Conceptual Physics Question

[mrSHEiK124]

Got into a huge argument in Physics class, nobody can agree on an answer

The questions:

You have an ice cube floating in a glass of water. After the ice melts, what happens to the water level? Choices: Falls, rises, remains the same.

You have an ice cube floating in a glass of water. The ice cube has a small piece of iron embedded in it. What happens to the water level when the ice melts? Same choices.

You have an ice cube floating in a glass of water. The ice cube has many air bubbles in it. What happens to the water level when the ice melts? Again, same choices.

 

[pcy]

Hi,

The water level is unaltered in the first and third cases; in the second case the level drops.

Ignoring second order effects like surface tension or the change in air density (the bubbles warm up from 0C to room temperature when the ice melts in the third case) or the change of water density caused by the change in water temperature when the ice melts.


The key point is that the ice floats in all three cases. As a direct result the volume of water displaced must be exactly that whose mass equals the ice cube plus air/metal.


In cases 1 and 3 the water melts so (ignoring the mass of the air bubbles) that volume will be repalced by the equal mass of water, which by definition is the same volume as the displaced volume.


In the second case the displaced mass, and hence the displaced volume, si slightly larger because of teh piece of metal. The metal sinks, and not all of the displaced volume that matched it's weight is recovered.


Peter

 

[mrSHEiK124]

Crap

And I was sure I was right because ice is less dense than liquid water and thus occupies more space.

 

[Paperdoc]

pcy is eaxctly right. They key is: if it is FLOATING, the water displaced has a MASS equal to the MASS of the floating object. Case 2 is different AFTER it melts because the iron piece is NOT floating, so it only displaces a VOLUME of water equal to its own VOLUME, and that is less MASS of water than the iron's. In Case 3 after melting the mass and volume of the air are NOT in the glasss any more - we are left with only the mass of the water / ice in the original cube. And in the original cube, the total mass was almost entirely from the mass of the water - the air part effectively contributed no mass (well, so small it can be treated as zero) to the mass of the cube.

 

[silverpig]

Originally posted by: mrSHEiK124
Crap

And I was sure I was right because ice is less dense than liquid water and thus occupies more space.

Yes, that is correct, but part of the ice sticks up above the surface of the water. As that ice melts, it's sort of like pushing all of the ice underneath the surface of the water with a needle. The water would rise. However, as the ice melts it occupies less volume. This negates the rise of the water such that there is no net change in the water level.

 

[DrPizza]

It's odd that your physics teacher wouldn't have helped you come to an agreement... Look up archimedes principle. Some info here

 

[pcy]

When I was at school none of my physics teachers knew much about physics (or about anything else for that matter).


So I gess that's anothe rbig bunch of no change then.



Peter

 

[Smilin]

So you have an aquaduct, basically a water canal where the canal has to cross some gap so a 'bridge' is built to hold the canal.

If a boat drives across the aquaduct, what happens to the amount of weight resting on the aquaduct support columns?

 

[Yossarian451]

Another Fun One:

You have a sealed box sitting on a scale. Inside the box is a bird resting on a perch. The bird flies off of the perch and stays in flight inside the box. What happens to the scale reading.

 

[pcy]

Hi,

Answers:


The canal:

Basically no change. The Boat displaces water which distributes itself along the entire length of the canal.

So in fact there is a very slight rise in water level in the entire canal, caused by the wieght of the boat, and hence a tiny increase on the load on the aquaduct. I actually does not matter (unless you start to bring in the resistance to flow caused by the water's viscosity, which means the boat would cause a non-uniform increase in the level of the water along the length of the canal) where in the canal the boat is.



The bird:

No change. The bird in flight is supported by air pressure differences accross its wings, and those pressures mut propogate creating a net downward force over the inside of the cage equal to the weight of the bird.


If it were an open cage, not a sealed box...





Peter

 

[bendixG15]

Why doesn't someone challenge these wrong answers ??

 

[pcy]

If you think they are wrong, why don'y you challenge them?


Peter

 

[MetalStorm]

PCY you are wrong.

The water will go down in all cases.

This is because the ice is less dense in all of the scenarios. As a result, when the ice melts, its volume is reduced and so the waterlevel is also reduced. The best way envisage this would be to get a glass beaker and put a small amount of water in to it, then put in a lot of ice... mark where the water comes to on the beaker. Let the ice melt and there you go.

