Which will keep my food colder longer, draining the melted ice water, or leaving it in the cooler?

Question

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?

Answer 1

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.

Coleman agrees:

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.

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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.

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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 ΔT: 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.

Answer 2

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.

Answer 3

Never remove cold water from a cooler so long as the water is cooler than the outside temperature.

1. 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.
2. 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.
3. 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.

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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.

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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.

Answer 4

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.

More here

Answer 5

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.

• NRFW - cooler packing

Answer 6

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.

Answer 7

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.

EDIT:

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.

Answer 8

Ok, put away your calculators people. Draining the cool water causes the warm air to enter the cooler. Cool water is COOLER than warm air, so leave the cool water in the cooler. If I had a compass, some graph paper a protractor handy I'd draw you a picture.