MadSci Network: Physics


Area: Physics
Posted By: Steve Czarnecki, senior technical staff member, systems engineering, Lockheed Martin
Date: Wed May 21 15:50:48 1997
Area of science: Physics
ID: 862499080.Ph
Cool question  :-)

As in many of life's conundrums, all we can say is, "it depends".  
Nonetheless, I will attempt a response and tell you what assumptions I made
to eliminate variables and get to an answer (i.e., I couldn't solve
your problem so I made one up that I could.  I'll argue that, in this 
case, describing
the thought process is more important than the actual result).

First off, I'm going to assume that the contents of the chest are 
waterproof, so we're interested in draining the chest only to better
keep the contents cold, not to   
keep the egg salad sandwiches dry.

Second, I'm going to assume that the chest is reasonably well insulated,
so that rate of heat transmission through the chest walls is small 
compared to the rate of heat transmission from the inner chest wall to 
the contents, be it through air or water.  This permits me not to
worry about the insulating properties of air inside the chest versus

Third, I'm going to assume that both ice and food have been added
a long time ago, and thermal equilibrium has been reached.  That means
all ice, water, and food inside the chest are at 32 deg F (0 deg C).
Therefore, we don't generally have to worry about hot spots or cold
spots, or significant convection within the chest.

Fourth, to keep my mental model simpler, I'm going to assume that the
chest has been placed in an open area exposed to free ambient air, and
protected from the sun.  For example, it's sitting in the shade of a 
tree.  Also, let's assume that we're dealing with ordinary 
summer day temperatures, not the surface of Mercury.  
This allows a simple model of heat transfer to apply.

Let's consider what happens when there's lots of ice in the cooler.
The temperature will remain 32 deg F, regardless of how much water
is there; draining water won't make things colder.  Will draining water
cause the ice to melt slower or faster?  Neither -- the rate of heat
transfer into the chest depends on the temperature difference between
the interior and exterior of the chest, i.e., the ambient air temperature.

Every BTU of heat entering the chest is offset by the melting of a 
smidgeon of ice (and if I had a CRC handy, I'd tell you exactly how much;
this is the so-called heat of crystallization.)
Melting ice remains at exactly its melting point until it's all melted,
as do all (?) other materials that undergo a well-defined phase change
(conversion from solid to liquid).

So it's only this temperature difference, and not the amount of water in
the chest, that matters.  So I claim, under the assumptions given, that 
from a "conservation of ice" standpoint it doesn't matter if you
drain the chest while the ice is melting.

Now let's consider what happens when it's time to add ice.
Of course, you will need to drain the chest eventually
to make room for more ice, but that's another matter entirely. 

There are two conditions to consider: the first is that the ice 
you are going to add has reached 32 deg F and is already melting (say,
you had brought along extra to replenish the cooler).  Under this 
condition, it doesn't matter if you drain the water upon adding more
ice.  The water won't melt the ice faster (the water is already at 32 deg

The second condition is that you are going to add ice directly out of a
freezer.  This ice will be much colder than 32 deg F, and it will not 
have begun melting.  In this case, it's slightly advantageous to have
water in the chest.  Upon adding this very cold ice to the chest, it will 
warm up to 32 deg, reach equilibrium, and we have the case previously 
consider.  So we need to consider how this fresh ice reaches 
equilibrium in the way that best conserves the "extra BTU's of cold" in 
the ice because it's below its melting point. 

If liquid water is present when this very cold deep freezer ice is added,
a small portion will freeze as it gives up heat to
warm the ice to its melting point.   Heat transfer into the chest will 
remain approximately constant, assuming that there's sufficient water to  
allow the ice to
warm and the water to freeze quickly so that 32 deg F equilibrium is
reached quickly.  This effectively stores the excess cold as a bit of
additional ice.  This is because the heat necessary to warm the deep
freezer ice up to the melting point comes from heat in the liquid water,
a bit of which freezes. (I'm assuming a couple of things here: 
first, that the rate of this heat transfer far exceeds heat transfer
through the chest walls, so this effect dominates any increased
heat transfer due to colder contents.  Second, I'm assuming that we're 
adding about as much, or less ice than we started with; if you add *lots* 
of deep freezer ice there's potential to freeze all the water, but I 
think this to be unlikely in practice.  If it should happen, then my 
argument below applies).

