To calculate the standard enthalpy change for the fusion of 50 grams of ice at -10C to liquid water at 10C, 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 10C.First, let's calculate the moles of ice:The molar mass of water H2O is 18.015 g/mol.moles = mass / molar mass = 50 g / 18.015 g/mol = 2.776 molesStep 1: Heating the ice from -10C to 0CThe specific heat capacity of ice is approximately 2.093 J/gC.q1 = mass specific heat capacity of ice change in temperature q1 = 50 g 2.093 J/gC 0 - -10 C = 50 g 2.093 J/gC 10C = 1046.5 JStep 2: Melting the ice at 0Cq2 = moles enthalpy of fusion q2 = 2.776 moles 6.01 kJ/mol = 16.681 kJ = 16681 JStep 3: Heating the liquid water from 0C to 10Cq3 = mass specific heat capacity of water change in temperature q3 = 50 g 4.184 J/gC 10 - 0 C = 50 g 4.184 J/gC 10C = 2092 JNow, let's sum up the enthalpy changes for all three steps:H = q1 + q2 + q3 = 1046.5 J + 16681 J + 2092 J = 19819.5 JThe standard enthalpy change for the fusion of 50 grams of ice at -10C to liquid water at 10C is 19.82 kJ rounded to two decimal places .