Question

An electric kettle containing 600 cm³ of water at 9⁰C and with a heater

coil resistance equal to 16 Ω was left connected to the mains. In how

much time will all the water in the kettle boil away? The voltage in the

mains is 120 V and the efficiency of the kettle 60%.

Answer 1

$V = 600cm^3$

$\rho= 1\frac{g}{cm^3}$

$m= \rho V=600g=0.6kg$

$\text{The amount of energy required to heat water to }100\degree$

$\Delta T =100\degree-9\degree= 81\degree$

$c=4200\frac{J}{kg*C\degree}$

$Q_1= cm\Delta T$

$Q_1= 4200*0.6*81=204120J$

$\text{The amount of energy required to convert water to steam}$

$L =2300000\frac{J}{kg}$

$Q_2 = Lm$

$Q_2=2300000*0.6=1380000J$

$Q=Q_1+Q_2$

$Q=204120+1380000=1584120J$

$\text {according to Joule-Lenz Law:}$

$Q_3=I^2Rt$

$I =\frac{U}{R}$

$Q_3=\frac{U^2t}{R}$

$\eta=0.6$

$Q =Q_3*\eta$

$t = \frac{QR}{U^2\eta}$

$t=\frac{1584120*12}{120^2*0.6}=2200s$

$\text{Answer: }2200\ s$