If the polar ice caps were to melt .

[warfreak131]

If the polar ice caps were to melt...

wouldn't the sea levels decrease?

Water ice is less dense than liquid water, so the volume (and displacement of liquid water) is greater when its frozen. But when you melt that, the volume decreases, and some of the water that was initially displaced by the ice would flow back to its original position.

 

[Dotini]

wouldn't the sea levels decrease?

Water ice is less dense than liquid water, so the volume (and displacement of liquid water) is greater when its frozen. But when you melt that, the volume decreases, and some of the water that was initially displaced by the ice would flow back to its original position.

You must remember that the ice at Greenland and Antarctica is mostly sitting on continental crust, and not floating.

Respectfully submitted,
Steve

 

[SW VandeCarr]

Yes. The melting of floating ice does not raise sea levels. The melting of all land based ice could raise the sea level by 65-70 meters (about 215 - 230 ft).

http://www.usatoday.com/weather/resources/askjack/2004-11-21-melting-polar-ice_x.htm

 

[billiards]

wouldn't the sea levels decrease?

Water ice is less dense than liquid water, so the volume (and displacement of liquid water) is greater when its frozen. But when you melt that, the volume decreases, and some of the water that was initially displaced by the ice would flow back to its original position.

No. As already mentioned the most important point is that the ice is not floating.

However you have misunderstood the physics, if the water and the ice have the same composition then the water level will not change when the ice melts. However, if floating ice of fresh water composition melts in a denser saline solution the water level will RISE.

Consider a mass of ice floating in sea water. Archimedes principle states that the ice displaces its own weight in sea water. When the ice melts it converts all of its weight into fresh water which is less dense than sea water -- this is the key point -- the same weight of fresh water takes up more volume than seawater. This causes the sea level to rise slightly. This effect is mostly ignored in sea level budgets, however, it is not a negligible effect.

 

[Andre]

There are several more factors to consider, for instance the change in gravity patterns as the ice sheets are still a big mass that adhere to Newtons law. Furthermore there is supposed to be isostacy/eustacy, the land previously below the ice sheets, rebouncing. But imo it's more complicated.

 

[SW VandeCarr]

However you have misunderstood the physics, if the water and the ice have the same composition then the water level will not change when the ice melts. However, if floating ice of fresh water composition melts in a denser saline solution the water level will RISE.

When land based water or water ice enters the sea, it is adding mass to the sea and sea levels will rise, discounting evaporation. The floating ice (of any composition) is part of the mass of ocean and displaces a volume of water equal to its weight. So its melting does not cause the water level to rise.

You can do a simple experiment: The water level in a glass of water that already contains a floating ice cube will not change as the ice melts. If you add an ice cube to a glass of water, the water level will of course rise.

 

[Greg Bernhardt]

When land based water or water ice enters the sea, it is adding mass to the sea and sea levels will rise. The floating ice (of any composition) is part of the mass of ocean and displaces a volume of water equal to its weight. So its melting does not cause the water level to rise.

So the concern comes from say the Greenland and Arctic ice caps?

 

[SW VandeCarr]

So the concern comes from say the Greenland and Arctic ice caps?

From the Greenland ice cap, but not the floating Arctic ice pack. By far, the biggest mass of land based ice is the Antarctic ice cap.

 

[256bits]

No. As already mentioned the most important point is that the ice is not floating.

However you have misunderstood the physics, if the water and the ice have the same composition then the water level will not change when the ice melts. However, if floating ice of fresh water composition melts in a denser saline solution the water level will RISE.

Consider a mass of ice floating in sea water. Archimedes principle states that the ice displaces its own weight in sea water. When the ice melts it converts all of its weight into fresh water which is less dense than sea water -- this is the key point -- the same weight of fresh water takes up more volume than seawater. This causes the sea level to rise slightly. This effect is mostly ignored in sea level budgets, however, it is not a negligible effect.

In that case, ice, or an iceberg, will float higher up in saline seawater. And when the ice melts the sea level will rise proportionally. I guess the extreme experiment to prove this would be to float an ice cube in a glass of liquid mercury ( denser than the ice). The liquid level ( mercury/water ) when the ice melted would be higher than the liquid level ( mercury ) with the ice floating on top.

 

[billiards]

When land based water or water ice enters the sea, it is adding mass to the sea and sea levels will rise, discounting evaporation.

Agree, and never expressed otherwise.

The floating ice (of any composition) is part of the mass of ocean and displaces a volume of water equal to its weight. So its melting does not cause the water level to rise.

