Say you have a cooler of frozen food and ice to last for several days or weeks of river-trip / car camping. To keep things as a cold as possible for as long as possible, is it better to leave the cold melted ice-water in, or to drain it out on a regular basis?
From a thermodynamics point of view, I'd say you should leave the water in.
Temperature is a measure of the active kinetic energy of the molecules in a substance. Warming up is essentially the surrounding environment imparting some of its kinetic energy into the object being warmed up. Simply thinking about that, the more you have that needs warming, the more energy it requires to warm, and so the slower its temperature will rise (given the same rate of exchange of thermal energy - the water submerges at least some of the other contents, so if anything this is an overestimate). Now consider that water has a relatively high specific heat, meaning that it takes more energy to warm it up. The food inside the cooler won't be warmed up faster than the surrounding water, so since it takes more energy to heat the food and water than just the food, so the food will stay cooler longer.
The Igloo (yes, the cooler company) FAQ supports this view:
During use, it is not necessary to drain the cold water from recently melted ice unless it is causing contents to become soggy. The chilled water, combined with ice, more readily surrounds canned and bottled items and will often help keep contents colder more effectively than the remaining ice alone.
Don’t drain cold water – Water from just-melted ice keeps contents cold almost as well as ice and preserves the remaining ice much better than air space. Drain the water only when necessary for convenient removal of cooler contents or before adding more ice.
I think the key point a lot of people forget is that what matters isn't how long the ice lasts, but how long the contents remain under some temperature.
Assuming convection is fairly significant relative to the rate of energy input from the outside (a good assumption, I think), it doesn't matter how good an insulator air is, the inside temperature will be the same throughout, necessitating that the rate of heat energy coming into the cooler is the same in both cases, so the ice will melt at the same rate, and once the ice is gone, the cooler with water will take much longer to warm. I'm still working on the no-convection theory (which would provide at best an extreme overestimate), but in the meantime if anyone wishes to posit that convection is tiny enough to bridge the enormous gap (assuming I find the upper bound does eclipse water), please explain why you believe so.
Some math/physics to back this up for the quantitatively inclined. (This would be so much easier with the MathML markup from the math and physics sites.)
The cooler will be very near perfect convection, the heat is entering the cooler slowly enough that the contents - air, water, and ice - are at the same temperature (namely 32°F/0°C/273.15K). Heat conduction,
H, as far as our coolers are concerned depends only on
H = kAΔT/x, where
k is the thermal conductivity of the cooler,
A is the area of the cooler through which thermal energy is flowing,
x is the thickness of the cooler, and
ΔT is the temperature difference between the inside and outside of the cooler (
T_out - T_in). Notice that all these are the same for both coolers. Now, melting the ice requires energy of
Q = Lm where
Q is the total energy required,
L is the (latent/specific) heat of fusion, and
m is the mass of ice in the cooler. Since the total energy is
Q = Ht, we can calculate the time required to melt all the ice:
Q = Ht = kAtΔT/x -> t = Qx/(kAΔT) = Lmx/kAΔT. Since all variables are the same for both coolers, it will take the exact same amount of time for the ice to melt.*
*: We are ignoring the air replacing the ice, which would actually give a (very) slight advantage to retaining water. Draining the water in that cooler requires adding heat - the excess heat of the air that replaces the melted water. Fortunately, that excess heat is fairly easy to calculate:
m = Vρ -> V = m_ice/ρ_ice = m_air/ρ_air -> m_air = m_ice * ρ_air/ρ_ice. The air comes in with an excess energy of
Q = m_air*C_air*ΔT = m_ice*(ρ_air/ρ_ice)*C_air*ΔT. This reduces the energy required from the cooler heat influx, reducing the time required:
Q = Lm_ice = Q_cooler + Q_air = kAtΔT/x + m_ice*(ρ_air/ρ_ice)*C_air*ΔT -> t = (L/ΔT - ρ_air/ρ_ice)*C_air)mx/kA. The fractional difference this will make ends up being about
4e-6*ΔT, or about
0.016% on a rather hot (40°C) day, coming to 2min 18s over 10 days. So we were right to ignore it.
