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 ` = 3.35 xx 10^(5) J kg^(-1)` and latent heat of steam ` = 2.256 xx 10^(8) 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 ") `
` = 3.35 xx 10^(5) J kg^(-1)`
latent heat of steam , `L_("ateam") `
` = 2.256 xx 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") Delta T_(1) = (3 kg) (2100 J kg^(-1). K^(-1)) [0 - (-12)]^(@) C = 75600 J`
`Q_(2)` = heat required to melt ice at `0^(@) C` to water at `0^(@) C`
` = m L_("f ice ") = (3 kg) (3.35 xx 10^(5) J kg^(-1))`
` = 1005000 J`
` Q_(3)` = heat required to convert water at `0^(@) C` to water at `100 ^(@) C`.
` ms_(w) Delta T_(2) = (3kg) (4186 J kg^(-1) K^(-1)) (100^(@) C)`
` = 1255800 J`
` Q_(4)` = heat required to convert water at `100^(@) C` to steam at `100^(@) C`.
` = m_("steem") = (3 kg) (2.256 xx 10^(6) J kg^(-1))`
` = 6768000 J `
So, Q = `Q_(1) + Q_(2) + Q_(3) + Q_(4)`
` = 75600 J + 100500 J + 1255800 J + 6768000 J`
` = 9.1 xx 10^(6) J`