How much heat energy is gained when 5kg of water at `20^(@)C` is brought to its boiling point (Specific heat of water = `4.2 kJ kg^(-1^(@))C^(-1)`

Explanation

`Q _(2) = mcd T + mL = 2000 xx 1 xx 25 + 2000 xx 80 = 2.1 xx 10 ^(5) cal. Q _(1) = Q _(2) T _(1) //T_(2)`
`Q _(1) = 2.1 xx 10 ^(5) xx 298 // 258 = 2.42 xx 10 ^(5) cal. ` Energy needed `= Q _(1) - Q _(2) = 3.25 xx 10 ^(4) cal= 3.25 xx 10 ^(4) xx 4.2 //3600 = 37.92 Wh.`