Question

Two 20.0-g ice cubes at -14.0 degree C are placed into 295 g of water at 25.0 degree C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Answer 1

NOTE: We consider rise or fall in temperature. So X ^oC rise or fall = X kelvin rise or fall

Two 20.0 g ice = 40g

Molar mass of water = 18 g/mol

So, 18 g of water = 1 mol

40 g of water = (40/22) mol = 2.22 mol

Heat required to heat ice from -14 ^oC to 0 ^oC

= Specific heat(H2O solid) * Mass * Temp rise

= (37.7 J mol^-1 K^-1) * (2.22 mol) * (14 K)

= 1171.72 J

Heat required to melt ice at 0 ^oC to water at 0 ^oC

= Mass * Enthalpy of fusion

= (2.22 mol) * (6.01 kJ mol^-1)

= (2.22 mol) * (6010 J mol^-1)

= 13342.2 J

Call final temperature X.

Heat required to heat water at 0 ^oC to X ^oC

= Mass * specific heat H2O (liquid) * Temp rise

= (2.22 mol) * (75.3 J mol^-1 K^-1) * (X K)

= 167.2X J

Total heat gained by ice = 1171.72 J + 13342.2 J + 167.2X

                                       = 14513.92 + 167.2X

Total heat lost by water = Mass * Specific heat * Temp Fall

                                       = (295/18 mol) * (75.3 J mol^-1 K^-1) * (25 - x)

                                       = (16.39 mol) * (75.3 J mol^-1 K^-1) * (25 - x)

                                       = 1234.2(25 - X)

                                       = 30855 - 1234.2X

Total heat gained by ice = Total heat lost by water

14513.92 + 167.2X = 30855 - 1234.2X

167.2X + 1234.2X = 30855 - 14513.92

1401.4X = 16341.08

X = 11.66