What will be the time required to heat a 15 l bucket full of water from `20^(@)C` to `40^(@)C` using a 1500 W immersion heater?

Explanation

Mass of 15 l water, m = 15 kg,
specific heat of water,
`s = 1000 cal cdot kg^(-1) cdot^(@)C^(-1)`
Now, `W = JH or, Pt = Jmstheta`
where P = power of the heater = 1500 W, t = time required,
`theta` = rise in temperature = 40 - 20 = `20^(@)C`.
So, `t = (Jms theta)/(P) = (4.2 xx 15 xx 1000 xx 20)/(1500) = 840 s = 14 "min"`.