The problem isn’t the little bit of gas that has leaked out of the battery, it’s the pressurized gas in the battery. Batteries are designed to slowly vent the hydrogen gases that are generated during the charging cycle. If the battery is under a heavy charging load, such as while on a good battery charger or while being jumped, the gases build up much faster than can be vented. The deader the battery, the more gases that are generated. Add a spark and kaboom. I use to have an 18 volt, 400 amp battery just for jumping dead batteries, I also blew up a couple of regular 12 volt batteries with it when I was a bit careless. Many tow truck carry this type of battery just for that purpose. Look in the cab of any tow truck and you will find boxes of baking soda just in case a battery does explode, the baking soda will neutralize the battery acid.
But how could there be oxygen to support combustion in the inside of the battery? Pressurized hydrogen and oxygen inside could easily catch, but what chemical process is producing oxygen?
I don’t doubt Rick, but it seems to me that it’s more likely to be something else that caused the battery to explode rather than hydrogen catching on fire.
I did a little reading up on it, and according to what I found it’s electrolysis of water in the electrolyte that’s the source of hydrogen, and if that’s correct it would of course be a source of oxygen as well.
My ignorance on this matter has been successfully fought, it would seem.
Battery electrolye is H2SO4 and H20. (it’s late and I am too tired to do sub tags. deal with it. 
) Anyway as the battery discharges the SO4 ions are deposited onto the plates and the peroxide ions are released into the electroyte. The hydrogen ions combine with the peroxide ions and water is the result. As a battery discharges its specific gravity gets closer and closer to that of water. When you go to recharge the battery these reactions are reversed. However under some conditions, the water will disassociate instead of the reaction reversing. This give you two parts hydrogen to one part oxygen in and around the battey.
Add a spark and you get a rrather large boom. :eek:
Well, I’m glad I read this thread. I always figured this was a miniscule risk, since hydrogen released from the battery would most likely dissipate. But obviously hydrogen and oxygen inside the battery are an entirely different matter. I’ll make sure to stick the jumper cables on the right spots from now on.
Just one more story about exploding batteries; during my VW mechanic days, I managed to explode a battery. In my case, the battery was sitting on a work bench and I was connecting a battery charger. I didn’t get acid in my eyes, but I did ruin a set of coveralls. I also had to endure a very cold hosing off administered by another mechanic.
IANAEE, but this can’t be right. Using your logic a #10 gauge wire with US house voltage of 120VAC will generate more heat than a tungsten filimant in a light bulb at the same voltage. If this were the case then the walls of my house catch fire when I turned on the lights.
It will generate more heat if you connected a #10 gauge wire directly across a 120V source, which is what is happening to a tungsten filament. The reason why the walls of your house don’t catch fire is that your house wiring is in series with high-resistance loads, so the current is limited by those loads and not by the #10 wire.
Or to put it another way, the voltage drop across the #10 wire leading to your bulb is not the full 120V but is only a fraction of a volt, while the voltage drop across the filament is practically the full 120V. So they aren’t equivalent situations.
But Chronos said “Given a constant (or nearly so) voltage source like a battery, less resistance generates more heat.” For any particular voltage drop, a 12 gauge wire (less resistance) will generate less heat than a 18 gauge wire (more resistance).
No, Chronos is right.
Consider a 100 watt light bulb and a 200 watt light bulb. The 200 watt bulb gives you more light, and so more heat, correct? Now, which one has the higher resistance?
If we assume the voltage drop for both is exactly 120 volts, we can back into the resistance. Since watts = volts x amps, we calculate the amperage draw as follows:
100 watt bulb = 120 volts x ? amps = 100 / 120 = .83 amps *
200 watt bulb = 120 volts x ? amps = 200 / 120 = 1.66 amps *
And since we know that resistance is Volts / amps, so we calculate the resistance as:
100 watt bulb = 120 volts / .83 amps = 144.5 ohms *
200 watt bulb = 120 volts / 1.66 amps = 72.3 ohms *
So we see that for a constant voltage drop (120 volts) a bulb with 72 ohms resistance gives us more light (and more heat) than a bulb with 144 ohms resistance.