What might be confusing people is that because ice is less dense, and essentially floats, part of it is above the water. However, that does not make a lot of difference to this as the critical part here is that when it melts its density is reduced (mass/volume).

The second case is more obvious because the weight inside the ice will pull it under the water and so the whole ice cube will be submerged.

The third case is exactly the same as the first. Ice already contains lots of tiny air bubbles, however if you were to have a lot of them, that would only reduce its density - the air would obviously escape when melted but the ice would just turn to water and you would be left with the exact same scenario as the ice in water.

 

[Nathonius]

Looks to me like PCY is right.

Assume:
Density of water is 1 kg/L (at 4 degrees C))
Density of ice is about 0.920kg/L
Density of air is about 0.001kg/L
Density of iron is about 7.8 kg/L
I apologize ahead of time for the metric units.

First case:

If we freeze 1000kg of water, we will have a block of ice 1000kg/0.920kg/L = 1087L in volume (rounded).

Now lets place that block of ice into a 10,000L tank of water at 4 degrees C.
In order for the ice cube to float, it needs to displace 1000 kg of water, which also happens to be 1000L.

The tank has markings on its side to indicate water level, in L.

Now the water level on the tank will read 11,000L (10,000L plus 1000L displaced by the ice). The cube of ice is floating with 87L of its volume above the water. (1087L ? 1000L)

When the ice cube melts, its 1087L volume is reduced back to 1000L {(1087L x 0.920kg/L)/1kg/L}, which happens to be the same volume it displaced to float, so the water level remains unchanged.

Third case (ice with air bubbles):

We?ll assume we still have that 1087L block of ice, but 20% of its interior volume has been removed and replaced with air. The block will now weigh:

(0.8 x 1087L x 0.920 kg/L) + (0.2 x 1087L x 0.001 kg/L) = 800 kg (rounded).
It has a density of 800kg/1.087m^3 = 0.736kg/L (rounded)


Now when we place it in the tank it will have to displace 800kg of water, or 800L to float. The tank will now have a water level of 10,800L.

When the ice melts, its 1087L volume is reduced back to 800L {(1087L x 0.736kg/L)/1kg/L} , which happens to be the same volume it displaced to float, so that water level remains the same at 10800L.

Second case (ice and iron)

To keep our ice/iron cube floating its density needs to be the same or less than that of water. With X as our ice fraction and Y as the iron fraction,

0.92kg/L (X) + 7.8kg/L (Y) < = 1.0 kg/L
and

X + Y = 1.0

So Y = 0.01163
X = .988372

So with a 1087L block we?ll have 1074L of ice and 12.64L of iron.

It will have a mass of: 1074L x .920kg/L + 12.64L x 7.8kg/L = 1087 kg

It is just barely floating, having exactly displaced its own mass in water. The level on the tank will be 11087L.

When the ice melts we?ll have 988L of water {(1074L x 0.920kg/L)/1kg/L} and 12.64L of iron, so the level is 10000 + 988 + 12.64 = 11001L (rounded), so the level drops.

 

[Fallen Kell]

Originally posted by: mrSHEiK124
Got into a huge argument in Physics class, nobody can agree on an answer

The questions:

You have an ice cube floating in a glass of water. After the ice melts, what happens to the water level? Choices: Falls, rises, remains the same.

You have an ice cube floating in a glass of water. The ice cube has a small piece of iron embedded in it. What happens to the water level when the ice melts? Same choices.

You have an ice cube floating in a glass of water. The ice cube has many air bubbles in it. What happens to the water level when the ice melts? Again, same choices.

Actually the answer to all of them are very simple. The water level stays the same in each and every case. Ask any scientist who studies global warming and the effects of melting ice. Any iceberg that is free floating in the oceans has already contributed its total mass to the water level. No matter how dense object is, the mass is the important part and the total mass is already affecting the water level once it is free floating in the water. (density/displacement of the object can also play a part, but that is when dealing with an object that is overall has a higher density then water, but contains large sections of it which has less density then the water, say something like an steal box that is water-tight and has air trapped inside the box, in this case the displacement of the object will be what affects the water level).

Edit: as I think more on it, the second case with the iron will cause the water level to potentially decrease. It will all depend on the size/mass of the iron. If the mass of iron is large enough, the floating ice itself will not float as high in the water because more water is needed to reach the displacement equallibrium point. And because ice itself is less dense then water, once the ice melts it may cause the water level to drop.

