# The Alkaline Battery inside My Pocket was Red Hot

1. ### McHeathen

32
I put a AAA alkaline battery in my pocket and later detected warmth. I took the battery out of my pocket and it was very hot. The only other items that I had in my pocket were a few coins and a key ring with keys and one of these plastic security fobs. I can only presume that some kind of reaction occurred between the battery and the security fob.

I would like to know the precise nature of how this reaction occurred, anyone?

### Staff: Mentor

It would have had to have been a conductive path between the ends of the AAA battery. It's a more common problem with 9V batteries, because both terminals are at one end of the battery. It had nothing to do with the plastic fob.

3. ### McHeathen

32
It was a 1.5v battery. The only things that could have caused the conductive path were my keys and my coins. How did the battery get so hot though?

### Staff: Mentor

I know; AAA alkaline batteries are 1.5V approximately. My point was about the terminal positioning.

So yes, your keys and coins had to have temporarily made a conductive path between the ends of the AAA battery. This shorted the battery, and a large current flowed. That's why it go hot. Try to avoid doing that again, obviously.

5. ### mgb_phys

8,952
V = IR
The resistance of some metal keys from one end of the battery to another is very low, so the I is quite high.
The limit to the current is the internal resistance of the battery, luckily it probably only shorted for a second or two or the battery might have burnt.

6. ### McHeathen

32
If I was high, then why were the metalic items in my pocket not hot?

7. ### mgb_phys

8,952
P = I R
The resistance of the keys is small so the power dissipated in them is small, also they will quickly cool down.
The battery has a larger internal resistance and so more heat is generated, also there was enough heat to damage the battery so more heat will have come from chemical reactions in the battery - the battery burning if you like.

8. ### Dickfore

The battery has a fixed voltage V. If the total resistance in the circuit is R, then the current flowing through the circuit is:

$$I = \frac{V}{R}$$

The total power released in the resistor is:

$$P = R \, I^{2} = R \, \frac{V^{2}}{R^{2}} = \frac{V^{2}}{R}$$

Notice that the power is inversely proportional to the resistance. The maximum power will be released when the resistance is minimal. We cannot make the resistance in the circuit to be less than the internal resistance of the source. This happens when we connect the ends of the source by a wire with negligible resistance. This is called short-circuit. The keys and coins essentially create a short-circuit.

When you say the battery was red hot, I presume it was no more than 60 oC, since anything above that would melt or burn your clothes. The AAA batteries have dimensions 44.5 mm in length and 10.5 mm in diameter. This gives a surface area of the side equal to:

$$A = \pi \times 10.5 \, \textup{mm} \times 44.5 \, \textup{mm} = 1.47 \times 10^{3} \, \textup{mm}^{2} = 1.47 \times 10^{-3} \, \textup{m}^{2}$$

According to Stefan's Law, the intensity of radiation is:

$$I = \sigma \, (T^{4} - T_{a}^{4})$$

where $T = 333 \, \textup{K}$ is the absolute temperature of the surface and $T_{a} = 293 \, \textup{K}$ is the ambient temperature. Here, $\sigma = 5.670400(40) \times 10^{-8} \, \textup{W} \, \textup{m}^{-2} \, \textup{K}^{-4}$ is the Stefan Boltzmann constant. From these figures, we get an intensity of:

$$I = 2.79 \times 10^{2} \, \frac{\textup{W}}{\textup{m}^{2}}$$

$$P = 0.41 \, \textup{W}$$
A current of $0.41 \textup{W}/1.5 \, \textup{V} = 0.27 \, \textup{A}$ could supply this power and for this, the total resistance in the circuit has to be $1.5 \, \textup{V}/0.27 \, \textup{A} = 5.5 \, \Omega$. This indeed corresponds to a short circuit.