Ice And water


by Harshatalla
Tags: water
Harshatalla
Harshatalla is offline
#1
Sep16-06, 09:00 PM
P: 3
Whar happens to the level of water when ice melts in a beaker
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
G01
G01 is offline
#2
Sep16-06, 09:15 PM
HW Helper
G01's Avatar
P: 2,688
since liquid water is more dense than ice, the water level would go down because more water would be able to fit in less space.
SGT
#3
Sep17-06, 04:29 AM
P: n/a
There is no change in the level. The ice floats because it displaces a volume of water which weights as much as the ice.
When the ice melts, it turns into water, that will occupy the same volume.
(I am assuming that prior to melting the water in contact with the ice will be at 0 degrees C).

Doc Al
Doc Al is offline
#4
Sep17-06, 05:28 AM
Mentor
Doc Al's Avatar
P: 40,878

Ice And water


Quote Quote by Harshatalla
Whar happens to the level of water when ice melts in a beaker
Is the ice floating?
DaveC426913
DaveC426913 is offline
#5
Sep17-06, 11:26 AM
DaveC426913's Avatar
P: 15,325
Quote Quote by Doc Al
Is the ice floating?
Is there any reason to think it wouldn't be?
rcgldr
rcgldr is offline
#6
Sep17-06, 01:50 PM
HW Helper
P: 6,925
Quote Quote by DaveC426913
Is the ice floating
Is there any reason to think it wouldn't be?
Could be a trick question involving the other stages of ice (frozen under high pressure), which I assume are more dense.
Hootenanny
Hootenanny is offline
#7
Sep17-06, 03:05 PM
Emeritus
Sci Advisor
PF Gold
Hootenanny's Avatar
P: 9,789
Or the ice could be physically held underneath the water.
DaveC426913
DaveC426913 is offline
#8
Sep17-06, 03:53 PM
DaveC426913's Avatar
P: 15,325
Quote Quote by Hootenanny
Or the ice could be physically held underneath the water.
Is there a reason to think it is?
Cyrus
Cyrus is offline
#9
Sep17-06, 05:16 PM
Cyrus's Avatar
P: 4,780
Ok, I will take a crack at it. Correct me if I am wrong Folks.

Let's take a differential element of water, where the top cube of the water lies at the free surface of the water. (i.e. the cube is just under the water).

This cube has dimensions, dx, dy, dz.

The volume of this cube is:

[tex] dV = dx dy dz [/tex]

the associated mass is:

[tex] m = \rho_w dx dy dz [/tex]

when water expands to ice, it increases by volume at about 9%, but the mass remains constant, therefore:

[tex] m = \rho_w dx dy dz = \rho_{ice}( 1.09 dx dy dz )[/tex]

So that means:

[tex] \rho_w = 1.09 \rho_{ice} [/tex]

We can now do a simple force balance to see our result: (weight of ice must balance buoyancy force)

[tex] dF_b = \rho_w g dV' = \frac {\rho_w}{1.09} g (1.09 dx dy dz) [/tex]

**Edit: Forgot the (1.09) infront of the dxdydz, thanks Gokul!

Now we simplify and get:

[tex] dV' = dxdydz[/tex]

Where dV' is the volume of displaced water by the ice.

Because dV' is equal to dV=dxdydz, (which was the original volume of differential water), the water level will stay the same when the ice melts.

I just wrote this proof down myself, so if its got a mistake point it out!
Doc Al
Doc Al is offline
#10
Sep17-06, 05:56 PM
Mentor
Doc Al's Avatar
P: 40,878
Quote Quote by cyrusabdollahi
I just wrote this proof down myself, so if its got a mistake point it out!
Your mistake is assuming that the ice floats yet is totally submerged. Yes, the volume of ice is greater than the volume of an equal mass of water--but only part of the ice is submerged and thus displacing water. (See SGT's post for a simple argument.)

On the other hand, if the ice is not floating but resting on the bottom of the beaker (exerting a non-zero force on the bottom of the beaker), then the water level will rise when the ice melts. In such a case the ice obviously weighs more than the displaced water, otherwise it would be floating.
Cyrus
Cyrus is offline
#11
Sep17-06, 05:59 PM
Cyrus's Avatar
P: 4,780
Not quite. I never assumed that. Take a second glance.

I started with a differential element of water just under the surface to see the volume it would occupy.

Then I compared that volume to the displaced volume of the ice (that had the same mass as the water).

