an ice cube floats in a glass of water. as the ice melts, what happens to the water level?

Melting ice and its effect on water levels

... or a fun exploration of volume, mass, density, floatation, global warming, and how to float in a swimming puddle.

by Jared Smith

Principles

Archimedes' Principles:

1. Whatsoever floating object displaces a volume of h2o equal in weight to the object's MASS.
2. Any submerged object displaces a book of water equal to the object's Volume.

Formula

Mass / Density = Volume

Melting ice cube

If you place water and an ice cube in a cup then that the cup is entirely full to the brim, what happens to the level of water every bit the ice melts? Does it rise (overflow the cup), stay the aforementioned, or lower?

The ice cube is floating, so based on Archimedes' Principle 1 above, we know that the volume of h2o being displaced (moved out of the way) is equal in mass (weight) to the mass of the water ice cube. And so, if the ice cube has a mass of 10 grams, and then the mass of the water it has displaced will be 10 grams.

We know the density (or compactness, weight per unit) of the ice cube is less than that of the liquid h2o, otherwise it wouldn't be floating. Water is i of the very few solids that is less dense than when in its liquid course. If you take a 1 pound bottle of water and freeze it, information technology will even so counterbalance 1 pound, but the molecules will have spread apart a bit and it will be less dense and take upward more book or space. This is why water bottles expand in the freezer. It's similar to a Jenga tower. When you showtime playing information technology contains a stock-still number of blocks, but as you pull out blocks and place them on acme, the tower becomes bigger, yet it still has the same mass/weight and number of blocks.

Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^three, every cubic centimeter liquid water will weigh one gram). Past the formula in a higher place (Mass / Density = Volume) and basic logic, we know that 10 grams of liquid h2o would take up x cubic cm of book (10g / 1g/cm^iii = 10cm^3).

So allow's say that our 10 gram ice cube has a density of just .92 grams per cubit centimeter. Past the formula in a higher place, ten grams of mass that has a density of .92 grams per cubic centimeter will take upwards nearly 10.ix cubic centimeters of space (10g / .92g/cm^3 = 10.9cm^three). Once again, the book of x grams of frozen water is more than the volume of ten grams of its liquid counterpart.

The floating ice cube has a mass of 10 grams, then based on Archimedes' Principle 1, it is displacing 10 grams of water (which has 10cm^3 of volume). You tin't squeeze a ten.9cm^3 ice cube into a 10cm^3 space, so the residue of the water ice cube (nearly nine% of it) will exist floating above the water line.

So what happens when the ice cube melts? The ice shrinks (decreases volume) and becomes more than dense. The ice density will increase from .92g/cm^3 to that of liquid h2o (1g/cm^iii). Note that the weight will not (and cannot) change. The mass just becomes more dense and smaller - similar to putting blocks dorsum into their original positions in our Jenga belfry. Nosotros know the ice cube weighed 10 grams initially, and we know it'due south density (1g/cm^3), and then let'south apply the formula to determine how much volume the melted ice cube takes. The reply is 10 cubic centimeters (10g / 1g/cm^three = 10cm^3), which is exactly the same volume as the water that was initially displaced by the water ice cube.

In short, the water level volition not modify every bit the water ice cube melts

Other oddities

Anchors away

Using this aforementioned logic, there are some fun analogies. Consider an aluminum boat in a swimming pool. If you put a 5 gallon bucket full of 100 pounds of lead or another metal into the boat, the gunkhole will get lower in the water and the additionally displaced water in the puddle will cause the pool level to rise. And based on Archimedes' Principle 1 for floating objects, it would rising by the volume of water equal in weight to the 100 pound atomic number 82 bucket. H2o weighs eight.3 pounds per gallon, and so the gunkhole will displace an boosted 12 gallons of water (12 gallons * eight.3 pounds per gallon = 100 pounds).

What would happen if you throw the bucket of lead overboard into the pool? Will the pool level increase, decrease, or stay the same?

When nosotros toss the bucket of pb overboard, the pool level goes downwards 12 gallons (the volume of water no longer displaced past the weight in the gunkhole). Merely when it enters the water, it will be submerged, so we at present demand to employ Archimedes' Principle 2 for submerged objects (it will readapt a volume of water equal to the object's volume). The water level volition and so become up by the volume of the pb saucepan, which is 5 gallons. And then, the cyberspace difference is that the pool level will go down by 7 gallons, even though the saucepan is still technically in the pool.

Only remember that mass and density don't matter for submerged objects. Volume is everything. Consider dropping a brick of clay and a brick of aureate into a bucket. The gold has more mass and is more dense than the clay, all the same if both bricks are the same size, both will displace the same amount of water.

