Question

For the following data: Warm water volume: 30.0mL; Warm water + ice volume: 51.5mL; Warm water initial temp: 62.8C; Final Temp (with ice): 1C;   

(A1) Calculate the volume of ice melted. (A2) Calculate the heat lost by warm water using 4.184J / g K as the specific heat of liquid water. (A3) Calculate the heat of fusion of ice in kJ/g and calculate the molar heat of fusion of ice in kJ/mol.

I already calculated ice melted to be: 51.5mL-30.0mL= 21.5mL. I think A2 is 7760J (3 sf), or 7.76kJ. Not sure for A3. We are not figuring for calorimeter, just water and ice. For A3- need solution for experimental values, not theoretical values.

Answer 1

A1) 51.5ml - 30.0ml = 21.5ml

A2) your answer is correct for A2

Volume of Warm water = 30.0ml

density of water = 1g/ml

mass of Warm water, m = 30.0g

Temperature difference ,∆T= 1℃ - 62.8℃ = - 61.8℃

Specific heat of water , C = 4.184J/g℃

q = m × ∆T × C

= 30g × (-61.8℃) × 4.184(J/g℃)

= -7757J

= - 7760J

So, heat lost by water is 7760J

A3) mass of ice = 21.5g

heat required to raise the temp of ice melted to 1℃

q= 21.5g × 1℃ × 4.184J/g℃

= 89.96J

Total heat released from warm water = 7757J

Heat used to melt ice = 7757J - 89.96J = 7667J

molar mass of ice = 18g/mol

mass of ice = 21.5g

no of mole of ice = 21.5g/18g = 1.1944

Heat used for 1 mol ice melting = (7667 J/1.1944mol)×1mol = 6419J = 6.42kJ

Therefore,

Molar heat of fusion of ice = 6.42kJ/mol

Heat of fusion of ice in kJ/g = 6.42kJ/18g = 0.357kJ/g