We (Kent and Deny) did an experiment in order to shed some light on this debate. We found that keeping the water in the cooler along with the ice kept the overall temperature of the cooler below 5 degrees Celsius for approximately 4 hours longer than when the water was removed.
Experiment. We filled a Coleman cooler with 12 341mL bottles of Waterloo Dark beer and 2.7 kg of store-bought ice cubes and sealed the cooler. Beer bottles were kept at 4 degrees Celsius in a refrigerator overnight before both experimental trials. The ice and thermometer were both kept in a freezer, which is at -10 degrees Celsius, before the experiment. Temperature was monitored using an Omega OM-62 temperature logger which recorded temperature every 5 minutes over the course of the experiment. The thermometer was kept in a Tupperware container to avoid water damage.
We setup two separate trials of the experiment. In the first, water was removed from the cooler approximately every 8 hours using a siphon. Excluding the first removal of water at 8 hours into the experiment when there was still little water in the cooler, approximately 450 mL of water was removed from the cooler every 8 hours.
In the second case, where all water was kept in the cooler,we opened the cooler for 2 minutes every 8 hours in order allow the same amount of warm air to enter the cooler, as it would in the case of draining the water.
Results. Temperature data is shown in the figure below.
From the figure, we can see a few points. Excluding the first draining, when there was little water, each time the water is drained, the temperature of the cooler immediately rises. After 24 hours, draining the water from the cooler preceded a linear growth in temperature, which grew in slope at each subsequent draining. Finally, we can clearly see that keeping the water in the cooler keeps the temperature below 5 degrees Celsius for approximately 4 hours longer.
We conclude that this experiment gives support to the argument for keeping the water in the cooler.
Never remove cold water from a cooler so long as the water is cooler than the outside temperature.
- Opening the lid allows more warm air in, but assuming the lid is on the top and air disturbance minimal, this could be a small loss of cooling / small entry of heat. Opening a drain will have to let warm air in to replace whatever cool water leaves the cooler.
- While ice remains, water will circulate and keep all the objects in closer thermal contact with the ice than will air alone - this will prevent a thin spot in the insulation from warming some of the food that stays in contact with the cool water. This won't have an effect on the long term temperature in the cooler, though.
- Once all the ice is melted, the cool water will continue to be a heat sink, absorbing more of the heat that is leaking in through the walls of the cooler. This keeps your actual contents inside the cooler more chilled than if you remove this cold mass from the system. Without the cold water, all that heat will go to the food and warm it up instead of warming both the combined mass of the food + cold water.
You don't need any fancy graphs when you look at the heat flow into the cooler. Keeping cool water in the cooler delays the melting of the remaining ice and once the ice is gone, delays the warming of the food by absorbing it's share of the incoming heat.
Say you have two coolers with one gallon of beer each in a perfect container that absorbs no heat - just keeps the beer together. Also, assume that you have one pound of ice in each cooler and the beer and cooler are chilled to 0°C (32°F). One cooler also has a gallon of 0°C (32°F) water.
So the cooler on the left has heat leaking into the insides - warming everything inside. It warms the air, the beer, the inside of the cooler and the beer's container.
The cooler on the right has the exact same amount of heat leaking inside. It warms the air (but less air due to the extra water), the beer, the inside of the cooler and the beer's container.
The only variable is does water absorb more heat per degree rise in temperature than air. The answer to that is yes - extremely so.
Air's specific heat is 1.007 J/(gK) - Joules is energy, gram is mass and K is degrees Kelvin. Water's specific heat is 4.18 J/(gK) - so for a fixed amount of Joules added, water goes up less than 1/4 a degree in temperature compared to a gram of air. If water and air were equally dense (they are not) then you have a 4 to 1 advantage. Water takes four times as long to warm up as air, so your beer stays cooler longer.
