To calculate the amount of heat energy required to change 100 grams of ice at -10C into liquid water at 20C, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the liquid water from 0C to 20CStep 1: Heating the ice from -10C to 0CQ1 = mass x specific heat capacity of ice x temperature changeSpecific heat capacity of ice = 2.1 J/gC approximately Q1 = 100 g x 2.1 J/gC x 0 - -10 C = 100 x 2.1 x 10 = 2100 JStep 2: Melting the ice at 0CQ2 = mass x enthalpy of fusion of iceFirst, we need to convert the mass of ice to moles.Molar mass of water = 18.015 g/molMoles of ice = 100 g / 18.015 g/mol = 5.55 mol approximately Now, we can calculate the heat required to melt the ice.Q2 = 5.55 mol x 6.01 kJ/mol = 33.36 kJ = 33360 J converted to Joules Step 3: Heating the liquid water from 0C to 20CQ3 = mass x specific heat capacity of liquid water x temperature changeQ3 = 100 g x 4.18 J/gC x 20 - 0 C = 100 x 4.18 x 20 = 8360 JNow, we can add the heat energy required for all three steps to get the total heat energy required.Total heat energy = Q1 + Q2 + Q3 = 2100 J + 33360 J + 8360 J = 44,820 JSo, the amount of heat energy required to change 100 grams of ice at -10C into liquid water at 20C is 44,820 Joules.