U-tube filled with water and inmiscible liquid

AI Thread Summary
A U-shaped tube contains water and an immiscible liquid, with heights h1 = 0.52 m and h2 = 0.16 m. The formula for calculating the unknown density is ρu = h1/h2 * ρk, where ρk is the density of water. Initial calculations yielded incorrect results due to unit errors and possibly not following the expected format for online homework submissions. The correct answer is approximately 0.563 g/mL, but there may be a typo or missing information in the problem statement. The atmospheric pressure on both sides of the tube does not affect the calculations as it cancels out.
xannaxiero
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Homework Statement



A U-shaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.52 m and h2 = 0.16 m, what is the density of the liquid?


Homework Equations



ρ_{u}=h_{1}/h_{2} * ρ_{k}

as in: unknown density = ratio of heights times known density

ρ_{k}= density of water = 1 cm^{3}/mL


The Attempt at a Solution



so I just plugged the info in, seems relatively simple...
ended up with

ρ_{u}= 52 cm/16 cm * 1 cm^{3}/mL = 3.25 cm^{3}/mL

but for some reason this is wrong.

just for kicks I also tried
ρ_{u}= 16 cm/52 cm * 1 cm^{3}/mL = .31 cm^{3}/mL

but this is also wrong.

any ideas?
 
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xannaxiero said:

Homework Statement



A U-shaped tube is partly filled with water and partly filled with a liquid that does not mix with water. Both sides of the tube are open to the atmosphere. If h1 = 0.52 m and h2 = 0.16 m, what is the density of the liquid?

Homework Equations



ρ_{u}=h_{1}/h_{2} * ρ_{k}

as in: unknown density = ratio of heights times known density

ρ_{k}= density of water = 1 cm^{3}/mL

The Attempt at a Solution



so I just plugged the info in, seems relatively simple...
ended up with

ρ_{u}= 52 cm/16 cm * 1 cm^{3}/mL = 3.25 cm^{3}/mL

but for some reason this is wrong.

just for kicks I also tried
ρ_{u}= 16 cm/52 cm * 1 cm^{3}/mL = .31 cm^{3}/mL

but this is also wrong.

any ideas?

Hi xannaxiero, welcome to PF! :smile:

Your unit for density is wrong.
It's not cm3/mL.

Btw, your formula assumes that the separation of the fluids is at the bottom of the U-tube.
I guess you have to make that assumption, because otherwise you do not have enough data.
 
Oops you're right, I'm sleepy >.< Its g/mL. My bad. That doesn't make any difference in the equation though does it? And yes, I'm assuming separation is in the bottom...this is supposed to be a pretty simple, standard u-tube problem, no tricks...
 
Beyond that, your answers are right.
Why do you think they are wrong?
 
Online homework...they're being graded as incorrect :'(
 
Oh those!
That usually means you did not follow the format that they expected.

Try rounding to 2 digits (since your input data is 2 digits each).
And try the unit kg/L or perhaps kg/m3 (adjusting of course the result to match).
 
The answer ended up being .563 g/mL. Any idea why this is? There's no explanation with the homework :(
 
Nope. No idea.

I can only guess that there is a typo in the problem statement, or that there is more information that is not given.
 
Do you think it could be because the tube is open to the atmosphere? Would that impact my formula at all?
 
  • #10
No, the atmosphere gives the same pressure of 1 atmosphere on both tubes.
It cancels out.

It would matter if the tubes had different diameters, or if the separation is not at the bottom.
 
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