# Which heated, partially filled bottle will explode first?

This is in reference to a pasteurization discussion on a homebrewing forum.

I have four closed bottles which will explode if containing too much pressure. Two of them are 50% full (A and B), and two of them are 75% full (C and D).

I completely submerge A and C in a water bath of 190 degrees so that I can bring their internal temperature up to 160 and hold it there for 10 minutes.

I but B and D in the same water bath, but they are only submerged up to the level of the liquid they contain.

I'm curious which is most likely to explode first. Here's my hypothesis: A will explode first since it has the most gas for expansion, and the gas will expand more than B because it's completely submerged.

If the above is correct, can a bottle completely filled be heated to a relatively high, or even very high temperature without exploding?

• When you say "75% full", presumably you mean 75% full of water, and the rest is air? – LarsH Sep 27 '12 at 18:22
• @LarsH correct. If it matters, in this case it would be some alcoholic solution around 5-10% alcohol, and CO2 – MStodd Sep 27 '12 at 18:33

Assuming you're using degrees Fahrenheit 160F is only (about) 70C so you won't be boiling the water. The increase in pressure will come partly from heating the air in the bottles and partly from the increased vapour pressure of the water.

The pressure of an ideal gas (air is near ideal in this temperature and pressure range) is:

$$P = \frac{nRT}{V}$$

where $n$ is the amount of gas, $R$ is the gas constant, $T$ is the temperature (in Kelvin) and $V$ is the volume. For any particular bottle the amount of air and the volume is constant, so we can make life simpler by taking the ratio of the pressure after heating to temperature $T$, $P_T$, and the pressure we started off with $P_0$ (the initial pressure, $P_0$, is just atmospheric pressure). So:

$$\frac{P_T}{P_0} = \frac{T}{T_0}$$

or:

$$P_T = P_0 \frac{T}{T_0}$$

and we need to add on the vapour pressure of water $P_w(T)$. As the notation suggests the vapour pressure of water changes with temperature. At 70C it's about 30% of atmospheric pressure. Anyhow our equation for total pressure is:

$$P_{total} = P_0 \frac{T}{T_0} + P_w(T)$$

Note that the volume of the air doesn't appear in this equation, so for the fully immersed bottles (where we can be sure of the air temperature) both bottles will burst at the same temperature.

When you only have the bottles immersed up to the level of the water the temperature of the air isn't well defined because it's heated by the water below it and cooled by the air above it. I would have guessed the air in the bottle with less air would be hotter simply because it sticks out of the water less. I'd expect these bottles to burst at a higher temperature than the fully immersed bottles, but how much higher is hard to say.

In the above I've assumed that you're raising the temperature gradually. Your question implies you're heating the water to 160F then throwing the bottles in. If you're doing that all bets are off because it depends on how quickly the contents of the bottle heat up. You can only do the experiment reproducibly if the put the bottles into cold water and raise the temperature slowly enough for the contents on the bottles to keep up with the temperature rise.

Finally note that this treatment only applies if you keep the temperature below the boiling point of water. Once the water starts to boil the pressure in the bottles will rise extremely rapidly.