Question

Calculate the standard enthalpy change for the fusion of 25 grams of ice at -10°C to liquid water at 20°C. The specific heat capacity of ice is 2.09 J/g°C, the specific heat capacity of water is 4.18 J/g°C, and the enthalpy of fusion of water is 6.01 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the fusion of 25 grams of ice at -10C to liquid water at 20C, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the liquid water from 0C to 20CStep 1: Heating the ice from -10C to 0Cq1 = mass x specific heat capacity of ice x temperature changeq1 = 25 g x 2.09 J/gC x  0 -  -10  Cq1 = 25 g x 2.09 J/gC x 10Cq1 = 523.5 JStep 2: Melting the ice at 0CFirst, we need to find the moles of ice:Molar mass of water = 18.015 g/molmoles of ice = 25 g / 18.015 g/mol = 1.387 molq2 = moles x enthalpy of fusionq2 = 1.387 mol x 6.01 kJ/molq2 = 8.338 kJ = 8338 J  converted to Joules Step 3: Heating the liquid water from 0C to 20Cq3 = mass x specific heat capacity of water x temperature changeq3 = 25 g x 4.18 J/gC x  20 - 0 Cq3 = 25 g x 4.18 J/gC x 20Cq3 = 2090 JNow, we add the enthalpy changes for all three steps to find the total standard enthalpy change:Total enthalpy change = q1 + q2 + q3Total enthalpy change = 523.5 J + 8338 J + 2090 JTotal enthalpy change = 10951.5 JThe standard enthalpy change for the fusion of 25 grams of ice at -10C to liquid water at 20C is 10,951.5 J.