One kilogram of ice at `0^(@)C` is mixed with one kilogram of water at `80^(@)C`. The final temperature of the mixture is (take: specific heat of water=4200 `J" "kg^(-1)K^(-1)`, latent theat of ice`=336" kJ "kg^(-1)`)

Options

A. `40^(@)C`
B. `60^(@)C`
C. `0^(@)C`
D. `50^(@)C`

Answer

C

Explanation

heat given out due to cooling of water from `80^(@)C` to `0^(@)C=H_(1)=msDeltat=1xx4200xx(80-0)`
`=336000J=336kJ`.
Heat required for the melting of ice from and at
`0^(@)C=H_(2)=mL=1xx336kJ`.
Since `H_(1)=H_(2)`, the final temperature of the mixture will be `0^(@)C` and it will be totally water.