An electric immersion heater of 1.08 kW is immersed in water. After it has reached a temperature of ${100}^{0}C,$ how much time will be required to produce 100 g of steam?

L is the latent heat of vaporization of water, the heat required for producing 1 g of steam. So 

$L=540cal=540\times 4.2=2268J$

Energy supplied is $1080J{s}^{-1}.$

Time required to boil 100 g of water is

$t=540\times 4.2\times 100/1080=210s$