SW VandeCarr said:
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.
Jack23454 said:
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?