# Water heating with a resistor

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).

Homework Helper
2022 Award
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).

Homework Helper
2022 Award
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.