how many lbs of ice will it take to bring down the temp of a 20000
gallon pool of water by 1 degreee

You would need 350 gallons of 32 degree water (ice) to bring 20,000
gallons of water from 90 degrees to 89 degrees, but keep in mind this
will only be a temporary change if it is hotter than 89 out. 
According to this article http://slate.msn.com/id/1003374/, you would
need about 875 pounds of ice.

32 degree water is not the same thing as ice. A block of ice will
absorb more energy than an equivelent volume of 32 degree water to
overcome its latent heat of melting.

Firmware - try an experiment.  Get a cup of water.  Put it in a 32
degree area (freezer) and come back in one hour.  Tell me what you
see.

I see I made an error, what I should have said is that a block of ice
will absorb more energy than an equivalent MASS of 32 degree water.
You MUST take account of latent heat when you answer this type of
problem - look here for more information on latent heat
(www.physchem.co.za/Heat/Latent.htm). Research_help_ga's answer is
wrong because he/she didn't do this.

This look suspiciously like a homework question, so I won't give the
answer. However, the correct approach is to derive an equation for how
much energy is required to first melt x lb of ice (assume it starts at
32 degrees), then raise the temperature of x lb of water to 89
degrees. Next work out how much energy is lost when the pool cools by
1 degree. You know that the energy lost by the pool cooling must all
be absorbed by the ice/cold water - so you will be able to relate this
to your equation and solve for x.

875lb is not too far out, but you will get a different figure if you do the math..

And who said how cold the ice is?  The stuff in my deepfreeze is -18°C, ca. 0°F.
I would venture to say that in practice ice is always several degrees
colder than 32°F.

And since we are playing with the question:  a part of the surface of
the ice will be absorbing energy from the air.  With square blocks of
ice, that surface would be a little over 1/6th,  with spheres of ice 
- the minimum surface area -  it would be less  ( nice additional
calculation).

But if we wanted to avoid this by submerging the ice, and made the
mistake of keeping on the bottom of the pool, that would  - I believe
- upset the effect of convection, the ice sitting on a level of
cooler, heavier water, slowing the absorbtion of energy by the ice.

Oh, maybe eventually that doesn't make any difference, as long as
there is no additional warming of the water from the air, the process
just takes longer.

Or not?

No one said how cold the ice is, which is why we must make an
assumption. Of course myoarin-ga is right that the ice would almost
certainly have a lower temperature, but by assuming the ice is at 32
degrees, we can work out the worst case scenario: i.e. the answer that
requires the most ice (and expense!)

Since we are only given one temperature for the pool we must assume
the pool is well stirred (using a mechanism that does not raise the
temperature), so the effects of convection can safely be ignored.

And as for the effects of the air... why assume it acts only on the
ice? Any warming, or cooling effect of the air will affect the whole
exposed surface of the pool. To take that into account we would need
to know something about the temperature of the air, speed of the air,
the humidity, the surface area of the pool etc. which makes the whole
problem far too complicated for a homework assignment ;-)

In order to give an exact answer, you must know the initial
temperature of the water and the ice.  Let's assume the ice is at 32
deg F, and the water is 70 deg F.  (Probably when you say degrees you
mean F, since you asked for lbs of ice).  Now I'm going to change
everything to SI units because that's what I'm used to.  Initial
temperature is 21.111 deg C, final temperature is 20.555 deg C, mass
of water is 75708 kg, heat capcity of water is 4180 J/kg/deg C, so
heat lost by water is 175.8 MJ.  That heat went into the ice, mostly
to melt it, but also partly to heat it up to 20 deg.  The latent heat
of fusion for water is 334000 J/kg, and, using the heat capacity of
water, heating water from 0 to 20.555 costs 85922 J/kg, for a total of
420000 J/kg.  Since total required heat is 175800000 J, that means
about 419 kg of ice, or 923 lbs.

So:  changing the temperature of a 20000 gal swimming pool from 70 deg
F to 69 deg F requires 923 lbs of ice.

Racecar-ga,
That is all over my head  - and seems to suggest that the correct
answer is beyond what could be expected for a school assignment in
gallons and lbs.

In the dim past of my high school physics, I seem to remember that
freezing water required more energy than that needed to bring water
down to 0°C (excuse any misuse of terms).  If this is true, then
wouldn't ice at 0°C  absorb more heat than water at 0°C?
Or is this reflected in your calculation  -  "latent heat of fusion of water"?

Hkapasia-ga,  
If this was a school assignment, it would be very interesting to learn
what was considered to be the correct answer.  :-)

Thanks, both of you,
Myoarin

Yep, latent heat of fusion is just the amount of energy required to
melt a given mass of, in this case, ice.  And yes, ice does most of
its cooling with latent heat (by absorbing energy in the course of
changing phase from solid to liquid).  To cool a 20000 gal swimming
pool from 70 deg to 69 deg, you would have to add over 500 gal of 32
deg water (4000 lbs, more than 4 times as much as if you add 32 deg
ice).

Wow!  Thanks very much Racecar-ga.

I wonder if the teacher knows that?

Myoarin