Submitted by terjeboe t3_10842e7 in askscience

Say I want to melt a block of water ice.

One option is to arrange for the melt water to drain away, thus reducing the mass that needs to be heater. However, the water that drains away will cary some heat out of my system. Which effect will dominate? Does it depend on the geometry of the ice block, or perhaps the method of heating?

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CrambleSquash t1_j3qceft wrote

Why not try it?

I would expect so.

Air is a poor heat conductor and so would do a bad job transferring heat into the frozen cube.

Liquid water on the other hand is quite a good conductor of heat.

If heat can efficiently transfer into the water, e.g. from the container the cube is sat in, I would expect this would increase the melting rate.

The water would also provide an additional surface for the heat from the air to flow into.

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Pablo-on-35-meter t1_j3qde9q wrote

  1. If the iceblock is under water, it is likely that the surface area is reduced. Then, drain. Because you need to maximise the surface area.
  2. Try to keep the meltwater but make sure the water has a big area (large tub), then several things are happening: evaporation of the water results in heat loss. And the bigger area increases the heat transfer.
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mfb- t1_j3qduup wrote

If your heat transfer rate is not high (e.g. you just have it sit in a room temperature environment, not over a fire) then both ice and water will be at the melting point while the ice melts. All the heat will melt ice, so it's only a matter of "collecting" as much heat as you can. Water provides a good contact to container surfaces, so keeping it in is probably a good idea in most cases.

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bigloser42 t1_j3qqg3u wrote

On one hand I agree with your premise, but you are overlooking the fact that the air will be much warmer than the meltwater. Not sure if the temperature difference would be enough to overcome the difference in thermal conductivity.

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JamesTKierkegaard t1_j3qru2w wrote

There is no right answer without knowing more about the system. All the previous answers hit on most of the salient details. One of the biggest factors will probably be surface area exposed to air. If the container is a poor conductor (e.g. glass) then the water filling the container might reduce that surface area and slow the system. If it's a metal container that would be less of a factor. Another factor to take into account is evaporation which will remove heat from the system, but how big a factor this will be depends on the temperature and humidity of the air. The ice can sublimate as well, but this is a much slower process than evaporation and I don't believe it's exothermic, but I could be wrong.

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pupeighkhaleuxpeh t1_j3qxdhb wrote

Depends on temperature of air, convection forces of both air and water, density/pressure of air, solutes in water, surface/shape/volume of ice. This isn't an exhausting list but it's the first tings that came to mind

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Captain_Poodr t1_j3rl36n wrote

Drain the water away and allow the ice to be in contact with air. Raising the temperature of water requires many times more input energy than raising the temperature of air. Also, air will flow over the exposed surface area of the block at a larger temperature differential than the melted water. This will encourage faster melting. When ice is melting in a glass of water, the temperature of the water itself remains close to 32F until the ice has melted. All of the energy being absorbed into the water from the environment is essentially going straight to phase changing the ice back to a liquid, but as mentioned above it requires far more energy to raise the temperature of water than air

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Londonforce t1_j3rnrn2 wrote

If it drains it will melt faster. The meltwater forms a layer of cooler liquid around the ice that partially insulates it and slows melting.

I know this because I recently watched this very experiment being conducted (as part of a series of experiments that has non-intuitive outcomes). I'll try to find the link for the video.

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januarytwentysecond t1_j3s7qm0 wrote

I know this! This was an elementary school science fair project of a good friend of mine - four equally measured bowls of ice, two in strainers, half with salt. The strainers lasted far longer than the bowls, the salt didn't have much effect. Empirically, not draining melts faster.

As for the why, there's plenty of pontificating here. My own thought is that heat transfer from ice to water is much faster than ice to air (see liquid-cooling), and the hemisphere of water had a larger surface of contact with the bowl below and the air above to absorb warmth from.

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Game_Minds t1_j3s8rl2 wrote

Yeah there would have to be a range of tests with different container shapes and materials, ice cube size and shape, ice composition and density, ambient temperature and humidity, air flow, water flow, a bunch of stuff would have impacts

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Chemomechanics t1_j3scs65 wrote

> Empirically, not draining melts faster.

