Two 20.0 g ice cubes at -20.0 C are placed into 225 g of water at 25.0 C

Question

Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf of the water after all the Ice melts.

heat capacity of H2O(s) = 37.7 J/ mol x K

heat capacity of H2O (l) = 75.3 J/ mol x K

enthalpy of fusion of H2O = 6.01 kJ/ mol

Answer 1

Heat gained by ice cubes = heat lost by water

Heat gained by ice cubes is the heat to raise them from -20º to 0º + heat to melt them + heat to equilibrate with the 225 of water

Since the constants are given on a mole basis, we'll convert grams to moles

20.0 g x 2 = 40 g x 1 mol/18 g = 2.22 moles of ice @-20º

265 g x 1 mol/18 g = 14.7 moles of water @ 25º

(2.22 mol)(37.7 J/molº)(20º) + (2.22 mol)(6010 J/mol) + (2.22 mol)(75.3 J/molº)(Tf-0º) = (14.7 mol)(75.4 J/molº)(25º-Tf)

1674 + 13,342 + 167Tf = 27710 -1108Tf

15016 + 167Tf = 27710 - 1108Tf

1275Tf = 12,694

Tf = 9.96º