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

Given, m= 3 kg
`c_("ice") = 2100 J kg^(-1) K^(-1)`
`c_("water") = 4186 J kg^(-1) K^(-1)`
`L_("ice") = 3.35 xx 10^5 Jkg^(-1)`
`L_("steam") = 2.256 xx 10^6 J kg^(-1)`
Change in temperature for ice at `-12^@C` to ice at `0^@C , Delta T_1`
` = 0 - (-12) = 12^@C`
Heat required, `Q_1 = mc_("ice") Delta T_1 = 3 xx 2100 xx 12 = 75600 J `
To convert ice at `0^@C` to water at `0^@C` , heat required is
`Q_2 = mL_("ice") = 3 xx 3.35 xx 10^5 = 1005000 J `
Change in temperature for water at `0^@C` to water at `100^@C` is, `DeltaT_2 = 100 -0 = 100^@C`
Heat required, `Q_3 =mc_w Delta T_2= 3 xx 4186 xx 100 = 1255800 J `
To convert water at `100^@C` to steam at `100^@C` , heat required is
`Q_4 = mL_("steam") = 3 xx 2.256 xx 10^8`
= 6768000 J
Total heat required,` Q = Q_1 +Q_2 +Q_3 +Q_4`
= 75600 J +1005000 J + 1255800 J +6768000 J `= 9.1 xx 10^6J`