No you're overlooking the fact that the density of fresh water is less than the density of sea water. It turns out to be important, for shelf ice at least.

If a floating body of ice has mass m we can do some simple calculations to consider the implications to sea level if it melts.

These relations are trivial and yet hidden within them is the truth.

m=\rho_{ice}V_{ice}

Upon melting to fresh water the mass remains the same.

m=\rho_{freshwater}V_{freshwater}

By Archimedes' principle, the weight of the ice is equal to the weight of the displaced sea water.

mg=\rho_{seawater}V_{displaced seawater}g

Divide by g.

m=\rho_{seawater}V_{displaced seawater}

Bring together the fresh water and seawater terms.

m=\rho_{freshwater}V_{freshwater}=\rho_{seawater}V_{displaced seawater}

Rearrange:

V_{freshwater}=\frac{\rho_{seawater}}{\rho_{freshwater}}V_{displaced seawater}

If seawater is 5% denser than freshwater then.

V_{freshwater}=1.05V_{displaced seawater}

Sea level is related to the volume of water in the basin such that the greater the volume of water in the basin the greater the sea level.

V_{filledbasin}(beforemelting) = datum + V_{displaced sea water}

V_{filledbasin} (aftermelting) = datum + V_{fresh water}

V_{filledbasin} (aftermelting) = datum + 1.05 * V_{displaced sea water}

Therefore:

V_{filledbasin} (beforemelting) < V_{filledbasin} (aftermelting)

 

[SW VandeCarr]

Agree, and never expressed otherwise.
No you're overlooking the fact that the density of fresh water is less than the density of sea water. It turns out to be important, for shelf ice at least.

If a floating body of ice has mass m we can do some simple calculations to consider the implications to sea level if it melts.

These relations are trivial and yet hidden within them is the truth.

I agree that 1 kg of fresh water occupies a larger volume than 1 kg salt water, other things being equal. The issue is whether the ocean level rises when floating ice melts. When fresh water ice enters the ocean, it will raise the ocean level immediately by displacing whatever volume based on its density. If liquid fresh water enters the ocean, it would "float" because it is less dense and raise the ocean level. So, as far as I can tell, there is no difference between a given amount of floating fresh water ice melting in the ocean and an equivalent amount of fresh water entering the ocean in liquid form. That's the point I was making.

EDIT: I did find some articles that agree with you, but they don't appear to be from peer reviewed journals. On one hand, it seems that as a large mass of floating fresh water ice melts, the released fresh water would occupy a larger volume than the salt water mass it displaces. In part, this would be compensated by the decreased volume of salt water displacement by the melting ice mass. In any case, how long does the melt water remain "fresh"? It seems that the fresh melt water would quickly lose its characteristics as it is dispersed in the much larger volume of salt water. The salinity of the oceans is maintained by a balance between evaporation and fresh water infusions.

Could you find a peer reviewed article that supports your argument?

 

[Jack23454]

If the ice is floating it is only displacing it's equal weight. If submerged it is displacing volume. If it has trapped air in the ice, the volume will decrease when it melts and the air is no longer contained. In that case the water level would be less. To answer the question empirically; submerge an ice cube with a magnet frozen in the center. Hold it in place near the bottom with a magnet. When the ice melts see if the water level changes, assuming evaporation is not a noticeable factor.

 

[Nik_2213]

Uh, IMHO, the relative expansions of salt and fresh water is a minor detail compared to the way the additional free water will arrange itself on the globe.

Also, you'll get 'primary isostatic rebound' as the land masses and associated 'moat' depressed by those huge ice-caps gradually pop up a kilometre or two, displacing more water, then 'secondary' effects as areas beyond the 'moat' sink slower...

 

[SW VandeCarr]

So the concern comes from say the Greenland and Arctic ice caps?

When you say "concern", yes there is concern about the melting floating Arctic ice pack because of its predicted effect on Arctic ecology and the world climate, but not because of its potential to add to rising sea levels. Rising sea levels will be mostly due to the melting of land based ice and the redistribution of water due to continental rebound as the ice melts.

The article I linked to treated rising sea levels is a long term issue, but that position has been disputed. Many worry about the possible instability of the Greenland and Antarctic ice sheets and that these ice sheets could become mobile if liquid water accumulates underneath them. That means that large volumes of ice could be dumped into the ocean in a relatively short time. It's just one of the several proposed theories.

http://www.sciencedaily.com/releases/2010/01/100116103350.htm

 

[billiards]

If liquid fresh water enters the ocean, it would "float" because it is less dense and raise the ocean level. So, as far as I can tell, there is no difference between a given amount of floating fresh water ice melting in the ocean and an equivalent amount of fresh water entering the ocean in liquid form. That's the point I was making.

