An electric kettle was marked `500 W,220 V` and was found to raise 1 kg of water to `20^(@)` to the boiling point in 20 minutes. Calculate the heat efficiency of the kettle. Sp. Heat of water `= 4200 J kg^(-1) K^(-1)`.

Answer

`0.595 %`

Explanation

Heat absorted by water ,if `= ms (theta _(2) - theta _(1))`
` = 1 xx 4200 xx (100 - 15) = 4200 xx 85 J`
Heat produced by electric kettle
`H = Pt = 500 xx (20 xx 60) J`
Heat effeciency `eta = (H')/(H) = (4200 xx 85)/(500 xx 20 xx 60)`
`= 0.595 %`