`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralization of `HCl` with `NaOH`, if find temperature was recorded to be `29^(@)C`.
`(rho_(H_(2)O)=1g//mL)`

Explanation

Total mass of solution `= 150 + 100 = 300 g`
`Q=` Total heat produced `= 300 xx (28.77 - 25.35) cal`
`=300 xx 3.42 = 1026 cal`
Heat of neutralisation `=(Q)/(150) xx 1000 xx (1)/(0.5)`
`=(1026)/(150) xx 1000 xx (1)/(0.5) = 13.68 kcal`
Since heat is liberated, heat of neutralisation should be negative.
So, heat of neutralisation `=- 13.68 kcal`