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Doug Miller

In article , wrote:
Doug Miller wrote:

... The objective is to remove heat from the can of beer as rapidly
as possible, and that will happen when the beer is immersed in a bath of
icewater. Not when it's immersed in a pile of ice.

I tend to agree.

And the reason for that is that water is a better conductor of heat than ice,

Water has a 0.596 W/mC conductance, vs 2.26 for ice. But it's hard to keep
the can completely in contact with ice, and that way, the heat transfer from
the ice is limited to the can surface. If the can is in a large well-stirred
bath of ice and water, the ice will have lots of heat transfer area to water,
and the can surface will be very close to the temp it would be if the entire
can surface were in contact with ice.

Well, maybe... that depends at least in part on the relative masses of the
ice, and the can of beer. Figure a can of beer at somewhere around 350 grams.
Pop a room-temperature beer into, say, 20 or 30 kg of ice at -20 deg C in a
well-insulated container, and the beer will eventually assume pretty near the
same temperature as the ice, long before any significant melting of the ice
can occur.

OTOH, put the same beer, and the same 20 or 30 kg of ice at -20 C into a water
bath, and the beer will *never* get colder than 0 C.

Which, of course, is why you add salt when you're making homemade ice cream:
so that the ice will melt at a lower temperature, giving you a water bath
that's much colder than 0 C.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.