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.

Answer 1

Since all the constants are given on a per mole basis, we will first convert the mass of H2O to moles.

2 x 20 g = 40 g x 1 mol/18 g= 2.22 moles H2O

Next, since the amount of heat lost by the warm (25º) must equal the heat gained by the ice, we can write:

heat gained by ice = heat lost by water

To find the heat gained by ice, we need several steps. First, it must go from -11º to 0º. Then @0º, it must melt. Then once melted it must equilibrate in temp with the warm water.

heat to raise temp from -11 to 0 = (2.22 mol)(37.7 J/molº)(11º) = 920.6 J = 0.921 kJ

heat to melt the ice @ 0º = (2.22 mol)(6.01 kJ/mol) = 13.34 kJ

Total heat to this point = 13.34 kJ + 0.921 kJ = 14.26 kJ

At this point we essentially have 40 g (2.22 mol) H2O @ 0º mixing with 265 g (14.7 mol) of H2O at 25º

14.26 kJ + (2.2 mol) (Tf - 0) = (14.7 mol)(75.3 J/molº)(25 - Tf)

14.26 kJ + 2.2Tf = 27.7 kJ - 1.11Tf

3.3Tf = 13.44

Tf = 4.07º

Of course you should check my math as I'm prone to errors when making this many conversions. But I think the idea and approach are sound.