What would happen to the temperature of a room and the water, if a room (with a bucket of water at inside room temperature of 30deg.C) is converted to vaccuum?
Jd
I am not quite sure what you are asking here but I think you are asking what would happen if we had a bucket of water in a room and then we instantly removed all of the air and created a vacuum?
If that is the case then it depends on the size of the room and the bucket. Most likely the water would nearly instantly evaporate and become colder. If the room were small enough compared to the bucket (I won’t hazard a guess as to the size) then some water will evaporate and some will remain in the bucket. Assuming the room is a closed system i.e. no heat transfer in or out the water and room will end up slightly colder. How much colder depends on the specifics of the room material and construction along with the amount of water.
If you hook up a vacuum pump to a room, and if the room happens to be airtight and strong enough to withstand the pressure differential, then as the pressure decreases the water starts to boil, and its temperature starts to drop. (Basically, you are allowing the faster-moving water molecules to escape, and the remaining molecules are the slower ones.) It may cool enough for some ice to form, but eventually even the ice will evaporate (sublimate) and nothing will be left in the bucket.
The room temperature will initially drop for the same reason, but not by much. (I use vacuum chambers at work all the time and never noticed any temperature drop during pumpdown.) After vacuum is achieved, walls do not receive heat from the vacuum, nor lose any heat to it. (It’s not like outer space where you can lose heat by radiation to the dark interstellar space; all the walls in the room are radiating and receiving heat to each other, so there’s no net heat transfer.) So the wall temperature will be whatever the temperature outside the room is.
I should add that once vacuum is achieved, the only temperature you can define is the radiation temperature - i.e. the equilibrium temperature an object would reach by exchanging radiation with the surroundings. This is basically the average temperature of the walls of the chamber/room.
I see treis gave a different answer, but only becaused we used different assumptions. I assumed that a vacuum pump is connected to the room and run continuously. This will eventually remove all water from the room. But if the pump were stopped after the air is removed, then as treis said, the room will be filled with water vapor, and some water and/or ice may remain in the bucket.
I agree with ** scr4 **'s analysis using his assumptions if the OP wants to give us more information we can give him a more specific answer. I would also like to add for the record that I love it when I barely beat someone to a post and hate it when I barely get beat 
.
Now I am confused. I threw in an imaginary question derived from the fact that the exhaust out of a LP turbine of a steam power plant actually condenses in a condenser which is at near vaccuum.
So what happens in a vaccuum to the water - condensation or evaporation?
jd
“Near vaccuum” does not mean “acts very nearly as a vaccuum”. I would posit that the temperature of air remaining in said near vaccuum is much lower than the temperature of the steam exiting the turbine.
If it’s connected to an exhaust, then it can’t really be a vaccum, can it?
Where did you read that a condenser operates at a near vacuum?
Lowering the pressure of water or any liquid for that matter will cause it or at least some of it to evaporate. Certainly if it is a gas already moving it into a near vacuum none of it should condense into a liquid.
Steam condensers do run at as much vacuum as possible and do condense steam to water.
http://www.engineersedge.com/heat_exchanger/large_steam_condenser.htm
If you inject steam into a vacuum chamber, it will expand and cool down. It would condense into liquid, unless the pump is powerful enough to remove the steam before it condenses. After that, if the water is allowed to draw some heat from the chamber walls, and if the pump is powerful enough, the water will evaporate again and be sucked out by the pump. I guess the pumps in steam condensers aren’t powerful enough to re-evaporate a significant amount of room-temperature water, or the continuous input of steam prevents pressure from dropping that low.
(Disclaimer: my only experience is with laboratory vacuum chambers. Una Persson is much better qualified to answer about steam condensers - I wonder if she does vanity searches?)
A vacuum is a lack of everything. It has no properties.
No properties? Volume is a property, isn’t it? What about pressure? Even a perfect vacuum has a very well defined pressure (zero). And as I explained above, a vacuum chamber (or any region of vacuum) has a temperature.
In practice the pressure isn’t exactly zero. It can be significantly higher, because the word “vacuum” can mean “lower than atmospheric pressure” (cf: vacuum cleaner). And even the best laboratory vacuum we can make has a measurable non-zero pressure.
Evev a perfect vacuum (no matter at all) has some pressure due, at least, to the Kasimir (sp?) effect.
Yes, and this level of pressure is defined as zero. 
Any less is negative.
Not quite. A vacuum has three space dimensions, electric permittivity (8.854210[sup]-12[/sup] Farad/meter), magnetic permeability (12.566410[sup]-7[/sup]Newton/Ampere[sup]2[/sup]. Cosmologists can address vacuum energy.
jd2838, I think that there are three important differences between your OP’s “room with a bucket of water” and an industrial steam condenser as described in yoyodyne’s link:
[ol]
[li]The continuous inflow of steam in the industrial condenser makes system a dynamic one, so the application and maintenance of a vacuum will be very different from your OP example.[/li].