Now - what about your typical 70 quart cooler? It is 66.24 Liters or 66,245 cm^3 in volume.
Air's density is 0.001275 g/cm^3 and water's density is 1.00 g/cm^3. Here's where water really owns the show in terms of cooling capacity since water is 784 times denser than air. That 70 quart cooler could contain either 84.5 grams of air or 66,245 grams of water (or 3 ounces of air versus 146 pounds of water).
Now we have water with a 4 to 1 advantage on heat absorption and a 784 to 1 advantage on packing mass into the same space, so water is over 3000 times better than air for absorbing heat while raising the temperature inside the cooler one degree. Whether you have a thimbleful of water or the cooler is mostly water, you want that amount of cold water staying inside the cooler to absorb its share of the heat leaking in.
Since it's easy to agree that the amount of heat entering the cooler is basically the same whether you drain or don't drain, leaving cool water inside the cooler slows the time it takes to both melt the remaining ice and warm the contents since that water will absorb heat if it is left in the cooler up to the point where the water in the cooler is equal in temperature to the outside air.
You do need formulas to calculate the rates of temperature increase, but the which is better scenario can be decided quite easily by considering where the heat leaking into the cooler goes and whether draining the water also has a side effect of allowing more heat in. Both of these fall on it being better to leave the container closed and with the melt water inside.
EDIT: The more I consider this, the ambient air temperature around the cooler is the largest factor. Replacing water with 95F (35C) degree air will have a much larger impact than replacing water with 40F (4.4C) degree air.
Actually, the answer is very simple because you asked longer, not colder.
If you drain all the water, then when the ice all melts...you're done. Because whether or not the ice is better than the water at keeping your food cold, the water is going to be better than nothing. The ice is going to melt (effectively) at the same rate either way, because you can drain the water, but you can't really dry the ice. So the ice will melt at roughly the same rate regardless of when you drain the water.
Also, when you drain the water, something has to replace it. That something will be air at the temperature of the environment you are in... which will destroy the equilibrium. Basically everything I read online ignores this point. It talks about the coefficients and thermodynamics, etc, and, neglects to even consider that the water is being replaced with something and that something is air which must then be cooled. When you drain the water, you are actively replacing cold with heat.
It should be noted that if you have infinite ice, you should probably drain the water. However if you have infinite ice it doesn't seem like this would matter.
I sometimes leave melted ice water in my cooler, which then gets into the food, making the food inedible.
If my "food" is a can of beer, fine, leave the water in the cooler. If it's a sandwich, drain the water if it might get the sandwich wet. Or have the cans of beer at the bottom and put the sandwich on top to keep it out of the ice water.
Another "outside the box" answer is if it gets cold enough at night, I'll leave the lid open. Leaving all the melted ice water inside is good thermal mass to keep it cool during the day. You have to remember to open and close the lid at the right time. If critters might get in this doesn't work so good.
Here is a different twist on the question of how to best use block ice. Before refrigeration northern states used to saw block ice fron frozen lakes and store for the summer in "ice houses", log houses that used only the thermal inertia of massive amounts of ice stacked together.
We just returned from a 6 day canoe trip in which we used block ice to keep fresh food cold. On Friday, july 19 we bought 200 lbs of block ice (20 blocks of 10 lbs). these almost completely filled two 12" x 12" x 30" low tech garage sale ice chests. We kept these two ice keepers drained of water because water is a conductor, not an insulator. Then we used single blocks of ice from these to supply as needed four smaller chests which we did not drain.
We then traveled from oklahoma City to the Current River in Missouri and floated 100 miles over six days and returned to OKC on Fri Jul 26 with about 15 lbs of block ice remaining. Eight days and 15 hours! We fed 18 Scouts and adults on fresh food that was previously frozen ar cooled for a week. I know that this is a completely unscientifec "experiment" , but is a practical example that worked for us. I hope this is useful for your wilderness trip.
