How does the melting of ice affect the oil-water interface in a fluid system?

[AdityaDev]

Homework Statement

A cube of ice is floating in water such that some part of the ice is submerged. Oil is poured on the water.( so water on the bottom, oil on top and ice in between). When the ice melts completely, the level of oil-water interface ______(rises/falls) and the top level of oil ______(rises/falls)

Homework Equations

None.

The Attempt at a Solution

The water level goes up a bit due to the ice cube. Also, the level of oil will also be higher than usual since the ice occupies a part of its volume.
Now, as the ice starts melting,
1)the volume of water increases but the level of water has a tendency to fall as volume of water displaced by ice decreases. <---- But I am not able to draw a conclusion here.
2)the top level of oil has a tendency to fall since the volume of oil displaced by ice decreases. And the the net increase or decrease in level of oil depends on the level of water. <---- I need to know case (1) to answer this.
(answer given: Rises,falls)

 

[Orodruin]

The last question you should be able to answer without any reference to the first. What happens to the total volume when the ice melts?

For the first question, try to figure out what happens to the ice cube when you add the oil. What happens to the water surface when the ice melts and there is no oil?

 

[AdityaDev]

The last question you should be able to answer without any reference to the first. What happens to the total volume when the ice melts?

For the first question, try to figure out what happens to the ice cube when you add the oil. What happens to the water surface when the ice melts and there is no oil?

Total volume of water increases. when there is no oil, the level of water can also stay constant because the 2 phenomena:
a] the ice displaces some water due to Archimedes principle. say, from 100ml, of water, the new level due to ice becomes equivalent to 110ml.
b] as the ice melts, the volume displaced by ice decreases so the level of water decreases.let 10ml of water be added as ice melts completely.
Now there is no change in water level.

 

[Suraj M]

Yes,so if you add oil, what happens to the volume of water displaced by the ice cube.

 

[Chestermiller]

Yes,so if you add oil, what happens to the volume of water displaced by the ice cube.

Yes. Good point. After the oil is added, the amount of the ice cube sticking out of the water will be greater than before the oil was poured.

Chet

 

[AdityaDev]

Yes. Good point. After the oil is added, the amount of the ice cube sticking out of the water will be greater than before the oil was poured.

Chet

Can you explain this situation? or can you help me figuring out the answer?

 

[Chestermiller]

Can you explain this situation? or can you help me figuring out the answer?

Sure. Let the side of the cube be s, and let f be the fraction of the cube volume above the liquid water surface. In terms of s and f, what is the volume of oil displaced? What is the volume of liquid water displaced? What is the buoyant force on the cube? From a force balance on the cube, what is f equal to?

Chet

 

[AdityaDev]

Sure. Let the side of the cube be s, and let f be the fraction of the cube volume above the liquid water surface. In terms of s and f, what is the volume of oil displaced? What is the volume of liquid water displaced? What is the buoyant force on the cube? From a force balance on the cube, what is f equal to?

Chet

Volume of cube in water = $(1-f)s^3$
volume of cube in oil = $fs^3$
Buoyant force due to water = $(1-f)s^3\sigma_1g$
Buoyant force due to oil =$fs^3\sigma_2g$

 

[haruspex]

Volume of cube in water = $(1-f)s^3$
volume of cube in oil = $fs^3$
Buoyant force due to water = $(1-f)s^3\sigma_1g$
Buoyant force due to oil =$fs^3\sigma_2g$

Right, so what equation relates those to the weight of the cube?

 

[AdityaDev]

Right, so what equation relates those to the weight of the cube?

$$(1-f)s^3\sigma_1g+fs^3\sigma_2g=s^3\rho g$$

 

[Chestermiller]

$$(1-f)s^3\sigma_1g+fs^3\sigma_2g=s^3\rho g$$

OK. So now solve this equation for f. How does this solution compare with the value of f you get when there is no oil on top of the water?

Chet

 

[AdityaDev]

OK. So now solve this equation for f. How does this solution compare with the value of f you get when there is no oil on top of the water?

Chet

$$f=\frac{\sigma_1-\rho}{\sigma_1-\sigma_2}$$ when oil is present
Taking similar situation with f part inside water,
$$f=\frac{\rho}{\sigma_1}$$

In presence of oil, the buoyant force is increased due to buoyant force by oil.
So with oil, the part inside water is lesser.

 

[Chestermiller]

$$f=\frac{\sigma_1-\rho}{\sigma_1-\sigma_2}$$ when oil is present
Taking similar situation with f part inside water,
$$f=\frac{\rho}{\sigma_1}$$

Actually, to be clear, your first equation for f is the part outside the water, and your second equation for f is the part inside the water.