I assumed the ice was some fractional distance from the surface, kdz, where k turns out to be 1/(1.09)^2.
DaveC426913
DaveC426913 is offline
#12
Sep17-06, 06:03 PM
DaveC426913's Avatar
P: 15,325
Simply put, the ice will float at a height where it displaces an amount of water exactly equal to its mass. If the volume of that block of ice happened to increase (for whatever reason), or even decrease (for whatever reason), without changing it mass, it is the volume above the waterline that will grow or shrink. The volume below the water line will not change, and thus the displaced amount of beaker water will not change. (It couldn't change! The volume displaced in the beaker water is directly created by the displacement from the mass of the block of ice, which hasn't changed in mass!) Since the amount of wtaer displaced in the beaker does not change, it has no effect on the water level.
Gokul43201
Gokul43201 is offline
#13
Sep17-06, 06:06 PM
Emeritus
Sci Advisor
PF Gold
Gokul43201's Avatar
P: 11,154
Quote Quote by cyrusabdollahi
Ok, I will take a crack at it. Correct me if I am wrong Folks.

Let's take a differential element of water, where the top cube of the water lies at the free surface of the water. (i.e. the cube is just under the water).

This cube has dimensions, dx, dy, dz.

The volume of this cube is:

[tex] dV = dx dy dz [/tex]

the associated mass is:

[tex] m = \rho_w dx dy dz [/tex]

when water expands to ice, it increases by volume at about 9%, but the mass remains constant, therefore:

[tex] m = \rho_w dx dy dz = \rho_{ice}( 1.09 dx dy dz )[/tex]

So that means:

[tex] \rho_w = 1.09 \rho_{ice} [/tex]

We can now do a simple force balance to see our result: (weight of ice must balance buoyancy force)

[tex] F_b = \rho_w g dV' = \frac {\rho_w}{1.09} g dx dy dz [/tex]
Error in the last line, on the RHS.

[tex]dF_{weight} = \frac {\rho_w}{1.09} g dV_{ice} = \frac {\rho_w}{1.09} g dV_{water} \cdot 1.09 =\rho_w g dx dy dz [/tex]
Cyrus
Cyrus is offline
#14
Sep17-06, 06:08 PM
Cyrus's Avatar
P: 4,780
Thanks Gokul, you got it!


Dang, stupid last line threw me off.....grrrr!

How did I do that! ...oops
Cyrus
Cyrus is offline
#15
Sep17-06, 06:27 PM
Cyrus's Avatar
P: 4,780
Quote Quote by DaveC426913
Simply put, the ice will float at a height where it displaces an amount of water exactly equal to its mass. If the volume of that block of ice happened to increase (for whatever reason), or even decrease (for whatever reason), without changing it mass, it is the volume above the waterline that will grow or shrink. The volume below the water line will not change, and thus the displaced amount of beaker water will not change. (It couldn't change! The volume displaced in the beaker water is directly created by the displacement from the mass of the block of ice, which hasn't changed in mass!) Since the amount of wtaer displaced in the beaker does not change, it has no effect on the water level.
I think my revision thanks to Gokul has put into equations what you have put into text. Though I think there is need of more equations to back up these arguments at the start of the thread. You guys were right, but it was too informal for my taste.

The way in which the differential element grows or shrinks is not needed in this analysis. All you need to worry about is how much volume the ice displaces, and compare that to how much volume that same element of ice would occupy as a liquid.

The transient is not of concern.
DaveC426913
DaveC426913 is offline
#16
Sep17-06, 06:46 PM
DaveC426913's Avatar
P: 15,325
See attached pic. It's so simple that it should be intuitively obvious that the water level doesn't change.

(If anyone thinks my diagram can be improved let me know. But I can't see anything else it needs.)
Attached Thumbnails
PF060917ice-water.gif  
Cyrus
Cyrus is offline
#17
Sep17-06, 06:48 PM
Cyrus's Avatar
P: 4,780
Not to idiots like me
DaveC426913
DaveC426913 is offline
#18
Sep17-06, 07:20 PM
DaveC426913's Avatar
P: 15,325
Quote Quote by cyrusabdollahi
Not to idiots like me
(Hm. Inb retrospect, my post may have sounded almost patronizing. I didn't mean to suggest it's simple to understand, merely that the diagram is simple.)

Does the diagram not work for you?


Register to reply

Related Discussions
Formula: final temperature when mixing water with water Introductory Physics Homework 5
Tap water flow and water diameter at end of it (Bernoulli eqzn...) Introductory Physics Homework 4
energy loss converting water to hydrogen then back to water General Physics 2
HELP! relationship between rate of flow of water and height of water column Introductory Physics Homework 1
Basic question regarding triple point of water and physical properties of water Introductory Physics Homework 9