It won't "float" in the sense of the ice, which clearly has some of it's head above the water. The fresh water will redefine sea level. When it is ice it redefines sea level by displacing saline sea water; when it melts it redefines sea level using its own fresh water. < That's the key.

I fully acknowledge that it is a small effect. But it is one that is almost always overlooked. It can safely be overlooked when we consider sea ice because sea ice is derived from the ocean itself and melts and forms every year, it cannot be safely overlooked for shelf ice which is the floating portion of the continental ice sheets. The point is that the melting of the shelf ice will act to slightly increase sea level by the effect I have described, it is not a zero contribution as is often assumed.

Could you find a peer reviewed article that supports your argument?

Haven't looked. But the point is that it is widely ignored, which is why I bring it up. I would be interested to see if someone here can debunk it.

If the ice is floating it is only displacing it's equal weight. If submerged it is displacing volume. If it has trapped air in the ice, the volume will decrease when it melts and the air is no longer contained. In that case the water level would be less. To answer the question empirically; submerge an ice cube with a magnet frozen in the center. Hold it in place near the bottom with a magnet. When the ice melts see if the water level changes, assuming evaporation is not a noticeable factor.

Good point. We can include this in the maths quite simply.


Upon melting to fresh water the mass remains the same except for the component of trapped air which escapes.

m=\rho_{freshwater}V_{freshwater}+\rho_{air}V_{trappedair}

Bring together the fresh water and seawater terms.

m=\rho_{freshwater}V_{freshwater}+\rho_{air}V_{trappedair}=\rho_{seawater}V_{displaced seawater}

Rearrange:

V_{freshwater}=\frac{\rho_{seawater}}{\rho_{freshwater}}V_{displaced seawater}-\frac{\rho_{air}}{\rho_{freshwater}}V_{trappedair}

Consider sea water 1% denser than fresh water, and freshwater 500 times denser than the trapped air (conservative numbers) and rearrange:

1.01V_{displaced seawater}-V_{freshwater}=0.002V_{trappedair}

Now how much trapped air would we need such that the volume of released fresh water is equal to the displaced sea water?

If V_{displaced seawater}=V_{freshwater}
then 0.01V_{freshwater}=0.002V_{trappedair}
that is 5V_{freshwater}=V_{trappedair}

Now the volume of freshwater is increased by about 10% when it freezes to ice.

5V_{freshwater}=5.5V_{ice}=V_{trappedair}

So on the back of an envelope I estimate conservatively that you need air to occupy 2 parts of volume for every 11 parts of volume occupied by ice (air occupies at least 15% of the total volume) such that when the ice melts in sea water the released fresh water does not raise sea level. Of course, the ice would need to be greater than 15% air for the sea level to drop, any less than 15% and the sea level will rise.

Now what is the volume of air as a percentage of the antarctic ice shelf?

 

[Studiot]

@Billiards

When considering the ice melting you need to include the dilution effect of the meltwater in your calculations.

The resultant seawater will be somewhat less saline due to the addition of fresh water.

This will change its density.

go well

 

[SW VandeCarr]

It won't "float" in the sense of the ice, which clearly has some of it's head above the water. The fresh water will redefine sea level. When it is ice it redefines sea level by displacing saline sea water; when it melts it redefines sea level using its own fresh water. < That's the key.

You didn't address my point that even if fresh water is released, it will be dispersed in the much larger volume of salt water so that its true effect in raising sea levels is negligible, even for large Antarctic icebergs. Many factors affect ocean salinity which reflects the balance between of fresh water input from all sources vs evaporation. That's why I asked for peer reviewed sources support the position that this is not merely a theoretical argument.

I think a much more important effect from major fresh water sources such as from a large river, is that it raises the freezing point of seawater over a large area since, being less dense, the fresh water will tend to remain on the surface. It's demonstrably true that sea ice forms regularly in the Gulf of St Lawrence, which is at the lowest latitude for seasonal sea ice formation in the North Atlantic.

 

[billiards]

@Billiards

When considering the ice melting you need to include the dilution effect of the meltwater in your calculations.

Why?

The resultant seawater will be somewhat less saline due to the addition of fresh water.

This will change its density.

Yes, but this is after the effect has already established itself.

