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

by Jared Smith

- Any floating object displaces a volume of water equal in weight to the object's MASS.
- Any submerged object displaces a volume of water equal to the object's VOLUME.

Mass / Density = Volume

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

The ice cube is floating, so based on Archimedes' Principle 1 above, we know that the volume of water being displaced (moved out of the way) is equal in mass (weight) to the mass of the ice cube. So, if the ice cube has a mass of 10 grams, 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 water, otherwise it wouldn't be floating. Water is one of the very few solids that is less dense than when in its liquid form. If you take a one pound bottle of water and freeze it, it will still weigh one pound, but the molecules will have spread apart a bit and it will be less dense and take up more volume or space. This is why water bottles expand in the freezer. It's similar to a Jenga tower. When you start playing it contains a fixed number of blocks, but as you pull out blocks and place them on top, 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^3, every cubic centimeter liquid water will weigh 1 gram). By the formula above (Mass / Density = Volume) and basic logic, we know that 10 grams of liquid water would take up 10 cubic cm of volume (10g / 1g/cm^3 = 10cm^3).

So let's say that our 10 gram ice cube has a density of only .92 grams per cubit centimeter. By the formula above, 10 grams of mass that has a density of .92 grams per cubic centimeter will take up about 10.9 cubic centimeters of space (10g / .92g/cm^3 = 10.9cm^3). Again, the volume of 10 grams of frozen water is more than the volume of 10 grams of its liquid counterpart.

The floating ice cube has a mass of 10 grams, so based on Archimedes' Principle 1, it is displacing 10 grams of water (which has 10cm^3 of volume). You can't squeeze a 10.9cm^3 ice cube into a 10cm^3 space, so the rest of the ice cube (about 9% of it) will be floating above the water line.

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

In short, **the water level will not change as the ice cube melts**

Using this same 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 some other metal into the boat, the boat will get lower in the water and the additionally displaced water in the pool will cause the pool level to rise. And based on Archimedes' Principle 1 for floating objects, it would rise by the volume of water equal in weight to the 100 pound lead bucket. Water weighs 8.3 pounds per gallon, so the boat will displace an additional 12 gallons of water (12 gallons * 8.3 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 we toss the bucket of lead overboard, the pool level goes down 12 gallons (the volume of water no longer displaced by the weight in the boat). But when it enters the water, it will be submerged, so we now need to apply Archimedes' Principle 2 for submerged objects (it will displace a volume of water equal to the object's volume). The water level will then go up by the volume of the lead bucket, which is 5 gallons. So, the net difference is that **the pool level will go down by 7 gallons**, even though the bucket is still technically in the pool.

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

Similarly, if your aluminum boat weighed 100 pounds, it would displace 100 pounds of water (12 gallons) when floating. But if the boat springs a leak and sinks, the pool level would decrease 12 gallons minus the volume of the aluminum in the boat. The boat's volume (amount of space comprised of aluminum metal) would be much less than 12 gallons. In fact, based on the density of aluminum (.1 pounds/in^3), we can determine using our formula that the volume of our 100 pound boat will be about 1000 cubic inches (100/.1 = 1000). There are 231 cubic inches per gallon, so the boat 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 same aluminum displaced when floating. In conclusion, **when our boat sinks, the pool level goes down by 7.7 gallons**.

As an experiment, fill a sink with 5 or 6 inches of water and note the water level. Next set a heavy 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 rise to make room for the empty glass and you'll note that it's difficult to get the glass to sink while also it is upright. The heavy glass displaces a lot of water because of the heavy mass of the glass (Archimedes' Principle 1), yet it still floats because of it's low density (don't forget about all the air inside the glass). The glass will feel lighter to you because of the buoyancy principle (the force of the displaced water pushing up against the weight of the object displacing it). It will float perfectly in the water when the weight of the glass equals the weight of the water it is displacing.

Now lay the glass down sideways and let it submerge in the sink. The water level will be only barely higher than the original level. It now displaces very little water because the glass has a very low volume (Archimedes' Principle 2).

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

Let's say we have the same ice cube as 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 above, we know the marble has to have a volume of .5 cubic centimeters (1g / 2g/cm^3 = .5cm^3). Obviously the marble would just sink if we tossed it in the glass because its density (2g/cm^3) is higher than water's (1g/cm^3). And we know when submerged it would displace .5cm^3 of water (Archimedes' Principle 2). But when embedded in the ice cube, what happens?

