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 capicity of water = `4186 J kg^(-1)K^(-1)`
Latent heat of fusion of ice = ` 3.35 xx 10^(5) J kg^(-1)`
and latent heat of steam = ` 2.256 xx 10^(6) J kg^(-1)` .

Explanation

We have
Mass of the ice, m = 3 kg
specific heat capacity of ice, `s_("ice") = 2100 J kg^(–1) K^(–1)`
specific heat capacity of water, `s_("water") = 4186 J kg^(*–1) K^(–1)`
latent heat of fusion of ice, `L_("f ice") = 3.35xx 10^(5) J kg^(–1)`
latent heat of steam, `L_("steam") = 2.256xx 10^(6) J kg^(–1)`
Now, Q = heat required to convert 3 kg of ice at `–12^(@)C` to steam at `100^(@)C`
`Q_(1)` = heat required to convert ice at` –12^(@)C` to ice at `0^(@)C`.
= m `s_("ice") DeltaT_(1) = (3 kg) (2100 Jkg^(-1)K^(-1))[0-(-12)]^(@)C=75600 J`
`Q_(2)` = heat required to melt ice at `0^(@)C` to water at `0^(@)C`
`= mL_("f ice") = (3 kg) (3.35xx 10^(5) J kg^(–1))`
= 1005000 J
`Q_(3)` = heat required to convert water at `0^(@)C` to water at `100^(@)C`.
`= ms_(w) DeltaT_(2) = (3kg) (4186J kg^(–1) K^(–1)) (100^(@)C)` = 1255800 J
`Q_(4)` = heat required to convert water at `100^(@)C` to steam at `100^(@)C`.
` = m L_("steam") = (3 kg) (2.256xx10^(6) J kg^(–1))`
=6768000 J
So,` Q = Q_(1)+Q_(2)+Q_(3)+Q_(4)`
=75600 J + 1005000 J + 1255800 J + 6768000 J
`=9.1xx10^(6) J`