An electric immersion heater of `1.08 kW` is immersed in water . After it has reaches a temperature of `100^(@)C` , how much time will be required to produce `100 g` of steam?

Options

A. `50 s`
B. `420 s`
C. `105 s`
D. `210 s`

Answer

D

Explanation

`L` is the latent heat of vaporization of water, the heat required for producing `1 g` of team. So
`L = 540 cal = 540 xx 4.2 = 2268 J`
Energy supplied is `1080 Js^(-1)`. Time required to boil `100 g` of water is
`t = 540 xx 4.2 xx 100 //1080 = 210 s`