If water is not present, the food will try to cool below 32 deg F as it 
gives up heat; if there's enough free water in the food a little
will freeze and the food temp will hold at 32 deg F as the ice warms and
eventually begins to melt.  This is the same situation as if we hadn't
drained the chest.  

If the moisture in the food is actually a solution (e.g., due to dissolved
salt or sugar), the freezing point will be somewhat lower and the ice will
warm only to this temperature below 32 deg F.  Once all moisture is frozen,
then the food will cool below the freezing point until equilibrium with 
the ice is achieved at some temperature below 32 deg F.  In these instances
I think we will wind up melting the ice sooner than if there were water in the 

This can be seen by considering what happens to the extra
BTUs of "cold" (that is, the heat it takes to warm deep freezer ice
up to the melting point).  I've explained before how freezing the
liquid inside the chest quickly captures and "stores" the "excess cold".

If there is no free water in the chest and the contents chill below 32 deg
F (for reasons described above), then the remaining warming of the 
deep freezer ice up to melting point must occur due to heat transfer 
through the chest.  If the rate of heat transfer were constant, then these
"additional BTUs of cold" would be lost at the same rate as if they first 
froze the water inside the chest, and the total amount of time needed to 
melt all of this ice would be the same as the case where the deep freezer 
ice is added to an undrained chest.  But the heat transfer rate will  
increase because the contents have reached a colder equilibrium 
temperature; therefore these extra BTUs will be lost more quickly.  
This means the total time needed for melt the ice (i.e., the time needed 
to transfer enough BTUs through the chest walls) may be a bit shorter.
My judgment is that difference will be small but not negligible. The heat
required to melt an ice cube at its melting point is very much larger 
than the heat required to raise that cube's temperature 1 deg when it
more than 1 deg below freezing, or fully melted, for that matter.  
(I think the factor is about 150 -- that is, it takes as much heat to melt
one ice cube once it's temperature is at equilibrium
it does to raising the temperature of the resulting liquid to 180 deg F. 
That's why ice cubes are so effective at cooling!  Conversely, it takes as
much cooling to freeze an ice cube from water at the freezing point as it
does to chill the resulting cube to -118 deg F.  In other words,
there's some, but not a huge amount of "excess cold" stored in deep 
freezer ice --  perhaps an additional 20% BTUs of heat are absorbed for 
ice at 0 deg F versus ice already at the melting point.  On the other 
hand, this implies that, properly conserved, this "excess cold" could
be used to chill the chest contents 20% longer.

Thus, this argues that one shouldn't drain *all* the water out of a chest
when replenishing the ice.  But this may happen naturally, for there will
always be residual water, plus foodstuffs kept in a cooler tend to 
have significant moisture content (who keeps potato chips in a cooler?)

Reviewer Tom Cull also pointed out that having water in the chest 
minimizes the amount of warm air that can enter the chest when ice is
added; draining water out of the chest will cause warm air to be sucked

Finally, let's consider what happens once the ice has all melted and
there's no more available to replenish the chest.  The temperature
inside the chest will begin to rise.  The heat transfer in the chest
continues to be proportional to the temperature difference between 
inside and outside.  For every unit of heat transferred in, the 
temperature rise in the chest will be inversely proportional to the
thermal capacity ("thermal mass") of the chest contents.  Keeping 
the chest full of water will be advantageous, for there's simply more
mass that has to be warmed up.  (Proof: consider inverse problem --
if you are going to cool off the chest from ambient temperature, common 
sense would say to remove all extraneous water and put the food in direct 
communication with the ice; no need to cool off the water remaining
from the last batch of melted ice).

So bottom line, from a "conservation of ice" point of view, under
the assumptions given, there is no difference on whether you drain the
cooler while the ice is melting, there is a some potential benefit to 
keeping the water in when replenishing with freezer-fresh ice, and there is
definite benefit for keeping water in once the ice is melted and you
are trying to keep things as cold as possible as long as possible.

Happy Summer!

Steve Czarnecki

Current Queue | Current Queue for Physics | Physics archives

Try the links in the MadSci Library for more information on Physics.

MadSci Home | Information | Search | Random Knowledge Generator | MadSci Archives | Mad Library | MAD Labs | MAD FAQs | Ask a ? | Join Us! | Help Support MadSci

MadSci Network
© 1997, Washington University Medical School