If the polar ice caps were to melt .

[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.

 

[SW VandeCarr]

:

When an object is floating in a liquid medium it displaces it weight in the medium. A thirteen ton boat will displace thirteen tons of water.

When that same object is totally immersed, it displaces its volume, not its weight. Since the rocks weigh more than the lake water, when totally immersed they will displace a volume that is less than the volume that they displaced by weight. The water level will be lower.

So when you throw rocks into a confined volume of water, the water level will drop?

 

[Evo]

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)

Noerdlinger demonstrates that melt water from sea ice and floating ice shelves could add 2.6% more water to the ocean than the water displaced by the ice, or the equivalent of approximately 4 centimeters (1.57 inches) of sea-level rise.

We're going to accept the above publications as the best answer, the difference being the consideration of fresh water added to salt water. If the icebergs were melting in freshwater, there would be no rise is water level.