A copper electrical kettle weighing 1 kg contians 0.5 kg of water at `20@C`. It takes 10 minutes to raise the temperature to `100^@C`. If the electric energy is supplied at 220V, calculate the strength of the current,assuming that 20% heat is washed. Specific heat of copper is 0.1.

Explanation

Here, mass of kettle, `m_(k) = 1000 g`,
mass of water, `m_(w) = 500 g` ,
sp. Heat of copper, `S_(Cu) = 0.1 cal g^(-1) C^(-1)`,
sp. Heat of water, `S_(w)= 1 cal g^(-1) .^(@)C^(-1)`
Rise in temperature `= theta_(2) - theta_(1) = 100 -20 = 80^@C`
Heat required = `(m_(k) S_(Cu) + m_(w) S_(w)) (theta_(2) - theta_(1))`
`= (1000 xx 0.1 + 500 xx 1) xx 80 = 48000 cal`
Heat produced,
`H = (VIt)/J = (220 xx I xx (10 xx 60))/4.2 cal`.
Useful heat produced = (100 -20)= 80%
`= 80/100 xx (220 xx I xx 10 xx 60)/4.2 cal`.
`:. 80/100 = (220 xx I xx 10 xx 60)/4.2 = 48000`
On solving, `I= 1.9A`