Question

What is the standard enthalpy change for the fusion of 10 grams of ice at 0°C to liquid water at 25°C? Given the standard enthalpy of fusion for ice is 6.01 kJ/mol and the specific heat capacity of water is 4.18 J/g°C.

Answer 1

To calculate the standard enthalpy change for the fusion of 10 grams of ice at 0C to liquid water at 25C, we need to consider two processes: 1. Melting of ice at 0C to liquid water at 0C2. Heating the liquid water from 0C to 25CFirst, let's calculate the number of moles of ice. The molar mass of water  H2O  is 18.015 g/mol.Number of moles =  mass of ice  /  molar mass of water Number of moles = 10 g / 18.015 g/mol  0.555 molNow, let's calculate the enthalpy change for each process.1. Melting of ice at 0C to liquid water at 0C:Enthalpy change  H1  =  number of moles    standard enthalpy of fusion H1 = 0.555 mol  6.01 kJ/mol  3.34 kJ2. Heating the liquid water from 0C to 25C:Enthalpy change  H2  =  mass of water    specific heat capacity of water    change in temperature H2 = 10 g  4.18 J/gC  25C  1045 J = 1.045 kJFinally, let's add the enthalpy changes for both processes to get the total standard enthalpy change.Total standard enthalpy change  H  = H1 + H2H = 3.34 kJ + 1.045 kJ  4.39 kJThe standard enthalpy change for the fusion of 10 grams of ice at 0C to liquid water at 25C is approximately 4.39 kJ.