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19 answers

I think it would be easier for ice to return to liquid form since they are more "condense" than the steam which can be any where in space.

btw lol @ vegasbill1963

2006-06-07 17:10:36 · answer #1 · answered by Anonymous · 0 1

I assume you mean the entire volume returning to its liquid state and that you are conducting the experiment at room temperature (25C).

In that case all of the steam would condense into water and only a small amount of the ice would melt during the same period.

Because for a given volume there would be a lot more water molecules present in the solid form than in the gaseous form, the amount of energy available to maintain the water as a gas (steam) would be far less than the energy required to melt the ice. If the ice and steam where present in equimolar quantities then the outcome might be a tie.

2006-06-08 00:20:23 · answer #2 · answered by uselessadvice 4 · 0 0

Its a heat transfer problem. You have to know the setup (materials used to enclose the system, outside temperature, the fluid surrounding the container and the condition of this fluid, moving or stationary, etc). This will involve engineering calculations which must be done with a background in heat transfer subject (one semester course).

So we cannot answer your problem as is. You have to supply the details. But even if you supply the detail we would still go through tables tabulating the properties of steam, ice and liquid water. We have to choose from a host of empirical equations to suit your problem, believe me there are lots of them, heat transfer course is not an exact science. Granting we have done the calculation, it is an engineering result, therefore its not exact. If this is done in the classroom the prof after the last digit is written on the board he will put a comma and add a comment: 20% plus minus allowance for error.

Interested?

2006-06-08 01:20:51 · answer #3 · answered by dax 3 · 0 0

Well it would be largely dependent on the volume actually. assuming that pressure is at standard pressure and temperature is room temp. because of course, freezing temp or temperatures well above 100 will cause them to either freeze or steam without the other turning liquid. Very small volumes would mean the ice had HUGE surface area/.volume ratio. It would melt intensly fast. But larger volumes would give the ice block a very very very small Surface Area/Volume ratio, giving the steam plenty of time to cool of and condense. I hope this helped.

2006-06-08 00:12:57 · answer #4 · answered by Tim-Timothy 1 · 0 0

Interesting. Remember it takes a lot of latent heat to turn ice to water and water into steam. If you are looking at equal volume then the steam will turn to water first

2006-06-08 04:11:30 · answer #5 · answered by bwadsp 5 · 0 0

Assume that steam is at 100 degree centigrade and ice at 0 degree centigrade and the pressure is one atmospheric pressure.

For the steam to change to water heat has to be removed where as to change ice to water heat has to be given.

If we place both of them in a thermally isolated place from the surroundings, then steam will give heat to ice and ice will take heat from steam.

Unit volume of steam has a mass of 0.6Kg and Specific Latent Heat of vaporization of steam is 2260000 J/ kg

Unit volume of ice has a mass of 920 Kg and Specific Latent Heat of Fusion is 335200 J/ kg.

To convert 0.6Kg steam to water it requires 0.6 x 2260000 = 135600 J.

This amount of heat is given by 0.4045346 kg of ice.

Thus, even after all steam is converted into water, 919.6 Kg of ice will remain as ice.

2006-06-08 06:16:53 · answer #6 · answered by Pearlsawme 7 · 0 0

Under what conditions? If you add heat to the ice too quickly you may just get steam... if you cool the steam to rapidly you may just get ice...I guess I don't understand the question.

2006-06-08 04:13:09 · answer #7 · answered by eggman 7 · 0 0

I think steam would return to liquid form faster.

But if you try this experiment in the Desert, I think the winner would be ICE.

2006-06-08 01:32:16 · answer #8 · answered by simba 3 · 0 0

Let us consider 1cc each of steam and ice
1 cc steam will weigh .6 g
The latent heat of evaporation of steam is 540 cal/g
So 1 cc steam will have to lose 540*(0.6) approx 330 cal
1 cc of ice will weigh .9 g approx.
The latent heat of fusion of ice is 80 cal/g
So 1 cc of ice will have to acquire80*(0.9) cal
so in my opinion the ice would return to its liquid state first

2006-06-08 05:12:52 · answer #9 · answered by raj 7 · 0 0

steam

2006-06-08 01:16:39 · answer #10 · answered by Anonymous · 0 0

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