For that container material and geometry and those environmental conditions, etc. Dropping these qualifies gives an unequivocal statement that's not convincing, considering the various factors discussed in this thread.

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VT_Squire t1_j3sd4rm wrote

Thermal conductivity of water is 0.598 W/m·K

Thermal conductivity of air is ~ 4.5 × 10^−2 W m^−1 K^−1

The disparity here is like... not even a contest.

Air and water of equal temp in the described scenario with controlled conditions leaves essentially no room whatsoever for the draining ice-cube to melt faster than the non-draining one.

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VT_Squire t1_j3sgdrh wrote

Think about it this way...

Scenario #1: You jump into water that is 33 degrees

Scenario #2: You walk around outside when it's 33 degrees.

You're wearing nothing but a bathing suit in both scenarios. In which environment are you going to induce hypothermia faster?

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H4zardousMoose t1_j3srwee wrote

I'd definitely be interested in the link because I'd imagine it wholly depends on the container, in which you keep the ice-water mixture. A well insulating container, with a small surface area for the water would probably be worse than having the ice in a sieve over the sink. Where as a wide metal bowl should be better. What matters is how much energy from the surrounding air can be transferred into the ice. Water is a great thermal conductor, so I don't see how it could insulate the ice. If you add energy to a mixture of water and ice it will always melt ice. That's why a mixture of water and ice is always at the melting point (given the energy transfer isn't too fast and there is some agitation).

Anyways that's what I'd expect from the experiments I remember from my physics classes.

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TerjiD t1_j3ssf5l wrote

Not a physicist, but here's a real-life example:

If you have two ice buckets at a party where bucket one has a raised bottom (so the water doesn't touch the ice) and the other one doesn't, the ice in the one with the raised bottom will last longer.

Ergo: Letting your cube stand in water will make it melt faster.

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VT_Squire t1_j3syxzc wrote

>However, the water and air may not be at the same temperature, no?

That's exactly what I am getting at.

Thermal-conductivity is like a speed-limit, so to speak.

Degree for degree, you'd need roughly 11,000 times as much air exposure as you'd need water to melt both cubes in the same amount of time.

Double the temp of the air, you still need 5,500 times as much air... and so on.

At that point, it's not about what melts ice faster, it's about how much assistance you need to give the air to compensate for the fact that water is better at it.

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Burstar1 t1_j3sz9b6 wrote

Remember that the melt water is also interacting with the warm air as well. It's not the temperature difference that matters so much as the huge difference in the frequency of interactions between gases and liquids which is why water will outpace air here.

FTR I've done this experiment myself using a consumer cooler and blocks of ice. The portion of the ice that is exposed to the water visibly melts/shrinks faster than the exposed ice. A cube of ice sitting on a sheet of plastic mesh will last much longer than the same sized ice cube sitting on the bottom of the cooler (the little puddle it starts to form almost literally gobbles it up).

Want to cool a can of soda fast? Don't plop it in a pile of Ice alone. Add water to the ice and even though you've definitely added heat to the system the rate of the soda's temperature reduction will be vastly quicker.

Edit: and by cool in my previous statement I don't mean literal measurable change in temperature. If the ice is melting both it and the water are at 0deg C. It is the Heat of Fusion that is being transferred that I'm referring to.

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Ambitious_Jelly8783 t1_j3t2qfn wrote

If you keep the water, all things equal, the water will protect the ice. The melted water will keep at 0*C, protecting the ice, where the air will be warmer, so if no water the air will apply a higher temp to the surface.

Water requires a lot of energy to change degrees. It is actually not a great conductor. Now if you have warm water, same reasonkng applies as it has lots of energy so it'll melt it very quickly.

Someone do the experiment and share it.

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ubik2 t1_j3t7k8m wrote

At some point, the extra surface area of the water/air interface means more heat is transferred from the ice cube to the air. If the surface area is small, the more important factor is that it slows the transfer from the ice.

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BuildANavy t1_j3tneoc wrote

You're making a big assumption that there is a significant heat source transferring heat to the water other than through the air. Otherwise in both cases you still rely on heat transfer from the air, it's just that in one case it has to heat up the water first. In a typical experimental setup this might be from a bowl sat on a countertop at ambient temperature or something, but if the bowl had a very low thermal mass and was supported by very thin members then the draining setup could easily melt the ice quicker.

