Question

Calculate the standard enthalpy change for the fusion of 100 grams of ice at 0°C to liquid water at 0°C. The specific heat capacity of water is 4.184 J/g°C, the standard enthalpy of fusion of water is 6.01 kJ/mol, and the molar mass of water is 18.015 g/mol. Assume all heat absorbed is used to melt the ice and the temperature remains constant.

Answer 1

To calculate the standard enthalpy change for the fusion of 100 grams of ice at 0C to liquid water at 0C, we can use the following equation:H = n * H_fusionwhere H is the standard enthalpy change, n is the number of moles of water, and H_fusion is the standard enthalpy of fusion of water.First, we need to find the number of moles of water  n  in 100 grams of ice:n = mass / molar massn = 100 g / 18.015 g/moln  5.55 molNow, we can calculate the standard enthalpy change:H = n * H_fusionH = 5.55 mol * 6.01 kJ/molH  33.36 kJTherefore, the standard enthalpy change for the fusion of 100 grams of ice at 0C to liquid water at 0C is approximately 33.36 kJ.