You have to calculate the heat for three separate processes:1) heat the ice from - 12°C to 0°C, 2) melt the ice at 0°C, and 3) heat the liquid water from 0°C to 27.0 °C.
1) Heating the ice from - 12°C to 0°C
Q1 = m * C * ΔT = 5.88g * 2.09 j/g°C * [0°C - (-12°C) ] = 147.47 j
2) Melting the ice at 0°C
Q2 = m * Δh fus
Convert 5.88 g to moles => 5.88 g / 18.0 g/mol = 0.327 moles
Q2 = 0.327 moles * 6.02 kj / mol = 1.96653 kj = 1966.53 j
3) Heating liquid water from 0°C to 27.0 °C
Q3 = m * C * ΔT
C = 4.1813 j/g°C
Q3 = 5.88 g * 4.1813 j/g°C * (27.0°C - 0°C) = 663.82 j
4) Total heat, Q
Q = Q1 + Q2 + Q3 =147.47j + 1966.53 j + 663.82 j = 2777.82 j ≈ 2778 j
Answer: 2778 j
