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So, I get the ad copy, the surface area to mass ratio results in more cooling, less dilution. But does this actually make sense?

Yes, total cooling is related to the mass/temp of the ice, not the surface area, but does the portion of the ice that is not in contact with the drink (above the waterline, or interior ice) actually provide any cooling effect on an instantaneous basis?

It seems like both the level of cooling, and level of dilution are directly related to the surface area in contact with the drink.

Almost by definition, the part that isn't melting, isn't providing any energy transfer either isn't it? (I'm remembering my high school phase change graphs, where the temperature stays constant while the ice melts and only goes up again after melting)

Perhaps that is the answer? Shape of ice doesn't matter until the drink makes it to 32 degrees - you had to melt the same amount of ice no matter what to get there. But after that it melts slower due to surface area? Wouldn't the large amount of ice exposed to the air in fact melt faster now since its outside of the now cooled drink (assuming the ambient air temp is not freezing)?

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The total cooling power of a block of ice in water (or other water-miscible liquid) depends only on two things: the mass of the ice, and the initial temperature of the ice. The shape is irrelevant, except in that it changes how fast the ice absorbs heat from the water.

If we assume that the cup of water and ice are insulated, and wait long enough for it all to come into equilibrium, then no matter what the initial shape of the ice the same amount of ice will melt, and the same amount of dilution will take place.

The only possible difference is if the cup is NOT insulated and is in a warm environment. If you put a single sphere of ice in one cup, and an equal weight of snow in a second cup, then the second cup's temperature will reach freezing much more quickly than the first. This will increase the second cup's absorption of heat, and thus more ice will melt and more dilution will occur in the second cup. Once the first cup cools to freezing, though, the difference will disappear and the two cups' melting/dilution will return to the same rate.

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  • $\begingroup$ Of course in the real world your "only possible difference" is exactly what the real situation is. $\endgroup$ Sep 18 '15 at 17:20
  • $\begingroup$ Yes, but you could get the same effect by just using less non-spherical ice. $\endgroup$ Sep 18 '15 at 18:01

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