I don't see how diffusion of salt will change the volume.

 

[Studiot]

Consider a beaker of salt water, density p₁

add a quantity of pure water, density p₀

This results in a quantity of salt water, density p₂

Since density is a function of salinity

p₁ > p₂ > p₀

Whether this effect is significant depends upon the relative quantities of original saline and meltwater.

go well

 

[billiards]

You didn't address my point that even if fresh water is released, it will be dispersed in the much larger volume of salt water so that its true effect in raising sea levels is negligible, even for large Antarctic icebergs.

I agree it is a small effect.

To put it into perspective consider the sea level, z in relation to the volume of the basin, V and the area, A of the basin V=Az.

V_{new}=V_{old}+\Delta V

Az_{new}=Az_{old}+\Delta V

Change in sea level is denoted \Delta z

\Delta z=z_{new}-z_{old}=\Delta V/A

\Delta z=(V_{freshwater}-V_{displaced seawater})/A

If for example V_{freshwater}=1.05V_{displacedseawater}, then
\Delta z=0.05V_{displaced seawater}/A

and V_{displaced seawater}=\frac{\rho_{ice}}{\rho_{seawater}}V_{ice}\approx0.9V_{ice}

\Delta z=0.045V_{ice}/A

So for example with these numbers a 1000 km³ volume of ice melting in a 100 million km² expanse of ocean basin will contribute about .45mm of sea level rise. Not much but I would not call this negligible.

 

[billiards]

Consider a beaker of salt water, density p₁

add a quantity of pure water, density p₀

This results in a quantity of salt water, density p₂

Since density is a function of salinity

p₁ > p₂ > p₀

Whether this effect is significant depends upon the relative quantities of original saline and meltwater.

go well

Right. But we are considering volume.

Is the volume of stratified p₁ + p₀ any different from well mixed p₂?

 

[Studiot]

As you rightly pointed out, the mass does not change just because the ice melts.

The volume is always governed by the equation volume = mass/density.

Since the mass remains constant, the volume changes with density.

 

[billiards]

As you rightly pointed out, the mass does not change just because the ice melts.

The volume is always governed by the equation volume = mass/density.

Since the mass remains constant, the volume changes with density.

The volume of what though? I don't think you've thought this one through.

 

[jambaugh]

This reminds me of a physics problem I had many many years ago. In the following case you are to determine whether the water level will rise, fall or stay the same after the ice melts:

A block of ice is floating in a beaker of water. Within the block of ice is frozen:
1.) Nothing, the ice is solid. (level stays the same)
2.) A ball bearing which will sink to the bottom when the ice melts.
3.) A ball of foam which will float when the ice melts.
4.) A sphere of air at ambient air density.
5.) A spherical void which is a vacuum.

 

[Studiot]

I don't think you've thought this one through.

?

If you float some pure ice in a beaker of pure water and let the ice melt, the level will not change after melting as Jambaugh has said.

However this statement is predicated upon the composition of the liquid being the same before and after melting.

If we live in magicland and the action of melting transforms all the mass of material to paraffin the level will be different.

A little less obviously, if the action of melting transforms the resultant liquid to a different composition the level will be different.

What I don't (readily) have is the information as to how much ice there is realative to the ocean mass so I can't make a judgement as to how significant the effect is.
As a geologist, you should be able to do this.

 

[billiards]

?

If you float some pure ice in a beaker of pure water and let the ice melt, the level will not change after melting as Jambaugh has said.

However this statement is predicated upon the composition of the liquid being the same before and after melting.

If we live in magicland and the action of melting transforms all the mass of material to paraffin the level will be different.

A little less obviously, if the action of melting transforms the resultant liquid to a different composition the level will be different.

What I don't (readily) have is the information as to how much ice there is realative to the ocean mass so I can't make a judgement as to how significant the effect is.
As a geologist, you should be able to do this.

Which bit of any of this really is relevant? It seems to me like you're taking a stab at me but you're not quite sure where you're putting the knife.

 

[Studiot]

It seems to me like you're taking a stab at me but you're not quite sure where you're putting the knife. QUOTE]

Perhaps if you were to address my point instead of posting insulting comments.

I have treated all your posts seriously but would argue, most strongly, my right to assert that

Two equal masses of liquids of different composition and therefore density occupy different volumes.

Therefore any attempt to assess the change in world ocean volume due to melting icecaps must take into account, or show to be negligeable, the change in composition due to the dilution of saline water by meltwater.