As a rafter I have taken part in lots of discussions about this and couldn't resist any longer so have set up an experiment to test this. I hypothesize that the drained cooler will hold ice longer due to the insulating effect of air--as Snitse has described above. Convection will reduce air's effectiveness but, as Snitse points out, it is still far superior to water. I have set this experiment up with identical styrofoam coolers loaded with equal amounts of crushed ice (in the second series of experiments I am going to substitute identical blocks of ice). The drained ice chest has a drain tube in place to allow it to drain continuously. These are placed side by side (with a gap between) at room temperature (in my lab in a large building where temperature is fairly constant). In the first replicate, the drained ice chest did, indeed hold ice longer, but not by much and the undrained ice chest contained water that reached room temperature at nearly the same time the ice melted! I am replicating this experiment and will begin the ice block experiment after replicating this three times.
This is a great topic deserving of some controlled experimentation. In general the system is non-linear, but can be analyzed in a piece-wise linear fashion the same way many other difficult problems are approached. Some thoughts on the matter;
conduction not convection, is the main transport mechanism,
cold does not exist, only the absence of heat,
heat transfer is a function of temperature difference between the inside and outside of the cooler, cooler thickness, thermal conductivity (w/m-k) of the cooler material, and surface area of the cooler.
You can use Fourier's law of heat transfer to calculate how much heat in watts enters the cooler through its walls. As ice melts, it takes 3.3x10^5 joules/kg to do so. Remembering that a watt is a joule/second you can then calculate how long a given mass of ice will last. Heat, like electricity, takes the path of least resistance.
Water, at 24X the thermal conductivity of air, provides a low thermal impedance path for heat to reach the ice, so if a block of ice is sitting in a water bath it should melt at a faster rate. I would like to propose and experiment where the ice and melt water in one cooler are kept segregated, and one where the ice and melt water are not.
At the time that the ice is completely melted, the two systems are the same in mass and phase (all liquid) but perhaps not temperature or even at the same time. Then its a different process, but with different initial conditions (temperature). There are some variables which have to be controlled in order to see the effects of draining vs not draining the water, namely ambient temperature has to be monitored and kept the same for both, admitting outside are into the system has to be eliminated, the coolers must be identical, the mass of the ice has to be measured.
Leave the system undisturbed, measure the inside water and air temperature at as many places as possible using thin thermocouples and a digital thermometer. Allow the experiment to run well past the expected lifespan of the ice. Some sort of viewing port capability would be helpful to examine the state of the ice without perturbing the system.
The thermal conductivity of air is 0.000057
The thermal conductivity of water is 0.0014
Therefore water is 24.5 times more conductive than air, and has a temperature above 32 degrees Fahrenheit. The reason this is a problem is that bacteria can grow in that water quite quickly (within 6 hours) and start to make your food unhealthy to eat.
Additionally all that warm 32+ degree water will assist in melting the remaining ice.
So do yourself a favor, drain the water that melts out the bottom of the cooler with the spout that is provided. It's safer, and your food will last longer.
See this fine article about cooler packing and draining.
As Dangeranger stated,
water is 24.5 times more conductive than air
The idea is that water will conduct heat energy throughout the cooler much more effectively than air. This means that water will conduct heat energy from the leaks and seams in the cooler to the ice more effectively than air.
Say we have two hypothetical coolers, Cooler A and Cooler B. Both leak heat at the same rate, and both started with a identical large block of ice in it, and equal air temperature occupying the rest of the cooler. Every hour you drain the water from A, and leave it in B. For the first hour they both have the same amount of water and the same amount of ice, and then you drain the water from A. Now, in the second hour because the water left in B conducts heat from the walls to the ice faster than the air in A, more of B's ice will melt than of A's ice.
