Question

What is the standard enthalpy change for the fusion of 50 grams of ice at 0°C to liquid water at 25°C, assuming constant pressure and no heat loss to the surroundings?

Answer 1

To calculate the standard enthalpy change for the fusion of 50 grams of ice at 0C to liquid water at 25C, we need to consider two steps:1. Melting the ice at 0C to water at 0C2. Heating the water from 0C to 25CStep 1: Melting the iceThe enthalpy of fusion for ice is 6.01 kJ/mol. First, we need to convert the mass of ice  50 grams  to moles.Molar mass of water  H2O  = 18.015 g/molMoles of ice =  50 g  /  18.015 g/mol  = 2.776 molesEnthalpy change for melting ice =  2.776 moles    6.01 kJ/mol  = 16.68 kJStep 2: Heating the waterThe specific heat capacity of water is 4.18 J/gC. We need to convert the enthalpy change to kJ.Enthalpy change for heating water =  50 g    4.18 J/gC    25C  = 5225 J = 5.225 kJTotal enthalpy change = Enthalpy change for melting ice + Enthalpy change for heating water = 16.68 kJ + 5.225 kJ = 21.905 kJThe standard enthalpy change for the fusion of 50 grams of ice at 0C to liquid water at 25C is 21.905 kJ.