 

[billiards]

Therefore any attempt to assess the change in world ocean volume due to melting icecaps must take into account, or show to be negligeable, the change in composition due to the dilution of saline water by meltwater.

If you add salt to water the volume doesn't change significantly. This is because salt is soluble in water.

At the risk of repeating myself, all I am saying is that when ice melts it turns into fresh water. The addition of this fresh water to the sea does more to increase the sea level than would the insertion of the ice into the sea. The maths is all there and I have made it very easy to follow for anybody interested in taking apart the logic.

What I think you are saying is that I have made an erroneous assumption in saying that the addition of the fresh water will increase the volume of the sea by exactly the volume of the fresh water. Instead the fresh water will take in some salt from the sea, and therefore it will become more dense. Up to here I would agree, but the next point I think is where we disagree.

You seem to argue that because the density of this water increases the volume must decrease to accommodate this increase in density. However, that is wrong because the increase in mass (by the influx of salt) is what causes the density to increase, the volume stays pretty much the same. I argue that the transfer of salt from one water body to the other makes no difference to the volume of the water, and therefore that this point is moot, and all my previous arguments stand unaffected.

This is my reading of your argument, I hope that I have made it clear enough for you to tell me where exactly I am wrong.

 

[Studiot]

This is my reading of your argument, I hope that I have made it clear enough for you to tell me where exactly I am wrong.

1) I have never stated that the volumes are additive. To do so would be to promote a falsehood.

2) I have stated (repeatedly) that the density of salt water is greater than that of fresh water.

3) That if you mix some salt water with some fresh water the resultant liquid will have a density intermediate between the two.

4) The the volume of the mixture will be equal to the mass of the mixture divided by its density.

Which of these statements do you disagree with?

 

[billiards]

1) I have never stated that the volumes are additive. To do so would be to promote a falsehood.

So if you have a 1 litre of salt water, and 1 litre of fresh water, put them together and shake them up, how much water do you have?

The answer may not be exactly 2 litres, but in my calculations I have assumed that it will be. Please can you clarify for me that this is your objection. If not what is your objection?

(If yes: How important is this effect? Anyone)

 

[Studiot]

So if you have a 1 litre of salt water, and 1 litre of fresh water, put them together and shake them up, how much water do you have?

I assert that my comment (4) is the only way to answer this, using experimental salinity-density curves which are published in geoscience texts. Temperature will also need to be accounted for.

I also said that I do not know if the variation is significant only proper calculation will show this.
However I do know that there has been discussion in this thread about the effects of the difference in density btween salt and fresh water. I also know that there is no maximum density at 4^oC in seawater as with fresh. The density/temperature cureve follows that of a normal substance, increasing steadily to the freezing point.

go well

 

[jambaugh]

OK Now, let's just do the math.

Put 1.025Kg of ice afloat in (V-1) cubic meters of sea-water with surface density 1.025 kg/m^3. The sea water then contains V-1 cubic meters of water = (V-1) kg and 0.025(V-1) kg of dissolved salt.

The ice displaces 1.025 kg = 1 m^3 of sea-water. So the containing volume below surface is (V-1)+1 = V cubic meters.

Now melt the ice. You have V+0.025 Kg of water containing 0.025*V kg of salt. The below-surface volume will have increased from V to V+0.025.

This assumes no significant change in volume of a fixed mass of water due to the amount of salt added. Remember that surface salinity is a bit less than at depth. There may be a slight affects due to changes in compressibility of water at depth but remember that water is supporting the same weight above it whether its ice or brackish water unless you significantly change the density at depth. That would cause slight additional increase in volume. Change in salt concentration may also change the compressibility of water and I'm not sure how. But I think all this is negligible relative to the 2.44% addition of the ice's mass to the below surface volume upon melting. That in turn is a small percentage of the total ice volume for floating ice.

Finally remember that this does not apply to water resting upon land e.g. the Southern ice-cap. Since the only water it displaces is the amount needed to fill the volume of ice which is below the sea-level.

 

[billiards]

OK Now, let's just do the math.

Put 1.025Kg of ice afloat in (V-1) cubic meters of sea-water with surface density 1.025 kg/m^3. The sea water then contains V-1 cubic meters of water = (V-1) kg and 0.025(V-1) kg of dissolved salt.

The ice displaces 1.025 kg = 1 m^3 of sea-water. So the containing volume below surface is (V-1)+1 = V cubic meters.

Now melt the ice. You have V+0.025 Kg of water containing 0.025*V kg of salt. The below-surface volume will have increased from V to V+0.025.

