A heating coil of `2000 W` is immersed in water . How much time will it take in raising the temperature of `1 L` of water from `4^(@) C "to" 100^(@) C`? Only `80%` of the thermal energy produced is used in raising the temperature of water.

Explanation

Here, `P = 2000 W , t = ?` (in seconds)
Volume of water is ` 1 L = 1000 cm^(3)`
Mass of water is ,
`m = volume xx density`
`= 1000 xx 1 = 1000 g`
Rise in temperature is ` theta _(2) - theta_(1) = 100 - 4 = 96^(@) C`
We know that the specific heat of water is
` mc ( theta_(2) - theta_(1)) = 1000 xx 1 xx 96 = 96000 cal`
Energy spent in heating the coils is `Pt = 2000 xx t`
Useful energy produced is `80% = 2000 xx t xx 80//100 J`
Useful heat produced is
`= (2000 xx t xx 80)/( 100 xx 4.2 ) cal`
As this heat is taken by water, `( 2000 xx t xx 80)/( 100 xx 4.2) = 96,000)`
or ` t = ( 96,000 xx 100 xx 4.2 )/( 2000 xx 80) = 252 s`