Calculate the heat required to convert 3 kg of ice at `-12" "^(@)C` kept in a calorimeter to steam at `100" "^(@)C` at atmospheric pressure. Given specific heat capacity of ice `=2100" J kg"^(-1)K^(-1)`, specific heat capacity of water `=4186" J kg"^(-1)K^(-1)`, latent heat of fusionn of ice `=3.35xx10^(5)" J kg"^(-1)` and latent heat of steam `=2.256xx10^(6)" J kg"^(-1)`.

Explanation

Mass of ice `m=3` kg
Specific heat of ice `S_("ice")=2100" Jkg"^(-1)K^(-1)`
Specific heat of water `S_(W)=4186" Jkg"^(-1)K^(-1)`
Latent heat of fusion of ice `L_(f"(ice)")=3.35xx10^(5)" Jkg"^(-1)`
Latent heat of vaporization of steam `L_("heat")=2.256xx10^(6)" Jkg"^(-1)`
`rArrQ_(1)=` heat required to rise temperature of ice from `-12^(@)C` to `0^(@)C`
`="mS"_("ice")DeltaT_(1)`
`=3xx2100xx[0-(-12)]`
`=6300xx12`
`:.Q_(1)=756000" J . . . (1)"`
`rArr Q_(2)=` heat required to convert ice of `0^(@)C` in to water
`=mL_("ice")`
`3xx3.35xx10^(5)`
`:.Q_(2)=1005000 " J . . .(2)"`
`rArrQ_(3)=` heat required to rise temperature of water from `0^(@)C` to `100^(@)C`
`=mS_(W)DeltaT_(2)`
`=3xx4186xx(100-0)`
`:.Q_(3)=1255800" J . . .(3)"`
`rArr Q_(4)` = heat required to convert water of `100^(@)C` in to steam
`=mL_("steam")`
`=3xx2.256xx10^(6)`
`:.Q_(4)=" 6768000" J . . .(4)"`
`rArrQ=` heat required to convert ice at `-12^(@)C` in to steam
`:.Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)`
From equation (1), (2), (3) and (4)
`Q=75600+1005000+1255800+6768000=9104400" J"`
`:.Q~~9.1xx10^(6)J`