Two 20.0 g ice cubes at −11.0 ∘C are placed into 265 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⋅K)

heat capacity of H2O(l) 75.3 J/(mol⋅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 to 0º + heat to melt them + heat to equilibrate with 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

265 g x 1 mol/18 g = 14.7 moles

(2.22 mol)(37.7 J/molº)(11º) + (2.22 mol)(6010 J/mol) + (2.22 mol)(75.3 J/molº)(Tf-0º) = (14.7 mol)(75.4 J/molº)(25º-Tf)

921 + 13,342 + 167Tf = 27710 -1108Tf

14263 + 167Tf = 27710 - 1108Tf

1275Tf = 13447

Tf = 10.5ºC