This is what I remember from my first year of Thermo.
OK..  
     -First find the Specific volume (v) = V/M
         where V = Volume of container
       M = Mass of water = Volume of water * rho of water
       ---> v
  
For rigid or seal container v1 = v2...  hence lookup steam table for v, then comparing the specific volume of steam at 100 degree Celcius(or what ever that is in Fereinheit)...  If the calculated Specific volume is greater than the Specific volume of the vapor then look at the supperheated steam table to get the pressure.
   Otherwise if the calculated specific volume value lies between the specific volume of both liquid and vapor state then use the pressure value at that 100 degree Celcius.

  Note:  Your temperature convention is in Fereinheit not in Celcius like mine, so use the fereinheit boiling point temp.

If you continue to add energy, the pressure and temperature will (theoretically) increase without limit.  A quick look at a steam table reveals the following:

At the critical pressure and temperature of water (3206 psia, 705 °F), the specific volume is 0.0503 ft^3/lbm.  Given that your vessel is 2/3 full of water, if it were single phase system, the specific volume would be (3/2)*(0.01603 ft^3/lbm) = 0.0240 ft^3/lbm.  Thus, the pressure and temperature would have to exceed the supercritical saturation point.

Power =Current x Voltage
Power = Current^2 x Resistance for DC circuits
You need to tell us the voltage and if its AC or DC
In the same system of units, Power =Heat
Heating of water, q (heat) = m x cp x (T1-T2)
m = mass of water
cp = specific heat for water
T1 and T2 are starting and stopping temperatures
How is the current used to heat the water? Just by the waters electrical resistance?
What temperature and pressure steam do you want?

You need to furnish more information. Hope the above helps you think about your problem.

Don Leffingwell

Reguarding
DonLeffingwellPE

Remember this

Q = mcp(T2-T1) = Power X Time

there fore

    Rate of Heat Transfer Rate =  Power

power = vi = i^2xr
for both ac and dc, when circuit is purely resistive ..i.e no active elements such as capacitor or inductance...then no phase angle exists... then 4500^2 multiply by the resistance is the power in watts ... resistance will change with temperature though..probably dramatically for this range of heating

All the info. I have recieved is great! Let me expand on the problem a little more.  In theory anyway, the submersible tank had a seal violation and water entered the enclosure approx. 2/3 of the water (this is from the visable water mark on the inside of the tank).  Someone came along and re-sealed the enclosure door and then put in approx. 1.5-2.0 psi of nitrogen.  The tank houses a low voltage 3 phase circuit breaker at 216 VAC and 4500A.  The breaker was re-energized and in theory the water heated and created enough pressure to bow out the sides and the bottom of this .25" thick steel tank.  Would this theory be possible?

drax: It's rather surprising that your vessel hasn't exploded or that you haven't experienced incredible shock; 4,500 amps is enough current to give even the most jaded GE Nuclear engineer pause. At any rate assuming that the vessel was filled to the 2/3 level at atmospheric conditions your starting conditions should be as follows;
                T1 = 68 deg F
                P1 = 16.2 psia ;     14.7 + 1.5
                v1 = 0.01608 cft/lbm;  specific volume of the water

Assuming that you ran current for approximately one hour; I assume you have some type of heating element in there, the amount of heat energy given to the water would be;       
                 Q = 3,320,000 Btu's or 972 KwHrs

Which is enough to raise the water temperature over 2,000 deg F. So conditions after one hour would be;
                  T2 = 2,128 deg F
                  P2 = 8,670 psia
                  v2 = 0.0241 cft/lbm
The pressure in the tank was extrapoted off the tables for superheated steam and it is assumed that all the water has become vapor. While I find your story rather improbable I was interested to find out what the results might be like and  I hope that you find them of some assistance to you. However what I can't understand is why none of the replying engineers didn't mention to you that a: the 4,500 Amp current sounds very unrealistic and b: there should be a relief valve on your tank so that the pressure never goes much beyond say 30 psi. about what you find in your average car tire. So no it's not at all surprising that your tank "bulged". I wish you all the best of luck with your tank and if I can be of any further assistance to you feel free to drop me a line at dsigel@pacbell.net
Sincerely,
Dave Sigel, ME

You forgot to acount for the latent heat of vaporization, i.e. the energy absorbed when the liquid water changes phase.  You can't just come up with a temperature increase.  That's not where all the energy went.