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U-tube filled with water and inmiscible liquid

  • Thread starter xannaxiero
  • Start date
10
0
1. 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?


2. Homework Equations

ρ[itex]_{u}[/itex]=h[itex]_{1}[/itex]/h[itex]_{2}[/itex] * ρ[itex]_{k}[/itex]

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

ρ[itex]_{k}[/itex]= density of water = 1 cm[itex]^{3}[/itex]/mL


3. The Attempt at a Solution

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

ρ[itex]_{u}[/itex]= 52 cm/16 cm * 1 cm[itex]^{3}[/itex]/mL = 3.25 cm[itex]^{3}[/itex]/mL

but for some reason this is wrong.

just for kicks I also tried
ρ[itex]_{u}[/itex]= 16 cm/52 cm * 1 cm[itex]^{3}[/itex]/mL = .31 cm[itex]^{3}[/itex]/mL

but this is also wrong.

any ideas?
 

I like Serena

Homework Helper
6,553
176
1. 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?


2. Homework Equations

ρ[itex]_{u}[/itex]=h[itex]_{1}[/itex]/h[itex]_{2}[/itex] * ρ[itex]_{k}[/itex]

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

ρ[itex]_{k}[/itex]= density of water = 1 cm[itex]^{3}[/itex]/mL


3. The Attempt at a Solution

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

ρ[itex]_{u}[/itex]= 52 cm/16 cm * 1 cm[itex]^{3}[/itex]/mL = 3.25 cm[itex]^{3}[/itex]/mL

but for some reason this is wrong.

just for kicks I also tried
ρ[itex]_{u}[/itex]= 16 cm/52 cm * 1 cm[itex]^{3}[/itex]/mL = .31 cm[itex]^{3}[/itex]/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.
 
10
0
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...
 

I like Serena

Homework Helper
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176
Beyond that, your answers are right.
Why do you think they are wrong?
 
10
0
Online homework...they're being graded as incorrect :'(
 

I like Serena

Homework Helper
6,553
176
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).
 
10
0
The answer ended up being .563 g/mL. Any idea why this is? There's no explanation with the homework :(
 

I like Serena

Homework Helper
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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.
 
10
0
Do you think it could be because the tube is open to the atmosphere? Would that impact my formula at all?
 

I like Serena

Homework Helper
6,553
176
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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