How long a straw could Superman use? To suck water through a straw, you create a partial vacuum in your lungs.  Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure.  This corresponds to drinking water through a straw about ten meters long at maximum.
By taping several straws together, a friend and I drank through a $3.07m$ straw.  I think we may have had some leaking preventing us going higher.  Also, we were about to empty the red cup into the straw completely.

My question is about what would happen if Superman were to drink through a straw by creating a complete vacuum in the straw.  The water would rise to ten meters in the steady state, but if he created the vacuum suddenly, would the water's inertia carry it higher?  What would the motion of water up the straw be?  What is the highest height he could drink from?
Ignore thermodynamic effects like evaporation and assume the straw is stationary relative to the water and that there is no friction.
 A: If I follow up on keenan pepper's suggestion, if the water is deep, and especially if you can mess with the topology of the straw you can go to almost unlimited height!
Consider a straw that is stuck very deeply into the ocean. Then coil the straw around at great depth many many times. I then blow very hard (I am superman afterall), and create a huge volume of airfilled straw at great depth. This configuration has a great deal of potential energy, so if we simply stop blowing we have the pressure at the great depth of the bottom of the straw accelerating water into and up the straw. Since by coiling the straw at great depth I can obtain an unlimited ratio of volume of the straw underwater to volume above water, the energy analysis allows me to reach an unlimited height. So the issue becomes if there is some other sort of limit. Can we get cavitation of water trying to enter the straw or something if the velocity gets too high? But, in any case you should be able to get really high, tens or hundreds of times the static limit, by preenergizing the system in this way. In the real world friction will limit how far you can take it.
A: Trick question, he'd use his super strength to bend the straw into an Archimedes' screw, then hold it at an angle to the surface of the water and rotate it about the axis.
This lets him draw it up to any height, and then he can drain the world's oceans to prove a point or do whatever other superdickery he's trying to do.
A: I am not fully convinced by this argument, but can't find a flaw in it.
Let's analyse a similar experiment, which I believe to be equivalent. Assume that initially, the water is already at the stable level $H=\frac{P_{atm}}{\rho g}$. The vacuum is already present.
Now, by some unimportant means, we lower the water to $h=0$, and then let it go up freely.
How much energy are we storing in the system by lowering the water? We can find out by calculating the work done. The work is done against pressure and in favour of gravity.
$$W= sP_{atm}H + \int^0_H{m(h)g\ \mathrm{d}h}$$
Substituting $H=\frac{P_{atm}}{\rho g}$, $m(h)=\rho s h$
$$W= \frac{sP_{atm}^2}{\rho g} + s\rho g\int^0_H{h\ \mathrm{d}h}$$
$$W= \frac{sP_{atm}^2}{\rho g} - \frac{1}{2}s\rho gH^2$$
$$W= \frac{sP_{atm}^2}{2\rho g}$$
Now when the water is released and allowed to rise, all this energy will used to make the water rise. No energy is assumed to be wasted on friction.
At the highest point, all the energy will be converted in gravitational potential energy. This can be expressed through the following formula:
$$U(h)=\frac{1}{2}s\rho gh^2$$
Therefore, at the top point, $W=U(h)$
$$\frac{sP_{atm}^2}{2\rho g} = \frac{1}{2}s\rho gh^2$$
Solving for $h$:
$$h^2=\frac{P_{atm}^2}{g^2\rho^2}$$
$$h = \frac{P_{atm}}{g\rho} = H$$
Therefore, the water will raise up to $H$, which it will reach with zero velocity.
A: I can not comment yet, so I will put it as an answer :-(   I like the energy approach, but why not to use Archimedes principle? First, substitute the air atmosphere by an extra ten meters of water around the straw, so that now the initial conditions are a vacuum straw inserted a lenght h in a fluid. The energy to produce such vacuum in the fluid you can see by Archimedes; and it is h/2 times g times the mass of the removed fluid. Let it move, and it can go up until filling a column of lenght h above the level, because the energy (now gravitational) of this column is, again, h/2 times its mass times g.  
A: Superman is going to manipulate the air pressure in the straw. To get the water to go up, he must provide a reduction in the pressure. It's clear that if this reduction is provided at a very slow rate, then he will not be able to significantly exceed 10 meters or so (as limited by atmospheric pressure).