Clearly, B's ice will be completely melted before A's ice is. Let's call the moment that ice B completely melts 'meltB'. At meltB ice A has x percent of its original mass remaining.
We can all agree that at meltB there is less heat energy in cooler B than in cooler A, since you've been replacing things in cooler A with other things that had more heat energy (replacing cool water with air temperature air).
Where it gets tricky is that after meltB, the heat energy in cooler B will increase faster than the heat energy in cooler A because of all that water conducting heat from the walls.
So from start until meltB, A gains energy faster than B, but after meltB B gains energy faster than A.
What this all comes down to is the exact numbers. If you want to eat your food at time a or b you are better off with cooler B, but if you want to eat your feed at time c or d you are better off with Cooler B.
As for how to calculate meltB, I have no idea. So I guess its somewhat impossible to figure out.
To clarify, the key point here is that different parts of the cooler are different temperatures. The wall is warmer than the area immediately around the ice. However, if the cooler is full of water, the water brings heat energy from the wall to the ice much more effectively than air would.
So the coolers leak at the same rate, meaning the walls gain energy from the outside at the same rate. However, if there is no water, the walls gaining energy does not necessarily mean that the ice will gain that energy and melt immediately. It will take time for that energy to get to the ice, and the time it takes depends on the medium the energy transfers through.
In summary, walls gain energy at same rate, ice in the middle does NOT gain energy at the same rate because of what is between them and the walls.
Removing water from the cooler always means the ice lasts longer - solid (ice) transfers heat slowest, then gas (air inside cooler), then liquid (melted ice). This is explained by the Zeroteh Law of thermodynamics. All three work to achieve equilibrium by becoming the same temperature. Remove the fastest heat-transferring part and your ice lasts longer. The classic experiment is placing an ice cube in a mesh bag and suspending it with string and lever over a clear glass containing cold water. Notice the ice melts faster where it is submerged than in the air. The air insulates the thin layer of melted ice above, but the liquid quickly exchanges heat with the ice to melt it, until all are in equilibrium. Using the drain to remove the melted water means more insulating air and solid ice in the cooler to maintain cold contents, and longer lasting ice.
It's really simple. The universe wants everything to be the same temperature so if we left the cooler in a sufficiently large, constant temperature environment, the cooler and all its contents would eventually end up at the same temperature, that being the ambient temperature of the environment.
Let's first begin by noting that we want to keep something cool rather than make it cooler. In that case, we want to stop it gaining heat energy. The heat energy is coming from the walls of the cooler.
If our food is packed in the cooler such that the ice touches the walls, then the ice is going to be subject to warming before the food. The ice touching the food will not impart any heat energy because it's the same temperature.
However, add some water to the system and there is path of conduction from the wall to the food since the water is at higher temperature than the food.
Air has 1000 times less heat capacity than water, so it's not actually that interesting to take into account replacement of cold air with ambient temperature air.
A simple fact: You conduct HEAT. You do not conduct COLD. Wrap your head around that.
2nd Simple Fact: Water conducts HEAT more efficiently than air.
Now, if you're trying to chill warm items, then use water to conduct the heat out of them and transfer it to the frozen water (ice). That will chill them quicker than just air.
However, if you're trying to keep chilled items cold, you want to prevent the transfer of heat from the environment (via the outside skin of the cooler, across the insulation or air gap, through the interior skin) to the items. In this case, air is the best choice; drain the water.
So, you catch a fish. Quench it quickly in one ice chest with ice and water to get it chilled, then put it into an ice chest with the ice kept separate from the contents and with the water drained.
