Question

Two 20.0-g ice cubes at –21.0 °C are placed into 295 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

heat capacity of H2O(s) 37.7

heat capacity of H2O (l) 75.3

enthalpy of fusion of H2O 6.01

Answer 1

Mass of two 20-g ice cubes = 2 x 20g => 40gm ice

Moles of ice = 40g/18.0 g/mol = 2.22 mol ice

Mass of water = 295 g

Moles of water = 295g/18.0g/mol = 16.4 mol water;

Heat capacity of H20(solid) = 37.7 J/(mol K)
Heat capacity of H2O (liquid) = 75.3 J/(mol K)

The heat needed for a 1 K change in temp is the same as for a 1 °C change so the unit J/mol*K will be the same for J/mol*°C

Enthalpy of fusion of H2O = 6.01 kJ/mol or 6010 J/mol

Use ΔH fusion to calculate the amount of heat needed to melt 40 g of ice at -21 ^oC to 40 gm of liquid at 0 ^oC.

By using equation, q = n C ΔT

Where ΔT = T_final – T _initial

Heat absorbed by ice = (2.22 mol*21°C*37.7 J/mol°C) +(2.22 mol*6010 J/mol) + [2.22 mol*(Tf - 0 °C)*75.3 J/mol °C]

Heat absorbed by ice cube = 1759.33 + 13355.56 + 167.33 Tf

   Heat absorbed by ice cube = 15114.89 + 167.33Tf

Heat released by water = 16.4 mol (Tf - 25 °C) x 75.3J/mol*°C
                                     = 1234.92 Tf – 30873

Heat absorbed + Heat released = 0

15114.89 + 167.33Tf +1234.92 Tf – 30873 = 0

-15758.11 + 1402.25 Tf = 0

      1402.25 Tf = 15758.11

                     Tf = 11.24 ^oC

Thus final temperature of water after melting ice is 11.24 ^oC