 

[Fallen Kell]

Originally posted by: MetalStorm
PCY you are wrong.

The water will go down in all cases.

This is because the ice is less dense in all of the scenarios. As a result, when the ice melts, its volume is reduced and so the waterlevel is also reduced. The best way envisage this would be to get a glass beaker and put a small amount of water in to it, then put in a lot of ice... mark where the water comes to on the beaker. Let the ice melt and there you go.

What might be confusing people is that because ice is less dense, and essentially floats, part of it is above the water. However, that does not make a lot of difference to this as the critical part here is that when it melts its density is reduced (mass/volume).

The second case is more obvious because the weight inside the ice will pull it under the water and so the whole ice cube will be submerged.

The third case is exactly the same as the first. Ice already contains lots of tiny air bubbles, however if you were to have a lot of them, that would only reduce its density - the air would obviously escape when melted but the ice would just turn to water and you would be left with the exact same scenario as the ice in water.

The problem with your scenario is that the ice is not "floating" in the water. What you did when you "put in a lot of ice" is that the ice was filled to the bottom and actually touching the bottom of the beaker. There was not enough water in the beaker to support the weight of the ice and thus it was not floating. Try it again, but make sure there is enough water in the beaker to keep the mass of ice floating.

The point you are missing is that ice will float at a specific level in water. There will be exactly the amount of ice above the surface of the water (and thus not displacing the waterlevel) as there is the difference in density between water and ice.

 

[pcy]

Originally posted by: MetalStorm
PCY you are wrong.

The water will go down in all cases.

Hmmmm



My answers in {bold italics}

This is because the ice is less dense in all of the scenarios. {true} As a result, when the ice melts, its volume is reduced {true} and so the waterlevel is also reduced {untrue - youare treating the part of the ice cube that is above the water level as if it were submerged and contibuted to the original water level. The un-submerged proprotion will always adjust in line with density because the weight of water displaced must be exactly equal to the weight of the complete icecube } . The best way envisage this would be to get a glass beaker and put a small amount of water in to it, then put in a lot of ice... mark where the water comes to on the beaker. Let the ice melt and there you go. {irrelevant - the original question specifed that the ice cube was floating}

What might be confusing people is that because ice is less dense, and essentially floats, part of it is above the water{true} . However, that does not make a lot of difference {true, but it makes exactly enough difference to cause the water level to remain unchanged. 1 .0 does not equal 0.999 just because 0.001 is a small number} to this as the critical part here is that when it melts its density is reduced (mass/volume) {true, but again irrelevant - the fundamantal principle here is that a floting object displaces an amount of fluid equal to it's weight, not to its total volume} .

The second case is more obvious because the weight inside the ice will pull it under the water and so the whole ice cube will be submerged {gibberish - the original question clearly stated that the ice cube floated in all cases. We must therefore assume that the piece of metal is sufficiently small that the cube still floats} .

The third case is exactly the same as the first. Ice already contains lots of tiny air bubbles, however if you were to have a lot of them, that would only reduce its density - the air would obviously escape when melted but the ice would just turn to water and you would be left with the exact same scenario as the ice in water. {true, so the water level remains unaltered as originally stated}

Peter

 

[pcy]

Hi,


Almost correct

Actually the answer to all of them are very simple. The water level stays the same in each and every case. Ask any scientist who studies global warming and the effects of melting ice. Any iceberg that is free floating in the oceans has already contributed its total mass to the water level. No matter how dense object is, the mass is the important part and the total mass is already affecting the water level once it is free floating in the water. (density/displacement of the object can also play a part, but that is when dealing with an object that is overall has a higher density then water, but contains large sections of it which has less density then the water, say something like an steal box that is water-tight and has air trapped inside the box, in this case the displacement of the object will be what affects the water level).

The total mass is indeed the important thing. Provided an Object is floating in a Fluid, its compostion and internal density variation is irrelevant and the followng are true:

Total Weight of Object = Total Weight of Fluid displaced
Total Volume of Fluid Displaced = Volume of Fluid whose Weight equals Total Weight of Object


When the metal is involved, after the ice cube melts the metal sinks (unless we are hypothesizing a metal whose specific gravity is less than 1.0). In consequence it is no longer floating, and a it displaces an amount of water equal to its volume. This by definition is less than the volume of water which has the same weight as as the piece of metal. Hence the water level drops.

