## Drinking Problem

Here’s a classic problem my dad tried to explain to me when I was ten. I didn’t understand what he was saying, but to my credit I was at least getting pretty damn close to being potty trained by then.

Superman is thirsty. He is flying above a lake, and he has a really long straw. He dips one end of the straw down to the lake and puts the other end in his mouth. How close to the surface must he get in order to drink? Remember he has superlungs.

### 6 Responses to “Drinking Problem”

1. Kenneth Finnegan Says:

Due to the weight of the water in the straw, he’d need to be <~10m to not be defeated by the lack of air pressure. The formula is p = ρgh, but the density of water escapes me. p = 101kPa, g = 9.81 m/ss, ρ is the density, and h is the height.
The problem I’ve always had with this answer is that you’re now left with a water-vacuum surface 10m up the tube. Vapor pressure wouldn’t allow it to be a true vacuum, so conceivably, using a condenser, you could draw water higher, it just wouldn’t be liquid.

2. sophismata Says:

“the density of water escapes me”

??? It’s the easiest of all: 1kg per liter, or 1000kg/m3.

3. Ken Clark Says:

What Finnegan said, but with a caveat, does superman have any forward motion, and if so, what velocity, what is the shape of the straw opening , and does the opening rest at surface or does it penetrate the surface, if so how far… [/pedant]

4. Nik Says:

Same as Kenneth and if the straw is dipped in the lake, you can increase the distance to account for capillary action.

h = p/ρg + 2(gamma)(cos theta)/ρgr
h = 1/ρg [p + 2(gamma)(cos theta)/r]

Gamma = 4.9 J/m^2
ρ = 1000 kg/m^3
g = 9.8 m/s^2
p = 101.3 kPa (assuming Superman is at sea level)

Don’t know theta and r.

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