Heating water with electricity
You can roughly calculate how much power is required to do this.

$$P = \frac{mc\Delta T}{t}$$ 

Where \$P\$ is power required, \$m\$ is mass of water, \$c\$ is specific thermal capacity of water, \$\Delta T\$ is temperature rise, and \$t\$ is time to change temperature (all SI units - i.e. \$^\circ\mathrm{C}\$ not \$^\circ\mathrm{F}\$). 

This is in fact a basic high school physics equation. Plug in your numbers and you get a rough estimate of the required power. The thermal capacity for water is roughtly \$4.184\space \mathrm{Jg^{-1}K^{-1}}\$. So doing a quick example, lets say you want to heat \$1\mathrm{l}\$ of water by \$60^\circ\mathrm{C}\$ (assuming water starts at room temp) in lets say \$1\mathrm{s}\$ as you give no time.

Plug all the numbers in, you get:

$$P = \frac{mc\Delta T}{t} = \frac{1000\mathrm{g} \times 4.184 \mathrm{Jg^{-1}K^{-1}} \times 60^\circ\mathrm{K}}{1\mathrm{s}} = 251\mathrm{kW}$$

Quite a lot of power there. Perhaps you should plug in some more reasonable values.

Whether or not it is possible will depend on whether or not you can transfer the required amount of power to the water in the required amount of time. I can't say whether that is possible or not as I don't know how much water you are trying to heat and how fast you are trying to heat it in.