Solving a Buoyancy Problem: Cube of Ice in Water & Ethyl Alcohol

[eri139]

Homework Statement

A cube of ice whose edge is 17.0 mm is floating in a glass of ice-cold water with one of its faces parallel to the water surface.

Ice-cold ethyl alcohol is gently poured onto the water surface to form a layer 5.00 mm thick above the water. When the ice cube attains hydrostatic equilibrium again, what will be the distance from the top of the water to the bottom face of the block?

Homework Equations

F_B=ρVg
W = mg
V = lwh

The Attempt at a Solution

Here is my work. I have checked and rechecked it, but for some reason it's still not correct! Please help!

mg = F_Balcohol + F_Bwater
m = ρ_alcoholV_alcohol + ρ_waterV_water
ρ_iceV = ρ_alcoholV_alcohol + ρ_waterV_water
934 x 17³ = 789 x 17² x 5 mm + 1000 x 17² x h
934 x 17 = 789 x 5 + 1000h
h = 11.933 mm

 

[Charles Link]

The $h$ in your 3rd to the last equation needs to be $L=17-5-h$ to compute the volume of water that is displaced. You basically computed $L=11.933$... which is how much of the cube is immersed in the water portion. $\\$ Edit: I read it too quickly: I was computing the $h$ from the surface of the liquid to the top of the block. Anyway, $L=11.93$ mm is immersed in water, and add $5$ mm to that to get their $h$. $\\$ Edit: Nope, I read it too quickly a second time: It says the surface of the water, and not the surface of the liquid. I think you may have it right, but the person who wrote out the problem didn't read his own words carefully enough. $\\$ Another idea: The problem could also be the sig figs in your answer. Try putting in 11.9.

 

[gneill]

Where did you get your ρ_ice value? What I've seen in the literature for ice at 0°C is more like $0.9150 gm/cm^3$.

 

[eri139]

Where did you get your ρ_ice value? What I've seen in the literature for ice at 0°C is more like $0.9150 gm/cm^3$.

Google. I typed in "ice density" and the first value that popped up was that one. Unfortunately, I've just figured it out that this density is wrong and is the reason why my answers have been thrown off. Now I've finally gotten it...glad to see it wasn't an issue with my understanding, though!

 

[gneill]

Google. I typed in "ice density" and the first value that popped up was that one. Unfortunately, I've just figured it out that this density is wrong and is the reason why my answers have been thrown off. Now I've finally gotten it...glad to see it wasn't an issue with my understanding, though!

Yeah, Never believe the first entry to pop up from Google. Always check the source to see if it's reputable. Too many non-academic searchers bubble disreputable sources to the top of the hit list. It's a sad thing, but its just the way it is. Glad you worked out your issue!

 

[rude man]

@eri139, Not a trivial problem. I see nothing wrong with the wording. What was your answer? I used symbols instead of numbers & won't do the arithmetic, as usual. The answer turned out to be pretty elaborate, involving all three densities plus the alcohol layer depth plus the cube's side.

(I also did a second computation to verify my answer.)

EDIT: the anser can be simplified to comprise just the three densities and one side of the cube. With the simplification the answer I got is 11.68mm using ice density at 934 kg/m^3. But post 3 is right: should be 917 kg/m^3 in which case the answer I get is 10.31mm.