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kilotesla t1_j3twpjl wrote

I believe that the 33° was supposed to be fahrenheit, for about 0.5° C. Perhaps you knew that and were making a joke, but I'm giving you the benefit of the doubt.

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kilotesla t1_j3txubj wrote

That's possible but not certain. If you had a pile of ice cubes on a grate, such that water could drain away, and warm air could filter through accessing that full surface area, that might result in higher heat transfer than the same ice in a bucket full of melt water where the surface area accessible to the air is the surface of the bucket and the surface of the water, and is smaller than the total surface area of all of the cubes.

But if you froze the ice in the bucket to get a roughly cylindrical block of ice, and put that on the grate with the water draining away, the surface area would start out similar to that of the bucket, but would gradually decrease as the ice melted, whereas if you kept it in the bucket with the melt water, the surface area would stay constant.

Topics left as an exercise for the reader are consideration of heat transfer by radiation and the possibility of putting it in a bucket with a hole in it such that the water drains out. And then the challenge of how to fix a bucket with a hole in it without access to a working bucket, which might be important for getting water necessary to sharpen tools for carrying out the repair operation.

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kilotesla t1_j3tz1rk wrote

That's good advice that the specifics matter. Since there will be heat transfer by radiation as well as convection, the emissivity of the surface of the container also matters, and if it's polished metal, the low emissivity would retard melting. Also, if the thermal mass of the container is significant, its starting temperature would matter.

When considered in comparison to air, glass is not a poor conductor of heat. It's conductivity is 33 times higher than that of air, and it's likely pretty thin, such that the heat transfer through the air will be the dominant thermal resistance. Air of course has the advantage that it is moving and so carrying heat by convection, not just conduction but it's still going to be the dominant thermal resistance.

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kilotesla t1_j3tzhd4 wrote

Those sound like valid results to me. One thing that I'd want to be a little bit careful of is the initial temperature of the bowl. If you had a bowl that started out at room temperature, and it was thick and heavy, it's thermal mass could contribute to the faster melting.

One other question is whether the ice was in the form of cubes, such that air could flow in between them in the strainer case, or whether it was perhaps frozen in the bowls so it was one solid hunk of ice

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root_over_ssh t1_j3u71oi wrote

Draining will be faster.

Let's assuming you have a glass of ice in a room with a constant air temperature.

The air will begin to melt the ice and soon you'll have a glass of ice water. The ice is no longer in contact with the air directly and is now submerged in a water that is only slightly above freezing temperatures. It takes a lot of energy at this point to keep warming, so now thay air has to heat the water first before there is a significant temperature gradient to melt the ice more.

Now, if you have the same amount of ice in a strainer, you have 2 changes going on - when the water is removed, they're now 2 separate systems in this constant air temperature exam, the ice doesn't impact the water and the water doesn't impact the ice. Since the water is being removed from the ice "system", you now have less mass to heat up as well and maintain a higher temperature gradient between the air and temperature of the ice.

While water is a better conductor of heat, heating ice with water that is heated by air is still far less efficient than heating the ice from the air directly.

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JamesTKierkegaard t1_j3ueydg wrote

It's not glass compared to air that's the issue so much as compared to water or ice. The air is the environment in this situation so if the system is in a glass container the heat transfer to the air will be negligible compared to the surface area of the water and ice (which again depends on the shape of the container greatly). If it's a thin metal container then water remaining will probably win in most configurations simply because it will act as a convective exchange surface. Realistically, radiation is going to be a meager source of heat loss, even hot water radiators to heat houses only supply about 5% of their heat contribution through actual radiation, and that's at higher temperatures.

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kilotesla t1_j3uolby wrote

>It's not glass compared to air that's the issue so much as compared to water or ice. The air is the environment in this situation so if the system is in a glass container the heat transfer to the air will be negligible compared to the surface area of the water and ice (which again depends on the shape of the container greatly).

