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 = 1 litre = 1000 c.c.
Mass of water, m = volume `xx` density
`1000 xx 1 = 1000` gram
Rise in temperature, ` = theta_(2) - theta_(1) = 100 - 4`
`96^@C`
We know sp. Heat of water, `s= 1 cal g^(-1) @C^(-1)`
`:.` Heat taken by water `= ms (theta_(2) - theta_(1))`
`= 1000 xx 1 xx 96 = 96000` cal.
Energy spent in heating coil = pt `= 2000 xx t`
Useful thermal energy produced `= 80 %`
` = 2000 xx t xx 80/100 = ( 2000 xx t xx 80)/( 100 xx 4.2)` cal
As this thermal energy is taken by water, therefore, `( 2000 xx t xx 80)/(100 xx 4.2) = 96000`
or `t (96000 xx 100 xx 4.2)/(2000 xx 80)`
`= 252 seconds`.