First, we need to determine whether the ice cube will sink or float now that it has the marble in it. To do this, we need to figure out the combined density of the ice cube AND marble. We know that the ice cube has a mass of 10 grams and the marble has a mass of 1 gram, for a combined mass of 11 grams. We also know that the ice cube has a volume of 10.9cm^3 and the marble has a volume of .5cm^3, for a combined volume of 11.4cm^3. Using the formula, we can determine that the combined density is .965g/cm^3 (11g / Density = 11.4cm^3, or 11/11.4 = .965). In other words, the small marble obviously increases the combined density, but it's still less than the density of water, so the thing will definitely float!

*NOT the author*

I have an interesting talent of being able to float on water. When I jump into a pool, I sink like a rock. I'm a pretty big guy (200+ pounds) of medium build. There's nothing particular about my body that would cause it to float. But if I lie on my back, extend my arms and legs, take a deep breath, puff out my chest, and flex all my muscles, I can float almost indefinitely without hardly moving a muscle. And you probably can too.

"How?", you ask. By making my volume larger, I decrease my density to just below that of the water. My mass doesn't change when I'm at the pool (heaven knows I wish it did). When you consider our formula, if my mass is fixed and I increase my body volume, by definition my density must decrease. Most anyone can float if they make themselves just a bit bigger, but not any heavier. Try it next time you go swimming.

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

**The water level will lower by .5cm^3 when the ice melts**.

At first this seems illogical until you 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, but with the heavier marble inside, the amount of water displaced by that ice cube is greater at the beginning (just like your lead bucket inside the boat). Once the marble is no longer floating, only its volume matters (just like tossing the lead bucket overboard or sinking the boat).

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

It's actually a very easy question to answer. Once submerged, we only need to look at volume. The marble takes .5cm^3 and the frozen ice cube takes 10.9cm^3 for a combined 11.4cm^3. Once melted, the ice cube's water takes up only 10cm^3 and the marble still (obviously) takes up .5cm^3, so the water level will decrease by .9cm^3 (11.4cm^3 - 10.5cm^3 = .9cm^3). **The water level will decrease almost twice as much as it did for the floating ice cube + marble**.

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 as a fresh water ice cube melts in it?

Let's assume the same ice cube as our first scenario (mass of 10 grams and volume of 10.9cm^3) and very salty water with a density of 1.05g/cm^3. Archimedes' Principle 1 applies, so we know the ice cube will displace a volume of salt water that weighs 10 grams. Using the formula we determine that 10 grams of salt water with density of 1.05g/cm^3 will have a volume of about 9.5cm^3 (10g/1.05g/cm^3 = 9.52cm^3). As before, you can't fit a 10.9cm^3 ice cube in a 9.5cm^3 space, so 1.4cm^3 (about 13%) of the ice cube will float above the surface. It makes sense that the ice cube would float higher in salt water because of the salt water's higher density.

Once melted to fresh water, the ice cube will take the same volume as before (10cm^3), but it was dispersing only 9.5cm^3 of water space when floating, so the water level will rise to account for the additional .5cm^3. This is a fairly small amount (only about 5% of the volume of the melted water), but it's notable.

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

What happens when you apply this to oceans and ice sheets? There's an estimated 1.3 billion cubic kilometers of ocean water. If put into a single cube, this water would be 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 volume of Earth is just over 1 trillion cubic kilometers, ocean water makes up about .1% of Earth's volume (though incredible it covers 70% of it's surface, which shows how shallow the ocean really is)!

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

If all of that ice were to melt, what impact would it have on the ocean levels? If both the ice and the sea water were both fresh water (or both salt water), it would have no impact at all (excluding all other factors, such as water temperature). But because of the difference in salinity (density) of the sea water and the ice, the increase in volume would be about 2.6% of the volume of the melted ice water, which when added to the volume of the oceans, would raise the ocean level only about 4 centimeters (1.5 inches). Details here.

Note that this only accounts for *floating* sea ice. The total amount of non-floating Arctic and Antarctic ice is about 50 times higher, and because this is not currently floating (and displacing sea water), if it were all to melt the sea levels would rise significantly.

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 author this. I welcome any comments or corrections.