U-tube filled with water and inmiscible liquid

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Homework Help Overview

The problem involves a U-shaped tube partially filled with water and an immiscible liquid, with specific heights given for each liquid. The goal is to determine the density of the immiscible liquid based on the height measurements.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of a formula relating the heights of the liquids to the density of the unknown liquid. There are attempts to calculate the density using the provided heights, but results are questioned as incorrect.

Discussion Status

Some participants have provided guidance on potential issues with units and assumptions made in the calculations. There is an exploration of the impact of atmospheric pressure on the problem, with some uncertainty about the correctness of the initial approach.

Contextual Notes

Participants note that the problem may lack sufficient information or contain a typo, which could affect the calculations. There is also mention of specific formatting requirements for the answers in an online homework system.

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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