This assumes no significant change in volume of a fixed mass of water due to the amount of salt added. Remember that surface salinity is a bit less than at depth. There may be a slight affects due to changes in compressibility of water at depth but remember that water is supporting the same weight above it whether its ice or brackish water unless you significantly change the density at depth. That would cause slight additional increase in volume. Change in salt concentration may also change the compressibility of water and I'm not sure how. But I think all this is negligible relative to the 2.44% addition of the ice's mass to the below surface volume upon melting. That in turn is a small percentage of the total ice volume for floating ice.

Finally remember that this does not apply to water resting upon land e.g. the Southern ice-cap. Since the only water it displaces is the amount needed to fill the volume of ice which is below the sea-level.

I think we agree then. I already did "the math" on page 1. Now we're discussing how valid that little assumption of 'no significant change in volume of a fixed mass of water due to the amount of salt added.'

 

[Studiot]

The ice displaces 1.025 kg = 1 m^3 of sea-water

Are you sure about this?

 

[jambaugh]

I think we agree then. I already did "the math" on page 1.

oops! Missed the math (I have posts listed by newest on top)

Now we're discussing how valid that little assumption of 'no significant change in volume of a fixed mass of water due to the amount of salt added.'

My guess is it is insignificant. I suggest you compare it to the actual volume of the equivalent undissolved salt plus the volume of pure water. But I'll leave it to you two to hash out.
Bye!

 

[Studiot]

I was under the impression the density of water was about 1000Kg/m^3

 

[billiards]

I was under the impression the density of water was about 1000Kg/m^3

True. But the maths is unaffected. Replace kg with Mg if you prefer, or even just the abstract symbol m.

 

[Studiot]

Surely the correct calculation is as follows

Take the total weight of world ocean (W) in Kg. Add to this the weight of ice (I)

Taking the salinity at 35g of salt per Kg of ocean water

So new salinity is 1000W(.035)/{W+I} g/Kg

Look up the density for this salinity at the desired temperature say p then volume = (W+I)/p

if you can show that I is insignificant in relation to W then you can discount the effect.

Edit: So far as I can determine I is somewhere between 2% and 3% of W say 2.5%

So new salinity is 35/1.025 ie approx 34 g/Kg - a reduction of a between 2% and 3%.

So all that remains is to check the effect of this on density. I suspect the effect is small. But it is nevertheless there.

go well

 

[jambaugh]

I was under the impression the density of water was about 1000Kg/m^3

Ooops again. I was writing kg but meaning metric ton.

 

[billiards]

I will attempt to address the salinity issue, and I will show that this effect works against the effect that I described earlier (of fresh water ice melting in salt water), however it is an order of magnitude smaller than that effect. Using the same numbers for volume of ice and sea water as before I find that this effect will lower the sea water by about 0.02 mm from what it would have been if the volumes were simply additive.

Here is my reasoning:

Consider two volumes of water V_1 and V_2, the question is, what is the volume V_3 if we mix them together, taking into account that both volumes have different salinities?

First consider the conservation of salt which gives us:

m_3 S_3 = m_1 S_1 + m_2 S_2

Now density of water is a function of salinity at a given temperature and pressure. I found this graph which suggests that the relationship is linear.

[PLAIN]http://www.marietta.edu/~biol/biomes/physsal.gif
http://www.marietta.edu/~biol/biomes/water_physics.htm

My reasoning will assume that density is a linear function of salinity:

\rho=\kappa S + C

rearrange:

S=\frac{\rho - C}{\kappa}

Substituting this into the salt conservation equation yields:

m_3 \frac{\rho_3 - C}{\kappa} = m_1 \frac{\rho_1 - C}{\kappa} + m_2 \frac{\rho_2 - C}{\kappa}

Use that \rho=m/V and multiply through by \kappa.

m_3 (m_3 / V_3 - C) = m_1 (m_1 / V_1 - C) + m_2 (m_2 / V_2 - C)

Use conservation of mass, i.e. m_3=m_1 + m_2 to cancel out C.

\frac{m_3^2}{V_3}=\frac{m_1^2}{V_1}+\frac{m_2^2}{V_2}

V_3=\frac{V_1 V_2 (m_1 + m_2)^2}{V_2 m_1^2 + V_1 m_2 ^2}

So now we have an expression for V_3 in terms of V_1 and V_2. The first thing to note is that is not simply additive, so my assumption was just that an assumption, so the real question is, how good of an assumption is it?

