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  #21 (permalink)  
Old 13th August, 2004, 05:57 PM
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Quote:
Originally Posted by Áedán
he decided it wasn't worth posting...
That's never stopped me. I post crap that's not worth posting all the time. Take this message, for instance. LOL
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Old 13th August, 2004, 06:34 PM
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Quote:
Originally Posted by gizmo
That's never stopped me. I post crap that's not worth posting all the time. Take this message, for instance. LOL
Sorry Giz.
I realized I was really not contributing to the thread... just describing the physics of it in a different way, so I deleted it.
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Old 13th August, 2004, 08:16 PM
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The thought about the total volume change was intriguing. I wonder what the pressure change would be? In a small system with a small amount of ice, I wouldn't expect it to change all that much, but I wasn't thinking through all of the ramifications to a totally closed system; I was thinking in terms of the ice floating in an open cup.
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Old 13th August, 2004, 08:29 PM
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My original thought had been that if you removed an certain volume of water and replaced it with an equal volume of ice - since ice is about less dense than water then the end result would be a lower volume. I'm guessing that you're thinking about removing a certain mass of water, and replacing it with an equal mass of ice, in which case end volume of the system would remain unchanged.

Ice is about 9% less dense than water, so in an equal-parts system with 100cc water and 100cc ice, the initial volume is 200cc and the final volume 191cc. In a closed system with fixed volume 300cc at 1bara, then P1V1=P2V2, the pressure after melting is 1*100 / 109 = 0.917 bara
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Old 13th August, 2004, 10:03 PM
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Yeah, we were approaching the problem from opposite ends, looks like.
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Old 13th August, 2004, 10:16 PM
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Put 100 engineers in a room and you'll get 100 different ways to reinvent the wheel
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