEAST GIVE YOU SEVERE ACID BURN DAMAGE. In short, don't be an idiot.

I think what all the other answers are leaving out is the internal resistance of the battery. When dealing with large currents and low resistance shorts, this becomes the significant factor limiting current.

An ideal battery can be modeled by a voltage source, but real batteries act more like a voltage source in series with a resistance.

$$V = IR$$

Let's imagine a hypothetical example. Say you have a 12 V battery. Now short the leads with a 0.1 ohm resistor. According to ohms law, you get 120 A.

$$V = I(R+R_{internal})$$

Now imagine that same battery, except with a 1 ohm internal resistance. With the same short, you get 1.1 ohms of resistance, or approximately 10.9 A. Big difference!

This should line up with everyday experience. When you directly short a battery, you don't get infinite current. You get it's voltage divided by it's internal resistance.

Your measurement of resistance is erroneous - The current flow will be limited by the resistance of the wrench and the internal resistance of the battery - both very low.

There will be a current flow in the region of, or more, than 1000A easily. If it were not so dangerous I would suggest that you try it, BUT I have seen batteries explode from this sort of thing...