In a series circuit with a very low resistance resistor, a medium value resistor, and a large resistor, fed by a voltage source, the voltage drop across the medium value resistor is affected a lot more by the large resistor than by the smallest resistor. If we have 1 ohm, 33 ohms, and 1000 ohms in series, the drop across the 33 ohm resistor is 3% of the source value, even if we drop the 1 ohm resistor to 0.1 ohms. We can't conclude that the 33 ohm resistor will have a lot of voltage drop because it is huge compared to 0.1 ohms. That doesn't work.

The glass outer surface will be very close to the same temperature as the water. The heat flow per unit area is determined by the temperature difference between the water and ambient. If the outer glass surface were at 1 C instead of zero, the temperature difference with respect to ambient would not change significantly. And the surface temperature wouldn't even be that high.

>Realistically, radiation is going to be a meager source of heat loss, even hot water radiators to heat houses only supply about 5% of their heat contribution through actual radiation, and that's at higher temperatures.

  1. 5% is way too low. Modern "radiators" have fins which enhance convection but not radiation, so convention is typically larger, but radiation is still about 25%, even just counting the outward facing surface.

  2. In the range of temperatures we are talking about, radiation is reasonably approximated by a linear function of temperature difference. Yes I know, that's counterintuitive with that fourth power, but it's T1^4 - T2^4 , not (T1-T2)^4. On the other hand, natural convection is nonlinear enough that it drops as a fraction of overall heat transfer when the temperature difference gets smaller.

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GiveItStickMan t1_j3uv1pl wrote

I agree with this. Melting ice outside of a bath has 2 functions, drying the thin melted layer and thermal exchange from thermal layer. I thought of this like fins on a heatsink and the melted ice on the exterior as thermal paste.

Also this has variables and would need a test to determine but this was my initial thought also.

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dnkushne t1_j3v1px8 wrote

This is not right still. Both the air and water will primarily transfer heat by convection. Thermal conductivity is the measure of heat transfer within the same substance, not between two substances.

Assuming room temperature air (i belive this is the OPs intention) earlier comments are right about how the melt water will remain cold and thus have mich slower heat transfer from the ice to the water, thus slower melt.

Your missing that heat transfer is a function of the delta T (difference in temp between two susbtances) times a resistance value (convection primarily here)

Also belive this is confirmed by ice in cooler during camping. Ice and drinks stay colder if you don't drain the water.

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VT_Squire t1_j3v81lr wrote

I'll put it to you this way by stealing an example straight from stack exchange.

Place a cocktail stick through an ice cube and lay it on top of a glass filled right to the top with (room temperature) water. The submerged half will melt quicker than that on top.

So what if we dial "room temp" down to 33 degrees? Same answer. What if we dial it up to 200 degrees? Same answer.

Air is an AWFUL medium for temperature to transfer. It'll take an hour to cool down a warm beer in the fridge. But it's like maybe 12 minutes submerged in water at the same temp as your fridge.

In a cooler you want the maximum of cold thermal mass but the minimum of heat transfer. THATS why draining a cooler makes the ice melt faster. You're literally allowing the ingress of warm air while actively displacing the cold thermal mass. It's got very little to do with wet vs dry like OP is getting at.

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H4zardousMoose t1_j3xyimx wrote

I think you are conflating thermal conductivity (how well thermal energy propagates through a material, measured in W/kg*K) and thermal capacity (how much energy a given amount of a material can hold, usually measured in J/kg*K).

Water has high values in both btw. So it takes a lot of energy to heat up and cool down, but it also exchanges energy quickly within itself and to it's surroundings.

Now instinctively I thought that water would have a higher thermal conductivity than ice, because Iglus insulate so well. But snow isn't ice (crazy I know:D) and it turns out ice has about a 4 times higher thermal conductivity than water at 0°C.

Therefore: If you cover the ice in a thin layer of water, this should slow down the melting. But if you put the ice into a decent sized container with water, where the total surface area of the mixture becomes more than 4x larger than the surface area of the ice it should speed up the melting process. This effect should also increase the further along the melting process you are, since the surface area of the ice will decrease (less of it left), where as the surface area of the mixture remains mostly equal (ice has a bit less density I know, but small effect).

So I learnt something: Solids generally conduct heat better than liquids. But the original point, where it depends on the container and it's conductive surface area still mostly remains valid.

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