If the power supply is 90 percent efficient, then it's 10% inefficent and will dissipate 1 KW. The total energy dissipated over the 1 minute is then 1 KW X 60 sec = 1000 Joules/sec X (60 sec) = 60000 joules.
If you neglect any dissipation of heat to the ambient, that's the same thing as saying that the tank is perfectly insulated. Therefore, all the energy will go into raising the water temperature. The thermodynamic relationship between the temperature rise of the water and the energy transferred to it is:
q = m*C* Delta T where
m = the mass of the water (in kg)
C = specific heat of water (make sure to use metric units)
Delta T is the temperature rise in degree C
So, rearrange the above equation and solve for Delta T. Check your units and make sure they cancel out with only temperature remaining.
This solution above is based on no water entering or leaving the 20 gallon tank. If you have watering flow into and/or out of the tank, you have a whole different problem.
Hope this helps.
Dave
