Determining the Depth of a Lake

[persephone]

Homework Statement

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.

Homework 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
(height)=2(surface tension)/(density)(g)(radius)
But neither of these equations seem to get me anywhere...

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.

 

[LogicalTime]

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

use
[itex]
P_1V_1 = P_2V_2
[/itex]
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

 

[persephone]

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

use
[itex]
P_1V_1 = P_2V_2
[/itex]
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

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?

 

[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 )

 

[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 :(

 

[Redbelly98]

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

 

[persephone]

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

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

 

[Redbelly98]

Yes.

 

[persephone]

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

Still not right...

 

[LogicalTime]

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

Still not right...

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

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

 

[Redbelly98]

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

Still not right...

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?