Take two vessels, identical in volume, and add an identical quantity of water to each one. Bring them to boiling point and release some steam from one, then seal it again. The other remains sealed the whole time. Keep both boiling at the same temperature- which vessel is under more pressure, the one which had some steam released or the one that was never open?
This smells of homework. Under the assumption that it is, I’ll start off simply by bolding the big hint in the problem:
Two identical vessels, sealed now, one with a bit less water (and air, unless you can filter for water vapour only) in it, same temperature. Which has the higher pressure? Hmmm…
Say, you don’t work for Tepco in Japan, do you?
Nope, not homework, honest. Actually an engineer pal posed it as a brainteaser and I’ve been going back and forth on it over the weekend. And I figured they both were the same- then I thought, hang on, I work in music and my science knowledge begins and ends with amps, reverb and mics. So… any ideas folks?
Well, it’s been a few years since my last physics class, but I’m going to go with “they’re the same, assuming that the volume of steam that escaped led to a negligible difference between liquid water volumes.”
And now I no longer have enough interest to write an explanation. Shoot.
The not-homework story is convincing enough for me…
I’m assuming the vessels have nothing in them except water. In this case, the boiling temperature of the water depends only on the pressure of the water vapor in the vessel. Since the problem states that the water is boiling in bother vessels at the same temperature, the pressure must be the same.
When one vessel is vented, it will boil off some of the water until the pressure has increased back to whatever the co-existence pressure is for water liquid and gas at the fixed temperature.
If the vessels have other things in them (air?), then the answer is less straightforward, but I don’t think this is what the problem is after.
Assuming the two are truly kept boiling at the same temperature, then it doesn’t matter whether the vessels have the same volume or not and it doesn’t matter if the two vessels start with the same amount of water or not.
The blue line in this graph is a property of water, full stop. Boiling occurs as you try to cross the blue line. During boiling (i.e., as long as there is coexistence of the liquid and the gas), the temperature and pressure are related by that line. So, fixed temperature means fixed pressure.
my thoughts: steam pressure is increased by raising temperature. since the volumes of the two pots have not changed substantially, and the temperature is being maintained between them, i would think there would be a negligible pressure change.
I’m wondering if it’s that simple though. The vented vessel contains water and mostly steam in its hullage space. The unvented vessel contains water, steam and air in its hullage space. Not sure if it makes a difference.
It is that simple under the assumptions I gave in my post:
Another note: I interpret “keep both boiling at the same temperature” to mean keep the two vessels are kept at the same temperature as one another, but not necessarily at a fixed temperature. Indeed, as the liquids boil, the pressure (and, thus, boiling point) also rises. The temperature must be continuously increasing throughout the “experiement”.
After the rest of the discussion in this thread, I’ll go with “the same” answer everybody else did, but I insist the problem has to be changed then. The part in red has to be “Return both to boiling at the same temperature”–otherwise, immediately after the pressure release, you’re not going to be able to “Keep both boiling at the same temperature”, right?
Wow- thanks everyone for the replies. I think I’ve got it, and I’m going with the same temperature-same pressure answer.
If anyone wants to know, I’ll post back with what my engineer friend thinks the answer should be- he’s (unfortunately?) quite fond of trying to catch us out with seemingly difficult physics questions that often have an obvious answer 