The same is true for the 12 gauge wire vs. the 18 gauge wire. For a given voltage drop, lower resistance means more current, more current means more power dissipated, more power means more heat.
The reason most people get confused on this is that you’re not used to thinking of the wire as a load. In a typical circuit, heavier wire (lower resistance) stays cooler because it has less resistance than the load in the circuit, so more voltage is dropped across the load and very little voltage is dropped across the wire. So the wire stays cool.
But decrease the wire size until it the resistance goes high enough to drop a significant voltage compared to the load, and wire starts to heat up. So you’ve increased the resistance and made the wire get hot.
But this isn’t the situation Chronos was talking about. By increasing the wire’s resistance, you’ve increased the voltage drop (and decreased the voltage drop across the real load) so now the wire heats up.
What Chronos applies only to a constant voltage drop. This doesn’t happen when you increase the resistance of the wire in a circuit where there is a load besides the wire itself.
In the situation of jumping a car, as the wires heat up, the resistance will increase, and the resistance goes up, and the current will start to drop off. As the current drops off, the cable will cool slightly. Eventually, the cable will reach an equilibrium temperature and current capacity, assuming it doesn’t melt first. But it’s not a runaway situation, rather it is a self limiting one.
Clear as mud?
(* Ok, it’s been nearly 30 years since I took this stuff in school, but I think all those numbers are correct.)
Without more information, it’s impossible to say whether the wire heating up and becoming more resistive will increase or decrease the amount of heat generated in the wire. In the case here, we’d need to know the resistance of the jumper cables and the how much other resistance is in the circuit. To say that less resistance generates more heat, Chronos and RJKUgly are assuming (whether they realize it or not) that the resistance of the jumper cables is larger than all the other loads in the circuit (the internal resistances of the good and dead batteries).
One of the things every EE is taught in their first circuits class is that maximum power to the load occurs when the load resistance matches the source resistance. For those of you who aren’t EEs, but want to play along at home, for a battery and a load (the jumper cables), the total resistance in the circuit is
R[sub]jumper[/sub] + R[sub]battery[/sub],
so current is
I = V[sub]battery[/sub] / (R[sub]jumper[/sub] + R[sub]battery[/sub]).
Energy turned into heat in the jumper cable is
I[sup]2[/sup] * R[sub]jumper[/sub] = V[sub]battery[/sub][sup]2[/sup] * R[sub]jumper[/sub] / (R[sub]jumper[/sub] + R[sub]battery[/sub])[sup]2[/sup]
It’s easy to see that there is zero energy loss in the jumpers in the limits R[sub]jumper[/sub] = 0 and R[sub]jumper[/sub] --> infinity. It’s not hard to show the energy loss is a maximum when R[sub]jumper[/sub] = R[sub]battery[/sub].
I don’t know what the internal resistance of a good or bad battery is, but my hunch is that they’re higher than the resistance of the jumper cables. If this is the case, Kevbo is right when he says that “Hot thin wire has even higher resistance than cold thin wire, so generates even more heat.”
It’s also possible that some of the strands inside the cable were broken. Most of the cables people own, in addition to being of smaller overall diameter, are made up of many to dozens of extremely small-diameter cables stranded together (like a rope.)
As the cables get older, flexing it weakens the individual strands, causing eventual breaking of some of those strands. As each strand breaks, the same amount of current still has to go from the good battery end to the bad battery end, but there are less individual conductors available to do this (a broken wire will not pass current.) More current through each individual strand will increase the heat generated in that strand.
As the cable gets older and more strands break, this process increases, and you’ll eventually get to the point where too much heat is generated for the cable to dissapate to the surrounding atmosphere.