Ice cube in a bucket.

A bucket is filled with salt water and then a large fresh water ice cube is put in it so that the level of the water is exactly to the brim.

Will the bucket overflow when the ice cube melts, will the bucket remain exactly full or will there be some space in the bucket for a drop more water? You will need to explain why your answer is correct rather than just guess at one of the 3 possibilities.
LVL 57
Who is Participating?
acerolaConnect With a Mentor Commented:
I couln't think the answer, so I calculated it:

Before the melting:

Vb = volume before

Vb = volume of the salt water + volume of the ice under water

Vb = Vsw + (Di/Dsw)*Vi

Di and Dsw are the densities of ice and salt water

Since V = M/D (M is mass)

Vb = Msw/Dsw + (Di/Dsw)*(Mi/Di)

-> Vb = Msw/Dsw + Mi/Dsw

After the melting:

Va = volume after the melting

Va = Vsw + Vfw (volume of salt water + volume of fresh water)

Va = Msw/Dsw + Mi/Dfw (mass of ice = mass of fresh water)


Va - Vb = Mi/Dfw - Mi/Dsw

Dsw > Dfw
so Mi/Dfw > Mi/Dsw
so Va > Vb

So the bucket overflows.
Dufo G. BelskiRetired bureaucrat/desktop supportCommented:
Hmmm.  The ice will melt.  The fresh water, being less dense than the salt, will float on the top.  The water, being less in volume than the ice, will lower the level slightly below the brim, allowing for a bit more water in the bucket. ???
Real Life Answer:

Long before the ice cube has melted someone will have kicked the bucket and some of the water will be spread across the floor. Consequently there will be plenty of room in the bucket once the ice cube has melted. Oh yes and someone will be shouting at you for leaving a full bucket of salty water in the living room.
Cloud Class® Course: MCSA MCSE Windows Server 2012

This course teaches how to install and configure Windows Server 2012 R2.  It is the first step on your path to becoming a Microsoft Certified Solutions Expert (MCSE).

andyalderAuthor Commented:
You can ignore some of the real world possibilities; evaporation, surface tension, drunks kicking the bucket etc. It will of course initially overflow when you put the monster icecube in the bucket of salt water and the cube will stick up above the level of the brim since it floats but it's what happens after that that matters.
Dufo G. BelskiRetired bureaucrat/desktop supportCommented:
The level will remain the same.  Normally it would be lower, but the fresh water, unable to mix with the salt, will overflow and leave the level even. ???
assuming the ice does not touch the bottom of the bucket,
after the ice melts the level will lower as the water warms to 4°C, then rise back as the water warms above 4°C
andyalderAuthor Commented:
The ice doesn't touch the bottom or the sides. I didn't think to mention temperature but you can consider the water to be a fraction above 0°C and the ice a fraction below 0°C and that the temperature doesn't change significantly during the melting.

I think it overflows like asyscokid says but some form of proof is needed.
is the water in the bucket fresh?
andyalderAuthor Commented:
No, salt water, like the sea. The icecube is made from fresh water though.

The same question could be phrased "if an iceberg melts in the sea will sea-level rise or fall or stay the same"
Richard QuadlingSenior Software DeveloperCommented:

The bucket is full of salt water and a fresh water cube is put in.

A fresh water cube will displace more than it contains. This is why pipes crack wihen they freeze.


But how much of the salt water has been displaced?

This really depends upon the size of the ice cube.

If the ice cube was put into the salt water and a wire grid was put over the top of the bucket to hold the cube totally submerged, then when the cube fully melts the water level will have gone down as the cube occupies MORE space than the melted water.

But, if the amount of the salt water being displaced will depend upon the temperature/pressure and density of the salt water.

If the water was VERY salty, then the cube would float HIGHER, displacing less water and I would guess that the melted fresh water would exceed the amount of the displaced salt water.

If the salt water is only a little salty, then the cube would displace MORE of the salt water and could, in theory lower the water level when the cube melts.

The chances of the water level remaining the same is, in my opion, very unlikely.

andyalderAuthor Commented:
There is no wire grid holdin it down, I already said that it is floating and that the tip would be above the level of the brim.
wow!    RQuadling      you are quick and you are quite correct on almost anything and everything.

andyalder there's your answer.
andyalderAuthor Commented:
>If the salt water is only a little salty, then the cube would displace MORE of the salt water and could,
in theory lower the water level when the cube melts.

I think that even if the water is just a little salty the water will overflow.
andyalderAuthor Commented:
I was expecting someone to have come up with the same conclusion I reached but nobody has therefore here's my logic. If anyone sees a flaw in the argument then please say so.

A. Start at the end of the experiment, with a bucket full to the brim with salty water and work backwards.
B. Remove a quantity of salty water from bucket and freeze it.
C. Put it back into the bucket in the form of ice.
D. Since a floating object displaces it's own weight of liquid and the cube weighs exactly the same as the quantity of water we removed in step 2 the water will be exactly at the brim.
E. Melt the salty icecube and we are back to the initial conditions of step A.

So far, so good but it's a fresh water cube so lets do it backwards again.

1. Start with the bucket of salt water full to the brim.
2. Remove a pint of water.
3. Remove the salt from this pint of water.
4. Freeze the pint of fresh water.
5. Put the fresh water icecube back into the water.
6. Put the salt in a weightless bag on top of the cube.
7. We have the same weight floating in our salty water as in step D above so the water in the bucket is exactly at the rim again.
6. Remove the bag of salt from top of the cube and the cube gets lighter, floats higher so there is some room in the bucket.
7. Add this salt to the already salty's the crux...a pound of fresh water takes up the same volume if you add a spoon full of salt to it.
7a. The freshwater cube floats even higher as the salt water is denser, the water is now well lower than the brim.
8. Melt the cube and return the missing pint of water to it and the bucket contains exactly the same as in step 1.

The same process ought to work in reverse (except for the need of a weightless-volumeless refrigeration unit to generate the icecube).

We started with a fresh water icecube in a bucket of salt water which is step 7a where there was space in the bucket for a few drops more water but our bucket was full at that stage so it will overflow when the cube melts.
The bucket will overflow when the ice cube melts

An ice berg will float higher in salty water than in fresh water.  In fresh water when the berg melts nothing happens, that which is displaced shrinks back and becomes water

But because more is being displaced in salty water there is more water being held in the bucket than before.  Hence it must overflow when it melts.
andyalderAuthor Commented:
Also dbrunton explaination is good so seperate question for you in this TA.
The ice cube displaces it's weight, not it's volume. As fresh water is less dense per volume, it displaces less salt water. The level of the bucket will therefore slightly lower as it melts.
"The ice cube displaces it's weight, not it's volume"

Correct. The ice cube only displaces its full volume if it is totally underwater:

ice cube volume under water = displaced salt water volume

displaced salt water mass = total ice cube mass

So, the mass of the displaced salt water is the total mass of ice cube:

Msw = Vi*Di

To know the volume of the displaced salt water, we divide by its density (volume=mass/density)

Vdisplaced = Msw/Dsw = Vi*Di/Dsw

So, the initial volume of the bucket is the volume of the salt water plus the displaced volume:

Vb = Vsw + Vi*(Di/Dsw)
Sorry, Msw should be Mdisplaced in both expressions.
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.