In presence of oil, the buoyant force is increased due to buoyant force by oil.
So with oil, the part inside water is lesser.

So, what does this mean with respect to what happens when the ice cube melts? How does the cube melting affect the depth of the water layer?

Chet

 

[AdityaDev]

So, what does this mean with respect to what happens when the ice cube melts? How does the cube melting affect the depth of the water layer?

Chet

As the ice melts, the volume of water increases, which tends to increase water level and the weight of ice decreases which tends to decrease the water level. But how do I calculate the net effect?

 

[Orodruin]

As the ice melts, the volume of water increases, which tends to increase water level and the weight of ice decreases which tends to decrease the water level. But how do I calculate the net effect?

What happens in the case where there is no oil? How does this case differ from the case of the oil in the original and final water levels, respectively?

 

[AdityaDev]

What happens in the case where there is no oil? How does this case differ from the case of the oil in the original and final water levels, respectively?

Let f part of ice be inside
$fV_i\rho g=V_i\sigma g$
So $f=\sigma/\rho$

So volume of water displaced is $V_i\sigma/\rho$

So apparent volume of water is $V+V_ik$, where k is ratio if density of water to ice which is less than 1.

And after I've has melted,
Volume of water is $V+V_i$
So there is a net rise in water level.

 

[Chestermiller]

As the ice melts, the volume of water increases, which tends to increase water level and the weight of ice decreases which tends to decrease the water level. But how do I calculate the net effect?

What is the volume of ice below the water surface before the ice melts (in terms of s)? What is the total volume of water that forms when the ice cube melts (in terms of s)? What is the difference between these two volumes?

Chet

 

[Orodruin]

So there is a net rise in water level.

In which case? The case where there is no oil? In that case this would be wrong. Note that the ice changes density to the water density when it melts!

 

[AdityaDev]

In which case? The case where there is no oil? In that case this would be wrong. Note that the ice changes density to the water density when it melts!

The value of k is greater than 1, because density of ice is less than density of water.
So the water level drops.

 

[AdityaDev]

What is the volume of ice below the water surface before the ice melts (in terms of s)? What is the total volume of water that forms when the ice cube melts (in terms of s)? What is the difference between these two volumes?

Chet

V_s volume of ice gives $V_sk$ amount of water, where k is the ratio of density of water to ice. This comes from conservation of mass.

 

[AdityaDev]

Now I am confused

 

[AdityaDev]

Here is what I know. Part of ice inside water is $f=\frac{\sigma}{\rho}V_s.$

 

[Chestermiller]

V_s volume of ice gives $V_sk$ amount of water, where k is the ratio of density of water to ice. This comes from conservation of mass.

You already determined that, before the ice melts, the fraction of ice under the water surface is $\left(\frac{ρ-σ_2}{σ_1-σ_2}\right)$. So the volume of the ice cube below the water surface before it melts is $\left(\frac{ρ-σ_2}{σ_1-σ_2}\right)s^3$. The volume of water that forms after the ice cube melts is $\frac{ρ}{σ_1}s^3$. Part of this melted volume of water goes into filling the void left by the part of the ice cube that was below the water surface. The rest of this volume adds to the depth of the water layer (if there actually is any left over). So from these relationships, is there any left over, and, if so, how much?

Chet

 

[AdityaDev]

You already determined that, before the ice melts, the fraction of ice under the water surface is $\left(\frac{ρ-σ_2}{σ_1-σ_2}\right)$. So the volume of the ice cube below the water surface before it melts is $\left(\frac{ρ-σ_2}{σ_1-σ_2}\right)s^3$. The volume of water that forms after the ice cube melts is $\frac{ρ}{σ_1}s^3$. Part of this melted volume of water goes into filling the void left by the part of the ice cube that was below the water surface. The rest of this volume adds to the depth of the water layer (if there actually is any left over). So from these relationships, is there any left over, and, if so, how much?

Chet

Water level does not change.
$mg=V_i\rho_wg$
$V_i=m/\rho_w$
When ice melts, the volume of water formed is same. So no change.

 

[Chestermiller]

Water level does not change.
$mg=V_i\rho_wg$
$V_i=m/\rho_w$
When ice melts, the volume of water formed is same. So no change.

That's only if these is no oil present.

Chet

 

[AdityaDev]

That's only if these is no oil present.

Chet

Finally... Water level rises when oil is present.

 

[AdityaDev]

Thank you sir. I have been sitting on this problem for a day. Thanks to @Orodruin as well.
Now I can study coordinate geometry for a while.