Ice bound
A glass of water has an ice cube floating in it. When the ice melts, will the water level rise, fall or remain the same?
Please ignore evaporation and other such sophistication.
Please ignore evaporation and other such sophistication.
Labels: funphysics
posted by Chris at
11:51 AM
8 Comments:
ice expands when frozen so the level would drop.
What about the portion of the floating cube which is above the water level ... once melted it would spread evenly and tend to raise the level. I guess it would depend upon whether the reduced volume from the melting is less or more than the level increase as described in my first statement. I don't know how to determine that.
It stays the same.
The same volume of water is displaced whether the cube is frozen or melted.
Convinced me ...
Another one minute wonder! You could also see it by realising that the ice cube will simply melt into water that has the same volume as the portion of the ice cube that was below the water.
archimede theory...
... thank you, now I understand.
Ice bound
Given:
-solid A floating in liquid B.
-solid A melts into liquid C
Assume:
when C melts into B it becomes well mixed with B
If liquid B has mass Mb and density Db then it occupies Vb=Mb/Db
If the glass has a constant cross sectional area of A, the height Hbo without displacement (the height without floating objects in it) is related to its volume by Vb=A*Hbo
Hbo=Vb/A
Hbo=Mb/(Db*A)
Similarly any melted liquid C would occupy a height
Hco=Mc/(Dc*A)
The height the combined liquid B+C would be displaced would be :
According to Archimedes
buoyant force = weight of liquid displaced
Let Vy be displaced volume, and Hy height due to displacement
Ma*g= Dmix*Vy*g
Ma*g= Dmix*A*Hy*g
Hy=Ma/( Dmix*A)
Dmix=(Mb+Mc)/(Vb+Vc) where Vmix=(Vb+Vc)
Call Mo the original mass of solid A and x the fraction of mass of A in liquid form
Ma=(1-x)*Mo
and from conservation of mass
Ma+Mc=Mo
Mc=x*Mo
Vc=Mc/Dc
Dmix=(Mb+Mc)/(Vb+Vc) where Vmix=(Vb+Vc)
Dmix=(Mb+Mc)/(Vb+Mc/Dc)
Dmix=(Mb+x*Mo)/(Vb+x*Mo/Dc)
Hy=Ma/( Dmix*A)
Hy=(1-x)*Mo*1/A*(Vb+x*Mo/Dc)/ (Mb+x*Mo)
Hbo=Mb/(Db*A)
Hco=Mc/(Dc*A)
Hco=x*Mo/(Dc*A)
Htotal=Hbo+Hco+Hy
Htotal=(1-x)*Mo*1/A*(Vb+x*Mo/Dc)/ (Mb+x*Mo)+ Mb/(Db*A)+ x*Mo/(Dc*A)
Find height when x=0
Htotal=(1-0)*Mo*1/A*(Vb+0*Mo/Dc)/ (Mb+0*Mo)+ Mb/(Db*A)+0*Mo/(Dc*A)
Htotal=Mo*1/A*Vb/Mb+ Mb/(Db*A)
Vb/Mb=1/Db
Htotal=(Mo+Mb)/(Db*A)
The change in height is
delta H=(1-x)*Mo*1/A*(Vb+x*Mo/Dc)/ (Mb+x*Mo)+ Mb/(Db*A)+ x*Mo/(Dc*A)- (Mo+Mb)/(Db*A)
delta H=(1-x)*Mo*1/A*(Vb+x*Mo/Dc)/ (Mb+x*Mo)+x*Mo/(Dc*A)- Mo/(Db*A)
delta H=1/A*((1-x)*Mo*(Vb+x*Mo/Dc)/ (Mb+x*Mo)+x*Mo/Dc- Mo/Db)
Substitute 1 into x for complete melting
delta H= 1/A*(((1-1)*Mo*(Vb+1*Mo/Dc)/ (Mb+1*Mo)+1*Mo/Dc- Mo/Db)
deltaH= 1/A*(Mo/Dc-Mo/Db)
deltaH=Mo/A*(1/Dc-1/Db)
if the density of liquid C is the same as liquid B there is no change,
if the density of liquid C is > liquid B the height drops,
if the density of liquid C is < liquid B the height rises,
density of salt water from ocean~= 1.025 g/cm^3
density of H2O~= 1.000 g/cm^3
Answer:
So for our glass of water:
-if the ice cube is made from water with the same density as it is floating in no change
-if the ice is made from water denser (e.g. salt water) than the water it is floating in the level will drop
- if the water the ice cube is floating in is denser (e.g. salt water) than the water melting off the ice cube the level will rise
Hopefully no mistakes.
Cam
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