How Does a Resistor Affect the Heating Time of Water?

[archaic]

Homework Statement

An electric water heater which is well-insulated warms 107 kg of water from 20.0°C to 45.0°C in 27.0 min.
Find the resistance (in Ω) of its heating element, which is connected across a 240 V potential difference.
How much additional time (in min) would it take the heater to raise the temperature of the water from 45.0°C to 100°C?
What would be the total amount of time (in min) required to evaporate all of the water in the heater starting from 20.0°C?

Relevant Equations

$$P=\frac{(\Delta V)^2}{R}=\frac{Q}{\Delta t}$$

1)$$R=\frac{\Delta t(\Delta V)^2}{Q}=\frac{\Delta t(\Delta V)^2}{mc\Delta T}=8.33\,\Omega$$2)$$\Delta t'=\frac{mcR\Delta T'}{(\Delta V)^2}=\frac{\Delta T'}{\Delta T}\Delta t=59.4\text{ min}$$3) I surfed the net a bit and have found a post on physics.stackexchange in which it is mentioned that the amount of water evaporated is related to the heat by $Q=Lm$, where $L=2.26\times10^6\,J/kg$. I used it to get$$P=\frac{(\Delta V)^2}{R}=\frac{dQ}{dt}=L\frac{dm}{dt}\Leftrightarrow\frac{(\Delta V)^2}{R}\Delta t=Lm\Leftrightarrow\Delta t=\frac{RLm}{(\Delta V)^2}$$but this surely is wrong (no taking account of the temperature, and gives a result less than the previous one). Any ideas please?
I really don't know much about thermodynamics.. I thought of $Q=mc\Delta T$ because it was shown in an example in the same chapter I am going through (Currents etc).

 

[hutchphd]

Not that I can see. Show the calculation (with units please). I believe the equation is correct.

 

[archaic]

Not that I can see. Show the calculation (with units please). I believe the equation is correct.

$$\Delta t=\frac{8.33[\frac{kg\,m^2}{s\,C^2}]\times107[kg]\times2.26\times10^6[J/kg=m^2/s^2]}{240^2[J^2/C^2=kg^2m^4/s^4]\times60[s/\text{min}]}=582.85\times\frac{kg\,m^2\,kg\,m^2\,C^2\,s^4\text{ min}}{s\,C^2\,s^2\,kg^2\,m^4\,s}$$I was wrong in my calculations earlier, but this is still wrong. The HW platform won't accept it (it wants it in minutes).

 

[hutchphd]

Seems reasonable to me. They want the " heat up" part too I think.

 

[archaic]

Seems reasonable to me. They want the " heat up" part too I think.

What bugs me is that the question says "starting from 20 degrees C", while my expression has no room for temperature.

 

[hutchphd]

Its two parts: heat the water to from 20 to 100C and then boil it!

 

[archaic]

Its two parts: heat the water to from 20 to 100C and then boil it!

where is my brain..
thank you!