On the other hand, if he reduces the pressure quickly, it's at least possible that the water could reach a higher height. How high can the water go under this assumption? The idea is to use the momentum of the water to get the water higher, so the figure of merit will be the maximum speed of the water at surface level.
Reducing the pressure cannot move the water faster than the air and the air speed is limited by the speed of the gas molecules in the air or about 330 meters per second. By equating kinetic energy $0.5 m v^2$ with potential energy $mgh$, water with that initial speed can reach a height of $h = 0.5 v^2/g = $ 2775 meters. The height is small enough to justify the assumption that $g$ is a constant. Maybe you should add 10 meters for the usual vacuum effect.
Hmmm. Ah, what the heck, I ought to just do the fracking calculation for how high the water goes in a wide straw when a vacuum is applied to it.
A: If we are looking at this from a purely suction related problem then superman the maximum height sumperman could lift water in a straw would be equal to the pressure being exerted on the water he is drinking.  If drinking from sea level then he could lift or suck the water about 10 m.  Theoretically he could create a complete vacuum in his mouth then the amount of lift is merely the differential pressures.   We run into this limit all the time with vacuum pumps.  However, if he is able to suck really quickly then the velocity of the air could allow for water entrainment beyond the maximum lift.  He could not suck the water as one big slug but rather as droplets carried up the straw due to the wind velocity within the straw.  In a more practicle case, vacuum pumps have been shown to lift water from over a couple of hundred feet above static water level using this method.  But the rate of recharge within the well must be suffuciently low to ensure that the water does not "clog" your suction pipe.  To put another way you need to have far more air than water going up your straw.
A: I think we can most easily consider the problem from the perspective of energy. For a unit area column the external energy put in equals the volume of the column times the air pressure. The gravitational energy is the mass of water raised times the average height of the water. So
as we pull the water up, we the water is gaining kinetic energy until the water reaches the static limit (roughly 10meters), but at this point the average water in the tube has only risen by half that amount, so the rest is kinetic energy of (upward) water motion. So the vacuum energy in dimensionless units is $h$, while the gravitational energy is $\frac{1}{2}h^2$. The solution is $h=2$.
When we reach 2 times the static limit (20 meters) then the gravitational energy in the water matches the "vacuum" energy we put in, so that would represent the high point of the oscillation. So I think we would get 2 times the static limit. The water velocity will be messy to solve for, as the amount of water moving in the column depends upon height, so just look only at net energy as a function of water height... Of course he will only get a sip of water, then the column would start to fall........
A: I have an argument that the water in the straw will rise to twice the equilibrium height.
David and Martin's answers consider the system of water in the straw.  I will consider the system of the water in the straw plus the water in the reservoir.
As water goes into the straw, the water level in the reservoir drops, and the atmosphere does work on the system.  If a volume $V$ of water enters the straw, the work done on the system is $PV$, with $P$ the atmospheric pressure.  Assume that the reservoir has a large surface area so that the level the reservoir drops is negligible.
When the water is at its peak in the straw, the kinetic energy of the system is zero, so the potential energy is $PV$.  The potential energy is also $\rho g V h/2$.  So the maximum height of the water is 
$$h = \frac{2P}{\rho g}$$
This answer is different from Martin and David's.  I think this might be because when the water starts moving, the pressure at the entrance to the straw may not be $P$ any more.
A: I went back and took a more careful look at this. I'm still not convinced it's correct, but I'm hoping this is at least better than what I had before.

Let $h$ be the height of the column of water inside the straw. As this height rises by an amount $\delta h$, the work done on the column is
$$\delta W = P A \delta h$$
where $A$ is the cross-sectional area. The change in potential energy is
$$\delta U = \rho A \delta h g h$$
so by conservation of energy,
$$\delta K = \delta W - \delta U = \left(P - \rho g h\right) A \delta h$$
This excess kinetic energy comes from two contributions: the added mass,
$$\delta K_1 = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A \delta h)\dot{h}^2$$
and any change in speed of the column of water,
$$\delta K_2 = \frac{1}{2}m(2v\delta v) = (\rho A h)\dot{h} \delta\dot{h}$$
Putting it all together, we get
$$P A \delta h - \rho A g h \delta h = \frac{1}{2}\rho A \dot{h}^2 \delta h + \rho A h\dot{h}\delta\dot{h}$$
If you assume (or prove) that $P$ is dependent on $h$ through Bernoulli's theorem,
$$P + \frac{1}{2}\rho\dot{h}^2 = P_0$$
Substituting in (and canceling the common factor of $A$), you get
$$P_0 \delta h - \rho g h \delta h = \rho \dot{h}^2 \delta h + \rho h \dot{h}\delta\dot{h}$$
which at least accounts for the mysterious factor of $\frac{1}{2}$ that appeared in previous versions of my answer.
