The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol ⋅ K. The conversion of 50.0 g of ice at 0.00 °C to liquid water at 15.0°C requires ________ kJ of heat.

Question

The heat of the fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol ⋅ K. The conversion of 50.0 g of ice at 0.00 °C to liquid water at 15.0°C requires ________ kJ of heat.

Answer 1

You are given values with units of kJ/mol and J/mol, so be sure to use the same units for both. Also, since the thermodynamic constants are on a per mol basis, we need to convert 50.0 g H₂O to moles.

50.0 g H₂O x 1 mol / 18 g = 2.78 mols

To melt 2.78 mols ice at 0º (phase change:

(mass)(∆Hfus)

2.78 mols x 6.01 kJ/mol = 16.71 kJ

To raise temp of liquid from 0º to 15º:

(mass)(specific heat)(∆T)

2.78 mols x 75.3 J / molº x 15º = 3140 J = 3.14 kJ

Total heat needed = 16.71 kJ + 3.14 kJ = 19.85 kJ = 19.9 kJ (3 sig. figs.)