To calculate the standard enthalpy change for the melting of 10 grams of ice at -10C to liquid water at 20C, we need to consider three steps:1. Heating the ice from -10C to 0C.2. Melting the ice at 0C.3. Heating the liquid water from 0C to 20C.Step 1: Heating the ice from -10C to 0Cq1 = mass x specific heat capacity of ice x temperature changeq1 = 10 g x 2.09 J/gC x 0 - -10 Cq1 = 10 g x 2.09 J/gC x 10Cq1 = 209 JStep 2: Melting the ice at 0CFirst, we need to find the moles of ice:Molar mass of water H2O = 18.015 g/molmoles = mass / molar massmoles = 10 g / 18.015 g/molmoles = 0.555 molNow, we can find the enthalpy change for melting the ice:q2 = moles x enthalpy of fusionq2 = 0.555 mol x 6.01 kJ/molq2 = 3.334 kJConvert to Joules:q2 = 3.334 kJ x 1000 J/kJq2 = 3334 JStep 3: Heating the liquid water from 0C to 20Cq3 = mass x specific heat capacity of water x temperature changeq3 = 10 g x 4.18 J/gC x 20 - 0 Cq3 = 10 g x 4.18 J/gC x 20Cq3 = 836 JNow, we can find the total enthalpy change by adding the enthalpy changes for each step:Total enthalpy change = q1 + q2 + q3Total enthalpy change = 209 J + 3334 J + 836 JTotal enthalpy change = 4379 JThe standard enthalpy change for the melting of 10 grams of ice at -10C to liquid water at 20C is 4379 J.