A sinking ship

Similarly, if your aluminum gunkhole weighed 100 pounds, it would readapt 100 pounds of water (12 gallons) when floating. But if the gunkhole springs a leak and sinks, the pool level would subtract 12 gallons minus the volume of the aluminum in the boat. The boat's book (amount of infinite comprised of aluminum metallic) would be much less than 12 gallons. In fact, based on the density of aluminum (.i pounds/in^three), we tin decide using our formula that the book of our 100 pound boat will be most 1000 cubic inches (100/.1 = k). At that place are 231 cubic inches per gallon, so the gunkhole is comprised of about 4.3 gallons of aluminum (1000/231 = 4.3) and thus displaces 4.3 gallons of water when submerged, much less than the 12 gallons that aforementioned aluminum displaced when floating. In conclusion, when our boat sinks, the pool level goes down past 7.seven gallons.

Experiment

As an experiment, fill a sink with 5 or 6 inches of water and note the water level. Next set a heavy drinking glass down into the sink while balancing it right side up (i.e., so it doesn't tip over and fill with water). The water level will notably ascent to make room for the empty glass and you'll note that information technology'southward difficult to get the glass to sink while as well information technology is upright. The heavy glass displaces a lot of h2o because of the heavy mass of the glass (Archimedes' Principle 1), yet information technology still floats considering of it's depression density (don't forget about all the air inside the drinking glass). The glass volition feel lighter to you considering of the buoyancy principle (the force of the displaced water pushing upwardly against the weight of the object displacing it). Information technology will bladder perfectly in the water when the weight of the glass equals the weight of the water information technology is displacing.

Now lay the glass down sideways and let it submerge in the sink. The water level will exist simply barely higher than the original level. Information technology now displaces very little water because the drinking glass has a very depression volume (Archimedes' Principle 2).

A marble in ice

Dorsum to our original scenario, what if the water ice cube had a small-scale marble embedded inside of it? When the ice melts, would the h2o level increase, decrease, or stay the same?

Permit'southward say we take the aforementioned ice cube every bit before (10g with a density of .92g/cm^3 and volume of 10.9cm^3) and a 1 gram marble with a density of 2g/cm^3. Using the formula higher up, we know the marble has to have a volume of .five cubic centimeters (1g / 2g/cm^three = .5cm^three). Evidently the marble would just sink if we tossed it in the glass because its density (2g/cm^3) is higher than water'south (1g/cm^3). And we know when submerged it would readapt .5cm^3 of water (Archimedes' Principle ii). Just when embedded in the water ice cube, what happens?

Will it float?

First, we need to make up one's mind whether the water ice cube will sink or float now that it has the marble in it. To practice this, we need to figure out the combined density of the ice cube AND marble. We know that the water ice cube has a mass of ten grams and the marble has a mass of i gram, for a combined mass of 11 grams. We also know that the water ice cube has a volume of 10.9cm^iii and the marble has a volume of .5cm^three, for a combined volume of 11.4cm^iii. Using the formula, we can determine that the combined density is .965g/cm^iii (11g / Density = 11.4cm^3, or xi/eleven.4 = .965). In other words, the small-scale marble obviously increases the combined density, but it'due south still less than the density of water, so the thing will definitely float!

Experiment

NOT the writer

I have an interesting talent of being able to float on water. When I spring into a pool, I sink like a rock. I'm a pretty big guy (200+ pounds) of medium build. There'due south nothing particular about my body that would cause it to bladder. But if I lie on my dorsum, extend my artillery and legs, take a deep breath, puff out my breast, and flex all my muscles, I can float well-nigh indefinitely without inappreciably moving a muscle. And y'all probably can too.

"How?", you ask. By making my volume larger, I subtract my density to but beneath that of the water. My mass doesn't modify when I'yard at the pool (heaven knows I wish it did). When you consider our formula, if my mass is fixed and I increase my torso volume, past definition my density must subtract. Most anyone tin can float if they brand themselves simply a chip bigger, but not whatsoever heavier. Try it side by side fourth dimension you get swimming.

Losing our marbles

Every bit noted higher up, the ice cube + marble has a mass of eleven grams, so it will initially displace 11 grams or 11cm^three of water. We've already figured out from above that the water from the melted ice cube will take upwardly 10cm^iii. One time the water ice cube has melted, the marble will submerge, and based on Archimedes' Principle 2 will readapt .5cm^3 (the marble's volume) of water. The combined book taken past the melted ice and submerged marble is 10.5cm^3, which is less than the 11cm^three initially displaced by them.

The h2o level volition lower by .5cm^3 when the ice melts.

Say what?!?