Now, I don't think we can simply assume that $\delta h \neq 0$ and divide it out, because if we do that, we get a factor of $\frac{\delta\dot{h}}{\delta h}$ which is undefined at the initial and maximum heights. (Roughly speaking, the variation $\delta h$ is second-order at those points whereas the variation $\delta\dot{h}$ is still first-order.) Instead, I'll divide by $\delta t$, which certainly should not produce any singularities, to get
$$P_0 \dot{h} - \rho g h \dot{h} = \rho \dot{h}^3 + \rho h \dot{h}\ddot{h}$$
At the initial and maximum heights, $\dot{h} = 0$, so the equation is trivially satisfied there. But consider the situation when displaced from either initial or maximum height by an arbitrarily small amount, such that $\dot{h}\neq 0$. Here we can cancel out $\dot{h}$ to get
$$P_0 - \rho g h = \rho \dot{h}^2 + \rho h \ddot{h}$$
Since $\dot{h}$ will be infinitesimally small around the maximum height, we can neglect the first term on the right, but not the second. So we're left with
$$P_0 - \rho g h = \rho h \ddot{h}$$
Note that this agrees with a simple analysis using Newton's second law. (The forces acting on the column of water at its maximum height are the pressure force $P_0 A$ acting upwards and gravity $\rho Ahg$ acting downwards, and the difference is equal to $ma = \rho Ah\ddot{h}$.) So the differential equation passes at least one basic consistency test.
Anyway, this equation no longer admits the solution $h = \frac{P_0}{\rho g}$. Instead we have
$$h = \frac{P_0}{\rho (g + \ddot{h})}$$
Unfortunately I can't think of a way to determine $\ddot{h}$ at maximum without solving the equation, so for now I'm limited to a numerical solution.
For a quick estimation, I plugged the full differential equation from above into Mathematica's NDSolve function. With boundary conditions $h(0) = 0$ and $\dot{h}(0) = 0$, it complained about undefined expressions, so I used boundary conditions at a nonzero time, 
$$h(\epsilon_t) = \epsilon_h$$
and
$$\dot{h}(\epsilon_t) = \epsilon_{\dot{h}}$$
for values of the various $\epsilon$ constants ranging from $10^{-3}$ to $10^{-8}$. In my tests, I get this graph, seemingly independently of the values of $\{\epsilon\}$ or the ratios between them:

Mathematica indicates that the graph peaks at $15.5\,\mathrm{m}$, so if this analysis is correct, that would be the maximum height. (FWIW I am still very suspicious of this calculation though)
A: You can manipulate the vacuum suction limitation of maximum height of 33.9 feet or 10 meter (14.7 psia or 0.1 Mpa) by using oscillating blow and suct. Use longer straw submerged deep enough into the water, blow it until the air almost reach the bottom of the straw then suck it! You will get height boost!!
A: Unless I'm missing something, this is simply the height of a water-based barometer, since it is really the atmospheric pressure that is pushing the water up the straw. at STP, the answer is 33 1/2 feet. If he were sucking Hg up the straw (not recommended for non-superheros!), the height would be ~30 inches.
A: If you immerse the staw in 10m of water, while holding the end closed, and then release it, the water will accelerate upwards past the level point and it will overshoot up to a height of about 6m. No sucking needed. Add some suction to this and you can go higher. A square section straw should minimize the friction loss allowing for better "spring back". Overall you can help the suction, by immersing the straw deeper and deeper. 
Note that it takes work do immerse the straw (displacing the water) and that is the energy conveted into ponetial energy that allows the water to rise. Close the end when the water reaches the maximum height and you can measure how high you can reach.
