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This is the problem:
A piece of ice floats in a vessel with water above which a layer of a lighter oil is poured. How will the level of the interface (oil and water) change after the whole of ice melts? What will be the change in the total level of liquid in the vessel?

I understood most of the answer to this solution, except this one part. He said the level of interface will change because the volume submerged is greater than the volume of melted ice. This part bugs me, as normally if there was no lighter oil on top, the volume submerged is equal to volume of melted ice (you can figure that out with equations). But here we have oil on top. The teacher justified his answer by saying that the volume submerged will be lower than that of the melted ice because there is a buoyant force acting on ice (by the oil). I don't get why.

How can oil exert buoyant force (there can't be any tangential force on ice because of law of hydrostatics so there can't be any force by oil in the upward direction). In fact I have an argument that says that there will be pressure exerted by oil on the ice from the top making volume of submerged ice to crease. Am I wrong?

Another doubt is intuitively there will be buoyant force on ice that is greater than the usual buoyant force on ice (i.e, in the absence of oil) but I can't find a way to justify with equations. Can someone solve this problem by actual equations and not word explanation so that I can have a better understanding? Can someone solve the specific heights to which interface will rise and liquid falls?

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  • $\begingroup$ What is meant by "interface"? Interface between what? $\endgroup$
    – Bob D
    Commented Dec 4, 2023 at 14:31
  • $\begingroup$ @Bob - between the layers of oil and water. $\endgroup$
    – mmesser314
    Commented Dec 4, 2023 at 14:39
  • $\begingroup$ @ Hammock the entire volume of ice is submerged, so the entire weight of ice, $mg$ has to be balanced by buoyant forces by the two liquids. Had there been no oil, you would have found that $mg=F1$ where $F1$ is buoyant force of water. That equation can be solved to get ratio of submerged volume and total volume as density of ice : density of water. Now that we have added the oil, $F2+F1=mg$ where$F2$ is buoyant force of oil. [To be continued] $\endgroup$
    – Stuti
    Commented Dec 5, 2023 at 18:10
  • $\begingroup$ [Continuing] If you assume that the edge of the ice block is $a$ units and the part of this length submerged in water is $x$ units , you get the equation:$$\alpha a²xg+ \rho a²(a-x)g=mg$$ where $\alpha$ is density of water and $\rho$ is the density of oil. Solve further to get: $$(\alpha-\rho)x+ \rho a= \gamma a$$ where $\gamma$ is the density of the ice ( I have assumed it cubical). You can find $a$ and $x$ from here. $\endgroup$
    – Stuti
    Commented Dec 5, 2023 at 18:18

3 Answers 3

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Suppose you have two setups. One has layers of oil and water. The other has "layers" of water and water.

In the second case, the ice floats higher than it would without the upper layer. So the upper layer exerts a buoyant force on the ice. It is indirect. The weight of the upper layer increases the pressure of the bottom layer, so the bottom layer pushes upward harder on the bottom of the ice.

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    $\begingroup$ Is it more like applying Pascal's law of undiminished pressure in all directions? $\endgroup$ Commented Dec 4, 2023 at 18:23
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Consider a vessel that is so big that the volume of the ice is negligible compared to the volume of water and of oil. This means that the height of the interface between water and oil is the same before and after adding the ice.

Let the subscripts $w,o,i,a$ refer to water, oil, ice and air. Then the considerations of the weight of fluid displaced and the weight of the ice itself gives $$\tag1\rho_wgV_w+\rho_ogV_o+\rho_agV_a=m_ig=\rho_igV_i=\rho_ig(V_w+V_o+V_a)$$ Now consider the sub-case whereby there is enough oil to completely submerge the ice. Then $V_a=0$ and the obvious only possible case that this happens is $0<\rho_o<\rho_i<\rho_w$ so that $$\tag2(\rho_w-\rho_i)V_w=(\rho_i-\rho_o)V_o$$ where every term here is chosen to be positive. You can consider the ice to be a thin cylinder and divide out by the uniform cross-sectional area $A$ and obtain the same relations for heights submerged, or even the fraction of total height of ice in each medium.

Now, considering the case of no oil, i.e. setting $V_o=0=\rho_a$ in Equation (1), we can figure out that the melted ice will have volume $V_i\rho_i/\rho_w$, whereas Equation (2) tells you how big $V_o$ is, compared to $V_w$; the combination of them allows you to work out the relative sizes of $V_i\rho_i/\rho_w<V_i=V_w+V_o$; it is clear that, since $V_o>0$, so $V_w<V_i$, but it comes down to the details for you to figure out if $V_w<V_i\rho_i/\rho_w$, in which case the melted ice will add to the volume of the water, threatening (in vain, per our current assumptions) to raise the interface level.

It is then up to you to relax the assumptions above and solve the problem for yourself. The algebra will quickly get messy.

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The thing that really matters is the relative pressure variations surrounding the ice, not the absolute pressure. When you put oil above the interface in place of air, the pressure relative to the interface pressure for the portion below the interface does not change, but, for the portion above the interface, the pressure on the surface of the ice relative to the interface pressure decreases.

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