Consider adding 10^3 km^3 of fresh water to an ocean of 10^8 km^3 of salt water.

Previously I assumed that the volumes were additive such that the total volume of water in the ocean would be (10^8+10^3) km^3=100001000 km^3.

But if I assume the ocean is 5% denser than the fresh water I actually find using my equation that the volume of water in the ocean would be 100000997.7324 km^3.

The difference between these answers is about 2.27 km^3, meaning that there is 2.27 km^3 less water in the ocean than I had previously assumed.

Now I showed earlier that the increase in water in the basin after melting of a floating ice body (assuming sea water 5% denser than melt water, and ignoring the effect of trapped air in the ice) was about \Delta V \approx 0.045 V_{ice}. This meant that if we melted 10^3 km^3 of ice, we added about 45 km^3 of water to the ocean, which I then estimated would lead to about 0.45 mm of sea level rise in an ocean of 10^8 km^2 area. This approximation was wrong because it assumed that the volumes were additive.

Taking into account the salinity effect (which means that the volumes aren't additive), the actual volume of water will be about (45-2) km^3 = 43 km^3. So sea level rises by about 0.43 mm instead of 0.45 mm; this makes a difference of 0.02 mm.

I completely acknowledge that we are talking about a negligible effect on top of a negligible effect. So what have we learned from this?

(1) All this added complexity is not relevant, but it is interesting.
(2) It looks as though if the relationship between the density of a substance and the percentage of dissolved matter is linear, then the volume will not be very much affected irrespective of the gradient of the linear function.

 

[Studiot]

That's proper science at work now.

I can contribute the following.

approx volume of oceans 1.3 x 10⁹ cu kilometers
approx volume of ice 3 x 10⁷ cu kilometers

source Wiki

The variation of density with temperature is interesting and shown in the attachment (after Harvey)

This is now a much more valuable thread.

 

[billiards]

Could you find a peer reviewed article that supports your argument?

I was trying to find some more realistic numbers to plug in so I could get a feel for the magnitude of the effect if all the shelf ice melted. I came across this:

http://onlinelibrary.wiley.com/doi/10.1111/j.1365-246X.2007.03472.x/abstract?systemMessage=Wiley+Online+Library+will+be+disrupted+6+Aug+from+10-12+BST+for+monthly+maintenance

This is exactly what I was talking about, and they have considered the salt mixing. They estimate that if all the shelf ice were to melt, the global sea level would rise by 4 cm due to this effect. However, they did not consider the air bubbles in the ice sheet, an effect that I might revisit as I think it could turn out to be important.

See also http://efdl.cims.nyu.edu/publications/refereed/grl_melt_floating_ice_07.pdf (open access). This paper talks of the temperature effect, the cooling of the ocean related to the melting of the floating ice increases density and lowers sea level a little bit. It is a little bit uncertain where this effect should be place, is it a melting effect or is it a thermal expansion (or in this case contraction) effect? In this case it is both, the authors argue that we should bulk it with the melting effect and then treat subsequent thermal expansion separately. Thermal expansion will eventually negate the cooling effect of melting. There is still no mention of bubbles!

 

[billiards]

The effect of bubbles (revised)

I've had a bit more of a think as to the importance of air bubbles and it turns out I made a mistake somewhere in my initial calculation.

I still think that we should think of bubbles as lost mass in the melting process.

m=\rho_{ice} V_{ice}=\rho_{freshwater} V_{freshwater} + \rho_{trappedair} V_{trappedair} = \rho_{seawater} V_{displacedseawater}

Now I consider the case where the displaced sea water is exactly balanced by the release of less dense fresh water (and ignore the salinity effect) so that there is no net change in sea level. I want to know, how much air do we need in the ice to cancel the effect?

V_{freshwater}=V_{displacedseawater}

In this case it is easy to show that:

\frac{V_{trappedair}}{V_{freshwater}}=\frac{(\rho_{seawater}-\rho_{freshwater})}{\rho_{trappedair}}

It is more geophysical to consider the volume of air in relation to solid ice in the glacier.

\frac{V_{trappedair}}{V_{solidice}}=\frac{\rho_{solidice}}{\rho_{freshwater}} \frac{(\rho_{seawater}-\rho_{freshwater})}{\rho_{trappedair}}

Let's plug in some reasonable numbers:

\frac{V_{trappedair}}{V_{solidice}}=0.9 \frac{0.02}{0.002}=9

In other words, you need about nine times more air than solid ice to stop the melting effect from increasing sea level (ignoring salinity and temperature changes). That's a glacier filled 90% with air! This is clearly unrealistic.

