How much heat is required to convert 5.88 g of ice at − 12.0 ∘c to water at 27.0 ∘c ? (the heat capacity of ice is 2.09 j/g∘c, δhvap(h2o)=40.7kj/mol, δhfus(h2o)=6.02kj/mol)?

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

2778 j

The answer is b metal denting