At first this seems illogical until you lot realize that the only influence of the ice cube on the water level is that it happens to float the marble. The ice cube itself neither increases nor decreases the water level, just with the heavier marble inside, the amount of water displaced by that water ice cube is greater at the beginning (but similar your atomic number 82 bucket inside the boat). In one case the marble is no longer floating, only its volume matters (just like tossing the atomic number 82 saucepan overboard or sinking the boat).

Bottom's upward

What if the marble and ice cube were instead submerged? A heavier/bigger marble would crusade it to sink once the combined water ice cube/marble density became greater than the water's. But let's say nosotros used the aforementioned marble embedded in the same ice cube as before, only used a magnet to forcefulness it to the bottom of the cup. When it melts, how much will the water level decrease? Will it decrease past more, less, or the same volume as when floating?

It's actually a very easy question to respond. Once submerged, we just need to look at book. The marble takes .5cm^3 and the frozen ice cube takes x.9cm^iii for a combined 11.4cm^3. In one case melted, the ice cube's water takes upwards just 10cm^3 and the marble nevertheless (obviously) takes upwards .5cm^iii, so the water level will decrease past .9cm^3 (11.4cm^3 - 10.5cm^3 = .9cm^3). The h2o level will decrease almost twice as much as it did for the floating water ice cube + marble.

An reverse scenario can occur if the water ice contains a notable corporeality of air bubbles or traps air between the liquid water and a layer of water ice. In this case the air causes the ice to float higher to a higher place the water surface causing lower displacement. When the ice melts, the bubbles go away and the increased volume of the floating ice joins the pitcher water volume and the water level may then increase.

Salt water

From above, we know that when an ice cube melts in fresh water, the water level stays the same. What if we used salt water instead of fresh water? Would the water levels change equally a fresh water water ice cube melts in it?

Permit's assume the same ice cube as our get-go scenario (mass of 10 grams and book of 10.9cm^iii) and very salty h2o with a density of ane.05g/cm^3. Archimedes' Principle 1 applies, so we know the water ice cube will displace a book of salt water that weighs 10 grams. Using the formula we determine that 10 grams of common salt water with density of one.05g/cm^3 will have a volume of about 9.5cm^3 (10g/1.05g/cm^3 = 9.52cm^3). Every bit before, you tin't fit a 10.9cm^3 water ice cube in a 9.5cm^3 infinite, then one.4cm^3 (almost 13%) of the ice cube volition float higher up the surface. It makes sense that the ice cube would float higher in salt water because of the salt h2o's higher density.

Once melted to fresh water, the water ice cube volition have the same volume equally before (10cm^iii), but it was dispersing only 9.5cm^3 of water space when floating, and so the water level will rise to account for the additional .5cm^three. This is a fairly small corporeality (just about 5% of the volume of the melted h2o), but information technology's notable.

NOTE: This does not account for the fact that the overall density of the h2o in the cup volition decrease a small amount as the fresh water mixes in with it. The result of this on things is rather miniscule.

Rising seas

What happens when you utilize this to oceans and water ice sheets? There's an estimated 1.iii billion cubic kilometers of body of water water. If put into a unmarried cube, this h2o would exist 1090 kilometers (675 miles) on each side and be 1090 kilometers high. It would fill a bath tub the size of Texas that is 30 miles tall! That's a lot of water, though when you consider that the book of Earth is simply over 1 trillion cubic kilometers, ocean water makes upwardly well-nigh .1% of Earth's volume (though incredible it covers 70% of information technology's surface, which shows how shallow the bounding main really is)!

There's an estimated 660,000 cubic kilometers of floating sea water ice. If put into one cake, it would be 87km (54 miles) on each side (nigh the footprint of the land of Delaware) and 87km high.

If all of that ice were to melt, what impact would it have on the bounding main levels? If both the water ice and the sea h2o were both fresh water (or both salt water), it would have no bear on at all (excluding all other factors, such as water temperature). But because of the difference in salinity (density) of the sea h2o and the ice, the increase in book would exist nigh ii.6% of the volume of the melted ice h2o, which when added to the book of the oceans, would raise the ocean level only nigh 4 centimeters (1.five inches). Details here.

Annotation that this only accounts for floating sea ice. The total amount of not-floating Arctic and Antarctic ice is about l times college, and because this is not currently floating (and displacing sea h2o), if it were all to melt the sea levels would rise significantly.

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Hopefully this has been a useful and thought-provoking presentation. Thank you to Doug who got me interested in this topic and inspired me to writer this. I welcome any comments or corrections.

andrewstross1967.blogspot.com

Source: http://www.smithplanet.com/stuff/iceandwater.htm

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