Glaciers Melting in a Teacup

Archimedes' Principles:

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.

Formula

Mass / Density = Volume Melting ice cube

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, 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 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 cubic 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.

Reference

The buoyant force felt by the floating ice is given by the weight of the water that is is displacing, i.e. the volume of the ice that is below the water line, times the density of water, times the gravitational field:

$F_\text{buoyant} = V_b \rho_w g$

This buoyant force is balanced by gravity. The strength of the gravitational force is the mass of the ice times the gravitational field. The mass of the ice is the total volume of the ice times the density of ice:

$F_\text{gravity} = M_\text{ice} g = \rho_\text{ice} V_\text{tot} g$

Because these forces balance ( $F_\text{gravity} = F_\text{buoyant}$ ), we have

$V_b \rho_w g = \rho_\text{ice} V_\text{tot} g$

or $V_b / V_\text{tot} = \rho_\text{ice} / \rho_w$ .

In other words, the fraction of the ice that is below the water is $V_b / V_\text{tot} = \rho_\text{ice} / \rho_w$ .

Now, what is the fraction of the ice that is above the water? It's $1 - V_b / V_\text{tot} = 1 - \rho_\text{ice} / \rho_w$ .

Therefore, the volume of the ice that's above the water is $V_\text{tot} (1 - \rho_\text{ice} / \rho_w)$ .

The original question we were interested in is whether when the ice melts, it will raise the water level in the glass. What I will show now is that the reduction in volume of the ice is exactly equal to the volume of ice that was above the water line.

What is the reduction in volume of the ice cube when it melts? Before it melts, it is equal to $V_\text{tot} = M_\text{ice} / \rho_\text{ice}$ . After it melts it is equal to $M_\text{ice}/\rho_w$ .

The difference is

$\begin{aligned} M_\text{ice} / \rho_\text{ice} - M_\text{ice}/\rho_w &= M_\text{ice}\left(\frac{1}{\rho_\text{ice}} - \frac{1}{\rho_w}\right) \\ &= \frac{M_\text{ice}}{\rho_\text{ice}} \left(1-\rho_w / \rho_\text{ice}\right) \\ &= V_\text{tot} \left(1-\frac{\rho_w }{ \rho_\text{ice}}\right) \end{aligned}$

If you compare this to the portion of the ice that was above the water line before it melted, the two quantities are exactly the same!

This means that the reduction in volume of the ice when it melts is equal to the volume of the ice that floated above the water level. The ice simply becomes water that occupies the same volume below the water line that it did to begin with.

The easiest way to explain the answer is to do the reverse experiment.

Put water in a glass (leaving some room at the top) and place it in a freezer. As the water freezes, the increase in volume will push some of the ice above the water's surface.

Now if you let this ice melt, the glass will go back to its original level and as it melts you will notice that the water level is always the same as the original level.

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

NICE, EASY SOLUTION (not too short though - for sake of better explanation)

PRINCIPLE: The weight of a floating body is equal to the weight of the displaced water (by weight we mean if we placed it on the scales).

CLARIFICATION: The displaced water is the water that would occupy the volume currently occupied by the submerged part of the body - it is SOMEWHAT abstract thing. It is NOT the water that has moved during the process of submerging a part of the floating body (see figure above).

EXPLANATION: If we (mentally) compress the floating body (no matter if it is ice or not) into the shape of its submerged part (green + yellow), maintaining uniform density:

• its mass stays the same (so does the gravitational force) and

• the buoyancy force upon it stays the same (no changes are made to the boundary of the water and body - as if the water "does not know" we changed the body at all).

Therefore, the body will remain in equilibrium. It will levitate (completely submerged, but not sinking). Note that such a body must now have the same AVERAGE density as the water itself:

• If it was denser - it would sink or

• If it was thinner - it would float to the surface,

but we know that it is impossible, since we know we have retained the equilibrium.

NOTE: We required UNIFORM DENSITY (during mental compression) to avoid the toppling of the body and leaving this geometrical configuration.

FURTHER EXPLANATION: For the case of ice, we do the same procedure. At the end, it will have the same density as the water and melting will not change anything - therefore, the level remains the same as when the ice was floating. (I know that you can not compress the ice like that - but we are not interested in the intermediate steps but only at the initial and the final states because we know that the way in which we melt the ice should not affect whether the water level will rise or lower).