|MadSci Network: Physics|
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 water. 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 F) 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 cooler. 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 in. 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
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