given:
-water (bag and heating volume) have an initial volume of ~10C
-amount of water to be heated is 1000g, give or take.
-max volume of the containment vessel is 1500 to 1600ml
-minimum air is contained in the containment vessel
query:
-Will the combine volume of liquid and gaseous water at 100C displace more than 1500g of liquid water at 10C?
Using Charles' Law of Gases:
V1/T1 =V2/T2
and making the following assumptions:
1. It works as an ideal gas.
2. At boiling the liquid phase still occupies ~1000mL of the container.
3. Pressure is constant at Standard Atmospheric Pressure (760 torr).
4. The bag will rupture at anything above atmospheric pressure.
5. The initial volume of gas vapor is 0.
Then:
V2=(T2V1)/T1
where
V1=0
T1=10C=283K (add 273 to 10)
T2=100C=373K (add 273 to 100)
V2=600mL (1600 total - 1000 liquid water)
So:
V2 = (373K*600mL)/283K = 790.8mL and the bag ruptures.
Now, lets figure out what sort of pressures the bag needs to hold this water/vapor.
Boyle's Law states P1V1=P2V2
In our case:
P1=760 torr
V1=790.8mL
V2=600mL
then:
P2 = (P1V1)/V2 = (760torr*790.8mL)/600mL = 1001.7 torr = 19.4 psi
So the bag would need to withstand at least 20psi to prevent rupture during this process. A good engineer would want a safety factor of at least 40% so figure a necessary rupture strength of 25psi.
Remember, this makes some assumptions about the property of water vapor. I did these calculations before my morning caffeine dosage so don't trust your life to them.
-Blast