Re: Ice versus water in cooler?
- From: bonomi@xxxxxxxxxxxxxxxxxxxx (Robert Bonomi)
- Date: Fri, 03 Oct 2008 18:40:00 -0500
Dusty's analysis is "mostly" right, and probably 'close enough for all
practical purposes'[1].
Also, your question, whether you realize it or not, is 'imprecise'. <grin>
There are at least two ways to measure 'efficiency' in this situation --
1) how much of the 'warmth' of the beer is absorbed by the ICE,
2) how fast the temperature of the beer reaches the desired point.
The new ice can absorb a _fixed_ amount of heat from the surroundings,
before things stabilize at any specified temperature.
Now, *IF* the old beer and water bath is _above_ that desired temperature,
you're going to have to cool the existing water as well as the beer (already
there and/or the 'newly arrived' stuff) to get the beer to the target
temperature.
Thus, from one viewpoint, it is 'more efficient' -- in this situation -- to
dump the water, and use the ice to cool -only_ the bear.
If the water bath is "at or below" the target temperature for the beer,
then, in -that- situation, it is desirable to _keep_ the water. Because
th 'excess' heat from the (new) beer will pulled into the water, and all
that heat _plus_ some from the water itself, will be pulled into the ice.
Complicating the issue is "how fast" the heat moves from one point to
another. This happens in different kinds of materials at different
rates. 'Stationary' air is 'medium lousy' (at best) at letting heat pass
through it, and can carry a fairly small amount of heat per unit volume.
Water is a much better 'heat conductor', _and_, has well over 1,000 times
the heat carrying capacity per unit _volume_.
Thus, "all else being equal" (but it isn't!) the rate of heat transfer out
of the beer container submerged in "ice water" (A water and ice mixture that has
stabilized at 32F) will be faster than one surrounded by 32F ice and air.
BUT, the rate of heat transfer is affected _both_ by the 'thermal resistance'
of _both_ materials involved, *and* the temperature difference between them.
'Straight' ice, direct from a freezer case, is going to be considerably
colder than 32F, "how much" colder depends on the set point of the freezer
case _and_ how long the ice has been out of it. Thus, for the areas of
the container in 'direct contact' with the 'below freezing' ice, the heat
transfer will be higher than that which is in contact with 'ice cold'
water.
Of course, the ice itself ISN'T in direct contact with the _entire_ container,
just small spots on it. The rest is 'insulated' by air between the ice
and the container. This slows down the heat transfer rate away from the beer.
(note: the air temp. is relatively immaterial. Because the air can absorb
so little heat relative to the ice and/or water)
*BUT*, a metal beer-can is a much better heat conductor than the beer
itself, so the can tends to be of "approximately" uniform temperature,
with the entire can surface functioning to transfer the heat from the beer
to the ice. Glass is a better conductor of heat than air, but that's
not saying much. <wry grin>
In case it isn't obvious, <grin> there are a _lot_ of things going on
in the situation you hypotheticate, with some of them working at cross-
purposes to each other.
To attempt "meaningful" analysis of 'which way is better', in any
particular situation, one has to know "how much of _what_" (old beer,
new beer, 'cool' water, 'old ice',if any, and new ice) and WHAT
TEMPERATURE each is at, as well as the 'target' temperature.
Some 'general' rules for maximum cooling 'efficiency':
1) get cans, not bottles.
2) keep the amount of water to the -minimum- needed to surround the
cans and ice. This minimizes the amount of 'extra' mass you need
to cool, to get the 'everything' to the desired temperature.
3) if the water has gotten 'warm enough' it _does_ make sense to throw
it -all- out, and start with just the "fresh ice".
Figuring out what is 'warm enough to throw out' for the water bath is a
non-trivial exercise.
I'm _guessing- that if the water has warmed up to halfway between the
target ('ice cold') and the room temperature (say 77F, to keep the math
easy :), new beer, _and_ the water volume is at least that of the beer
to be cooled, that it makes sense to dump the 'warm' water, and go with
ice alone.
Another guess is that if the water is still near 'ice cold', it is better
to keep the water (subject to rule #2, above), and just add ice.
In article <4I6dnRvStZeDs3vVnZ2dnUVZ_jSdnZ2d@xxxxxxxxxxxxx>,
Just plain \"Dusty\" <RV-dragger@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Damn! An O.T. (On Topic) post. Will wonders never cease...(:-o)!
"GBinNC" <GBinNC@xxxxxxxxxxx> wrote in message
news:o99ce4910ahoac5aebjrd0e7n90te7pqa9@xxxxxxxxxx
Okay, I have an on-topic, RVing-type question for you experts.
There's a friendly debate going on in this house right now about this. I
know I've seen the answer somewhere before but cannot find it now by
Googling, so I need some informed input.
Suppose you are camping (perhaps even flatspotting <g>) with a cooler
full of beer and ice. The ice gradually melts, of course, leaving the
beer sitting in fairly cold (but maybe not "icy" cold) water. Then a
friend shows up with a new bag of ice.
Which method is more efficient at maintaining the cold -- drain some or
all of the cold water from the cooler and then add the ice, or just dump
the ice into the cooler on top of the beer and the existing water?
And does your answer change if the friend has also brought some more
beer (currently not cold) to add to the cooler?
1) Any given volume of water contains several thousand times more heat (or
"lack-of-heat") than that same volume of air. So the water will always
yield more ?T (delta T = change in temperature).
2) In the unlikely event that the air is colder than the water, then
dumping the water is your best move. If not (the usual case) then adding
the ice and any warm beer directly is the best choice making sure to not
overflow the container--a waste of cold water, but better than adding ice
(and immediately letting it get warmer from the water) and then draining
excess water.
3) There is a "corner case" where it matters little given a nearly ambient
water bath, warmer beer, and colder ice. It would take a super-computer to
generate all of those cases...(:-o)! Me? I'd sip a few of the newly
cooling beers and it all work out...(:-{})!
A final note: Heat = energy. Cold = lack of energy. You heat things by
adding energy. You cool things by removing energy. You can not "add"
cold...(a very common misconception).
I'm looking for links to actual tests, if possible, or at least to
reasonably researched squatful theories. (bill, have you done any
testing along these lines?)
Sorry. I don't have any of those. I was simply regurgitating my
high-school physics. And I'm not certain of the amount of heat in air vs
water. I know it's a lot, and seem to recollect the number 1,000 was
involved in it. But my "thousands" of times may not be entirely accurate.
I don't have the value at hand (although, given my energy related work, I
should have!), and am just too lazy to Google it this AM...
My own POV is that dumping the water is a waste of "cold" and that the
slush mixture of ice and already-cold water is more efficient than just
ice at cooling and maintaining the cold. But that's just a supposition.
I don't have a lot of ego invested in it, and I'm open to changing my
mind -- at least on this <g>.
You've already got it! Well done!
So, how'd I do?
Dusty
.
- References:
- Ice versus water in cooler?
- From: GBinNC
- Re: Ice versus water in cooler?
- From: Just plain \"Dusty\"
- Ice versus water in cooler?
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