It is so not true that water mixed ice keeps food cold. Try your own experiment and see that it is true. This is exactly how you make ice cream, but just with some salt. Here is how you do it: 1) Buy your cooler according to size. Dont buy a cooler that it too big. 2) Fill the cooler with pre-chilled items. Try and reduce the air space as much as possible. This will dramatically extend the 'cold' 3) Top off with ice. 4) The Ice Melts. 5) Melted Ice turns to water and carries off heat. 6) Drain the warmer melted water. Heat goes right out the drain tube. Ice remains colder for MUCH longer. Half a day over my 3 day weekends. NOTE...This is a mostly closed system. Neither 'Dried' (Not Dry Ice...but not wet ice) nor water/ice slurry works well when kids are always leaving the lid open and warm air blows in.
The consensus around the people i travel with is varied and the best answer to this is "it depends".
Keep the water in - Keeping the water in does add thermal capacity but it's not as good as ice. Water absorbs about 1 BTU/lb/F. When all the ice has melted the cooler will start warming immediately, the water will absorb heat at that rate. I would call "Cold" somewhere below 45F, so 1lb of water will absorb about 13 BTU's while it's warms up from 32F. When ice melts, it absorbs about 144BTU/lb(enthlapy of formation) at a constant 32F temperature so keeping the water in the cooler gives another 9% thermal capacity before the cooler reaches 45F. There are a couple of factors that may lessen this 9% such as the ice will melt a little faster since it's sitting in a bath of water(the heat transfer from the remaining ice to the inside cooler wall will be higher when bathed in water) but it depends.
Also you will have cooler soup when you leave the water in. If this is your food cooler, i don't think it's worth keeping the water in. If it's your beer/drinks cooler than i don't think it's a bad idea to leave the water in.
Drain the Water - Well for one, you avoid making cooler soup. Also the ice will last a longer time for a couple of reasons.
Heat transfer from the ice to the cooler wall will decrease.(Warm air will displace the water however the effect minuscule, even really hot 90F, humid 90%RH, air will add about 0.04 BTU/lb of water displaced. Considering ice absorbs around 144BTU/lb it's about a 0.03% impact so i'm going to ignore that effect). Water has a huge thermal capacity and conductivity compared or air. Heat transfer from the ice soaking in water to the cooler wall is relatively perfect convection and conduction from the ice to inside of the cooler wall. Especially if the water is moving at all, like sloshing around on a boat or in a car. Air on the other hand, is a very poor conductor and has pretty terrible thermal capacity compared to water on a volume basis(a pound of air takes up about hundred gallons while a pound of water is only a pint).
How much longer the ice will last depends on the cooler, conditions and some really complicated heat transfer theory. Making an educated guess that draining the water would increase the insulating ability from the ice to the outside by a 0.5 r-value.(if anyone is interested and able to come up with a full scale transfer model of the inside of a cooler please do, i would love to refine this estimate) A typical 1 inch foam cooler(think $25-30 cooler) with an r-value of 5, the ice will last about 10% longer draining the water. However if you have a 2-3inch rotomolded cooler(think $300+ cooler) than you looking at only a 3-5% increase in ice lifetime. That's less than the 9% increase you get from warming up the cooler soup!!
Don't worry too much about exactly how much longer you can keep things cold since it really don't make a huge difference. I subscribe to the drain the water theory since it assures my cooler is always getting lighter and i don't have a cooler full of soupy water. Lugging around a warmish brew of old food and beer water effin sucks.
PS. I'm relatively confident in my numbers but if anything i have here is orders of magnitude wrong than please post and update.
Your ice will keep longer if you drain the water. If you want to prove it to yourself, get 2 cups. Fill one with cold fridge water and leave the other empty. Put an ice cube in each and see which one is melted first. Even with the ice cube in the water being surrounded by cold water it still melts much quicker than the other ice cube being surrounded by the hot ambient air.
One test I've done was to fill two cups with partly ice and emptied one cup every 5 minutes. The ice in the cup that was emptied of its water lasted longer than the ice that sat in the melted water. Showed easily the best way to make ice last longer. Seems it could be a different story if your aim is to keep the cooler temperature colder for longer. But if I'm keeping food cool I like to keep it relatively dry.