[li]The industrial condenser has very large pipes with cold flowing water, used as a heat exchanger. Beyond any consideration as to whether “a vacuum” will lead to condensation or evaporation, it’s these heat-exchange pipes that do most of the work of cooling the steam. They weren’t in the OP’s “room” analogy, and that’s bound to skew the answers.[/li].
[li]The word “vacuum” means different things to an industrial plant designer and a laboratory scientist, and your OP gave no indication as to which was meant. The former will consider a “good vacuum” to be 29 inches (see yoyodyne’s link), which corresponds to 736.6mm of mercury. Since scientists measure “pressure” not “vacuum” (i.e. a perfect vacuum will have 0 mm Hg of pressure), we have to subtract the “industrial vacuum” reading (736.6mm Hg) from atmospheric pressure (760mm Hg) to get the real pressure in the vessel. This gives us (760 - 736.6 =) 23.4mm Hg, which a lab scientist will also call 23.4 Torr.[/li]
No self-respecting lab scientist would consider 23.4 Torr to be a “vacuum”, however; it’s more of a low-pressure gas. As I’m sure scr4 will agree, even a poor-quality vacuum in a lab is measured in tens of millitorr (i.e. the base pressure of an oil-sealed rotary pump), and a good vacuum will be orders of magnitude better (i.e. lower pressure) than that. 10E-7 Torr is common, and high-vacuum experiiments routinely go down to 10E-11 Torr. How far you go depends on what you are studying.
On the other hand, the designer of the industrial condenser has huge quantities of steam coming into his/her “vacuum”, so 23.4 Torr (i.e. “29 inches”) is a worthy goal. The physics are the same, but the industrial designer and lab scientist have different expectations of what constitues a “good vacuum”.
[/ol]
treis and scr4 both answered perfectly correctly for a system as described in the OP, i.e. one in which there is no incoming flow of steam and no huge array of cold-water-carrying pipes.
>Yes, and this level of pressure is defined as zero.
Eh? Why isn’t zero pressure defined as zero force per area? Am I missing a joke here?
Permit me to disagree with the previous analyses. The partial pressure of water in a room containing a bucket should be completely independent of the partial pressures of nitrogen or oxygen. And in an ordinary room, the partial pressure of water is very low indeed. If you had some magical way of instantly whisking out all of the nitrogen and oxygen and other trace gasses but leave the water vapor in place, then the system would remain at equilibrium (assuming that it was at equilibrium before) with the same amount of water in the bucket as before and no change in the temperature of the water vapor remaining in the room. If you used a vacuum pump to very quickly remove all of the gasses in the room, then sufficient water would evaporate to bring the system back into equilibrium, but this amount of water would probably be negligible compared to the amount of water still in the bucket, so the temperature of the bucket-vapor system would drop only slightly.
Absolutely.
Here’s a definite source of disagreement. The vapor pressure of water at 30 degrees C (as per OP) is 31.8 mm Hg (aka 31.8 Torr). To anyone in a lab, the terms “enclosed space with 31.8 Torr of vapor” and “vacuum” are not synonymous, and the OP specified “vacuum”. If you attach a vacuum pump to the OP’s airtight, non-outgassing-walled, bucket-containing room and start to pump, the following will happen:
[ol]
[li]Initially, the pressure will drop linearly, limited by the throughput of the pump.[/li][li]Next, the pressure will fall following a negative-exponent curve, asymptotically approaching the pump’s base pressure (for an oil-sealed rotary pump, this is < 10E-1 Torr).[/li][li]If the water bucket is large enough, there will be a pressure plateau at ~32 Torr as water vapor is pumped out but replaced by evaporation of water from the bucket.[/li][li]Finally, the bucket will be empty, and the pressure will continue dropping as in step 2 until it reaches the pump’s base pressure.[/li][/ol]The above description is approximate, and the importance of the various stages will depend on the relative sizes of the room, the bucket, and the pump.
I am sure that this is what scr4 had in mind when analyzing the OP. He’s going from his real-world experience with vacuum systems, which sounds similar to my own. Any other analysis has to make broad assumptions of what the OP means by the term “vacuum”, and IMHO it’s a real stretch to allow “vacuum” to mean “vapor pressure=32 Torr”. [The industrial condenser designers get a pass from me because of the scale on which they operate.]
[BTW, Chronos, I’ve pumped down a vacuum chamber in a lab within a couple of miles of you – fortunately there were no water buckets involved, or it would have taken a lot longer!]