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 Homework Statement:

An electric water heater which is wellinsulated 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).
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).