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An electric immersion heator of 1.08 k W...

An electric immersion heator of `1.08` k W is immersed in water. After the water has reached a temperature of `100^(@)C`, how much time will be required to produce 100 g of steam?

A

`50 s`

B

`420 s`

C

`105 s`

D

`210 s`

Text Solution

Verified by Experts

The correct Answer is:
D

`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`
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