Determining the Depth of a Lake

1. Nov 23, 2008

persephone

1. The problem statement, all variables and given/known data
A 2.1 m long glass tube that is closed at one end is weighted and lowered to the bottom of a freshwater lake. When the tube is recovered, an indicator mark shows that water rose to within 0.43 m of the closed end. Determine the depth of the lake. Assume constant temperature.

2. Relevant equations
Honestly...I have no idea how to approach this problem.
The only relevant equations I know for this sort of problem are:
(Pressure)=(density)(g)(height)
and
But neither of these equations seem to get me anywhere...

3. The attempt at a solution
Like I said, I really don't have any idea how to approach this problem.
The furthest I got was to calculate the total height of the water in the tube: 1.67m
Using those equations I posted, I could calculate the radius of the tube or the initial pressure. But that information is irrelevant to the question. Any sort of information to help point me in the right direction would be greatly appreciated. Thanks in advance.

2. Nov 23, 2008

LogicalTime

how about calculating the water pressure at the bottom of the lake? that will probably get us somewhere.

use
$P_1V_1 = P_2V_2$
The pressure that the air inside comes to at the bottom is the same as the water pressure since at equal pressure that's where they would hit equilibrium

ah then use your first equation, look up the density of water, and make sure all your units are correct

3. Nov 23, 2008

persephone

Oooh I see where you're going with this...So calculate the pressure of the air after the tube is lowered, which would be the same as the water pressure? I'm not really sure what I'd use for P1 though. Would I use the pressure of air at sea level?

4. Nov 23, 2008

LogicalTime

Using SI units is probably best Standard atmospheric pressure is 101,325 Pascal, which is N/m^2
http://en.wikipedia.org/wiki/Pascal_(unit [Broken])

Last edited by a moderator: May 3, 2017
5. Nov 23, 2008

persephone

Nope...didn't work.
I tried P1=101,325 Pa, V1=(pi)(r)^2(h)=pi(0.008795)^2(2.1)=0.000528
P1V1=53.55
Then V2=(pi)(0.008795)^2(0.43)=0.000108
So P2=53.55/0.000108 = 494843
Plugging that into my first equation, 494843=(10^3)(9.8)(h)
h=50.5m

It wasn't correct :(

6. Nov 23, 2008

Redbelly98

Staff Emeritus
The water pressure is not rho*g*h at the bottom of the lake. Do you see what is missing?

7. Nov 23, 2008

persephone

Should it be rho*g*h + the atmospheric pressure?

8. Nov 23, 2008

Redbelly98

Staff Emeritus
Yes.

9. Nov 23, 2008

persephone

So (494843 + 101325)=(10^3)(9.8)(h)
h=60.8m

Still not right...

10. Nov 23, 2008

LogicalTime

If we equate the pressure due just to the water with the formula with density of water and gravity and height

$P_{water}=(494843 - 101325)= 393,518 \frac{N}{m^2}$
$393,518 \frac{N}{m^2}=(10^3 \frac{kg}{m^3})(9.8 \frac{m}{s^2})(h)$

Last edited: Nov 23, 2008
11. Nov 24, 2008

Redbelly98

Staff Emeritus
I think you're close.

Is the air in the tube (that we got the pressure for) located exactly at the bottom of the lake?