To calculate the total heat absorbed, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the water from 0C to 35CStep 1: Heating the ice from -10C to 0Cq1 = mass x specific heat of ice x TSpecific heat of ice = 2.09 J/gC approximately T = 0C - -10C = 10Cmass = 32 gq1 = 32 g x 2.09 J/gC x 10C = 669.6 JStep 2: Melting the ice at 0Cq2 = moles x enthalpy of fusionMolar mass of ice H2O = 18.015 g/molEnthalpy of fusion = 6.01 kJ/mol = 6010 J/molmoles = 32 g / 18.015 g/mol = 1.776 molq2 = 1.776 mol x 6010 J/mol = 10677.6 JStep 3: Heating the water from 0C to 35Cq3 = mass x specific heat of water x TSpecific heat of water = 4.18 J/gCT = 35C - 0C = 35Cmass = 32 gq3 = 32 g x 4.18 J/gC x 35C = 4697.6 JTotal heat absorbed q_total = q1 + q2 + q3q_total = 669.6 J + 10677.6 J + 4697.6 J = 16044.8 JThe amount of heat absorbed when 32 grams of ice at -10C melts into water and then further heated to 35C is 16,044.8 J.