Aconite said:
Is there some reason this homeless man (even delusional) wouldn't recognize ice, and keep thinking it's a diamond despite the fact it's 1) cold, and 2) floating, which are not characteristic of diamonds, but are of ice? What's his survival potential on the streets if he's that bad off?
Eh, I think you could
make it believable. For example, suppose the tramp's a little delusional, and he's obsessed with finding "treasure." So every morning, he takes his home-made "metal detector" (really just a mop handle, garbage can lid, and cast-off headphones) down to the mud flats by the lake and prospects. In the last ten tears, he
thinks he's found millions of dollars of Spanish doubloons, jewelery, bearer bonds, and the like -- really just flattened beer cans, plastic beach toys and litter.
So anyway, this particular morning, the tramp's out prospecting on the mud flats at the water's edge, and an early morning fisherman buzzes by out on the lake, sending some little waves rolling out over the mud flat. The tramp scampers out of the way and shakes his fist and the fisherman. He walks back down to the water's edge and... what's this? Washed up on shore, out of the water, is... Howard Hughes's wallet, and... a diamond! A huge diamond! The tramp's never seen a diamond that large. It's amazing how cold they feel.
Seem more plausible now?
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Anyway. Back to the original question in the first post. I know enough about thermodynamics to know that the answer to your question would be tough to calculate. In fact, I'm gonna prescribe an experimentalist approach.
Required:
One freezer, working order.
One empty gallon milk jug.
Water dispenser.
Twelve feet rope.
One utility knife.
One large-size cooler.
One dozen individually-packaged twelve-ounce liquid containers.
Digital cronograph.
One lake with dock.
Personal bio-thermometer.
Procedure
1. Fill gallon milk jug from water dispenser. Tie knot in one end of rope; dangle rope into top of jug so knot is in the center.
2. Place gallon milk jug in freezer. Be careful of wicking action up rope.
3. Remove jug from freezer. Unchip rope. Run jug under warm water. Use utility knife to cut away jug from ice (seriously, be careful here. Ice is slippery).
4. Place ice in cooler. Fill remaining space with individually-packaged twelve-ounce liquid containers.
5. Drive to lake.
6. Insert personal bio-thermometer ("big toe") into lake. Estimate temperature. Call this temperature T
L.
7. Tie rope to dock. Throw ice into lake. Note time on digital cronograph.
8. Consume liquid within individually-packaged twelve-ounce liquid containers as appropriate.
9. Occasionally check progress of ice. When ice reaches size of fist, note time on digital cronograph.
10. Drive home. Depending on consumption of dividually-packaged twelve-ounce liquid containers, you may wish to delay this step.
Calculations
1. The ice is, for the most part, 32 degrees. You've estimated the temperature of the lake (T
L). The difference (TL - 32) is Delta-T.
2. To get from gallon-jug-size to fist sized requires about 3" of ice be removed from each side of the jug.
3. The time this took was, of course, the difference between your two watch readings. Call this time t.
4. So you've experimentally shown that a temperature difference of Delta-T removes 3" of ice in a time t.
5. The time taken to melt any size chunk of ice should be proportional to thickness and temperature. In other words:
t = (Constant)*(T
L - 32)*(3")
Your experiment determined the constant. Now adjust the thickness and lake temperature to get what you want.
Make sense?