Edit: as I think more on it, the second case with the iron will cause the water level to potentially decrease. It will all depend on the size/mass of the iron. If the mass of iron is large enough, the floating ice itself will not float as high in the water because more water is needed to reach the displacement equallibrium point. And because ice itself is less dense then water, once the ice melts it may cause the water level to drop.

See above. Much simpler. It depends only on the fact that the metal sinks when no longer part of the ice cube.




Can I also plead for people to stop talking about mass. The entire concept of "floating" depends on the existance of a graviational field acting on mass to create weight and upon the pressure distributions created by thos weights. It's the weight that matters.

Mass and weight are closely related, but they are not the same thing. They are not even numerically identical. My weigt is a force and it is the weight shown on a pair of scales when I stand on them. This values is not equal to my mass multipled by the earth's gravitational field. It is less than that buy a small (but non-zero) amount namely the weight of the air my body displaces.

My weight if I were in a vaccum (and hence somewhat dead) would be exactly equal to the graviational pull on my body mass; but we are walking about on the surface of a planet equiped (conveniently) with an atmosphere, from which we all recieve lift, not to mention other benefits.

This is why the presense of the air bubbles in the third case is irrelevant. The weight of the ice cube is in fact it's weight in air; and the weight of the volume of water it displaces is the weight in air of that volume. The bubbles do not contribute to the weight in air of the Ice Cube nor are the present in the water after the ice cube has melted.


Of course, if you want to add in third order effects then the question becomes more difficult, e.g:

Variation of air pressure with altitude.
Surface tension creating a meniscus between the water and the ice cube


I think the effects of these two both cance themselves out, but they make the concepts of water displaced and weight in air rather more compex. A little thought is needed...




Peter

 

[Fallen Kell]

Originally posted by: pcy
Of course, if you want to add in third order effects then the question becomes more difficult, e.g:

Variation of air pressure with altitude.
Surface tension creating a meniscus between the water and the ice cube


I think the effects of these two both cance themselves out, but they make the concepts of water displaced and weight in air rather more compex. A little thought is needed...




Peter

Yes, the altitude/air pressure affect how the meniscus level is set, but if you remember back to your chemistry classes, when measuring the level of water, you measure the level in the center of the water (i.e. if the meniscus bends the water level up at the sides or down at the sides, it is the level at the center which is the important measurement point. The meniscus itself and the direction it points is directly related to the pressure of air.

 

[beansbaxter]

hmmm

 

[Tarzanman]

Kell. You are wrong about the second case.

-The water level goes down in the second case (the bit of metal inside the cube).

-Since water has a density of 1 g/cm3, in order to float the average density of the ice cube must be less than 1.00 g/cm. This means that there is a hard limit on the volume of iron that can be in the ice cube, if the conditions of the situation are to be met.

-Also, this means that the density of the ice cube can be anywhere between 0.92 g/cm3 and 1.00 g/cm3. (depending on the size of the piece of iron inside it)

-The extra weight in the ice cube will displace more water than the natural weight of the ice cube (i.e. instead of ~10% of the ice being above the surface of the water, somewhere between only 0%-10% will be above the water.

-This 'extra' displacement caused by the extra weight of Iron will drive the ice cube down and the water level up.

-Simplified, when the ice melts, the iron no longer exerts any force to 'unnaturally' submerge any objects in the water. Since ice is less dense than water... any extraneous H2O that was 'forced' underwater by the weight of the Iron takes up less space after melting... which lowers the overall level of the water.

 

[liquid51]

I'm suprised no one has mentioned the contraction or expansion of the iron piece itself

 

[pcy]

Because provided the ice cube with the metal in it floats, and the metal on its own sinks, it does not effect he answer.


Peter

 

[liquid51]

If the metal within the cube expands when heated (when the cube melts in the higher temp water around it), it will occupy more volume than when it was within the cube.

 

[pcy]

Hi,

Yes - true, but irrelevant.


When the metal is in the ice cube it displaces an amont of water whose weight is equal to the weight of the metal. This is because the metal id part of a floating object at this point.


When the ice melts the metal displaces an amount of watrer equal to it's volume, because it is no longer part of a floating object. The matal may have incresed slightly in volume, but not enough to equal or exceed the volume of watrer whose wight equals its own wight. This is self evident - the metal sinks.



Peter