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 fusion of ice `= 3.35xx10^(5) J kg^(–1)` and latent heat of steam `= 2.256xx 106 J kg^(–1)`.

Explanation

Q = heat required to convert 3 kg of ice at `- 12^(0)` C to steam at `100^(0)` C,
`Q_(1) ` = heat required to convert ice at `-12^(0)` C to ice at `0^(0)` C.
= m `S_("ice") Delta T_(1)`
`= (3 kg) ( 2100 j kg^(-1) k^(-1)) [ 0 - (-12) ]^(0) C `
= 75600 J
`Q_(2)` = heat equired to melt ice at `0^(0)` C to water at `0^(0)` C
= `mL_(iec) = (3 kg ) (3.35 xx 10^(5) j kg^(-1) ) = 1005000` J
`Q_(3)` = heat required to convert water at `0^(0)` C to water at `100^(@)` C
`= mS_(w) Delta T_(2) = (3kg) (4186 J kg^(-1) K^(-1) ) (100^(0) C ) `
= 1255800 J
`Q_(4)` = heat equired to convert water at`100^(0)` C to steam at `100^(0)` C
`mL_("steam") = (3kg) (2.256 xx 10^(6) J kg^(-1))`
= 6768000 J
So, Q = `Q_(1) + Q_(2) + Q_(3) + Q_(4) = 9.1 xx10^(6)` J