Now if we consider a glacier a bit under 10% filled with air (I suspect a safe over estimate of the real amount), we can see how significant the air bubbles are. In this case:

V_{freshwater} = \frac{\rho_{seawater} V_{seawater}}{(\rho_{freshwater}+0.1\rho_{air})}= \frac{1.03 V_{seawater}}{(1+0.0002)}=1.0298 V_{seawater}

Compare this to the equation with no bubbles:

V_{freshwater} = \frac{\rho_{seawater} }{\rho_{freshwater}}V_{seawater}=1.03 V_{seawater}

We're talking about an effect change of 1%. Two orders of magnitude down for ice filled ~10% with bubbles.

 

[DrClapeyron]

Wouldn't these massive cold water influxes change the amount of latent heat given off during ocean water evaporation, thus causing global cooling? Since the supposed phenomenon of man made global warming would heat the ocean's lower waters causing expansion, the influx of cold water from the north pole ice cap should cause a delay by cooling these waters. That is until the warm upper strata of ocean water, formed after the thermohaline has slowed down to a near hault, melts the Greenland and Antarctic ice.

So as temperature rises, it melts these ice caps, adding cold water to the oceans which then give off less heat as a result of being cooled. Eventually when all polar ice caps are completely melted the Earth will experience higher, more sustained temperature fluxations.

 

[Evo]

Let's stay on the topic of water displacement please. GW/CC are closed topics.

 

[Hush]

An astonishing, amazing thread. As an interested, in this case, noncontributing reader, I simply voice my appreciation to all participating and reading. Personal learning curve: infinite.

"That's proper science at work now." - Studiot
Yes.

Let this exemplary discourse and thread be viral and contagious - a welcome infection to any discourse in science. All here. Sincerity, integrity and honesty. A rarity. A benefit to anyone in search of and in the possession of luck to be their witness.

Bookmarked and contributors thanked.

 

[DoggerDan]

However, if floating ice of fresh water composition melts in a denser saline solution the water level will RISE.

Don't think so. Anything floating on the surface of the water, whether it's a sack of feathers or a battleship, will displace precisely what it weighs. The form doesn't matter. For example, the battleship could be floating on top of the water, or a molten heap at the bottom of the ocean. In both cases it'll displace precisely the same amount of water.

Same goes for ice, before and after melt.

Billiards, your math is extensive, but this is a very simple concept. No math is required.

 

[turbo]

If the Antarctic shelf ice were to melt, we'd probably not notice a change in sea level (apart from some ice that is anchored/partially supported by land-borne ice). If the ice-cap on solid land at Antarctica (and to a lesser extent Greenland) were to melt extensively, we would certainly notice.

 

[billiards]

Don't think so. Anything floating on the surface of the water, whether it's a sack of feathers or a battleship, will displace precisely what it weighs. The form doesn't matter. For example, the battleship could be floating on top of the water, or a molten heap at the bottom of the ocean. In both cases it'll displace precisely the same amount of water.

Same goes for ice, before and after melt.

Billiards, your math is extensive, but this is a very simple concept. No math is required.

Really? Try reading these peer reviewed papers.

http://onlinelibrary.wiley.com/doi/10.1111/j.1365-246X.2007.03472.x/abstract?systemMessage=Wiley+Online+Library+will+be+disrupted+6+Aug+from+10-12+BST+for+monthly+maintenance

http://efdl.cims.nyu.edu/publications/refereed/grl_melt_floating_ice_07.pdf (open access)

 

[billiards]

If the Antarctic shelf ice were to melt, we'd probably not notice a change in sea level (apart from some ice that is anchored/partially supported by land-borne ice). If the ice-cap on solid land at Antarctica (and to a lesser extent Greenland) were to melt extensively, we would certainly notice.

It is estimated that if all floating ice were to melt the global sea level would rise by about 4 cm. The Antarctic ice shelf makes up about 90% of the world's floating ice. This is a small but non-negligible contribution to sea level rise.

Consider an interesting secondary effect noted by Noerdlinger and Brower (2007). If a chunk of grounded ice sheet were to collapse into the ocean it may immediately displace enough sea water, for example, to provide a catastrophic 4 m of sea level rise. However, there is more, when the floating ice melts there will be an additional 10 cm of sea level rise. This is small compared with the initial 4 metres, but planners must be aware that there is an additional 10 cm to come.