Two 20.0 g ice cubes at −20.0 ∘C are placed into 215 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) h

eat capacity of H2O(l):75.3 J/(mol⋅K)

enthalpy of fusion of H2O:6.01 kJ/mol

Answer 1

For all the ice to melt, it will go through several stages (steps).

(1) raise temperature of ice from -20.0º to 0º (q = mC∆T)

(2) melt the ice at 0º (phase change; q = m∆H_fusion)

(3) raise temperature of the liquid formed from 0º to the final temperature (q = mC∆T)

q = heat

m = mass = 20.0 g x 2 x 1 mol H2O / 18 g = 2.22 moles H2O (ice)

∆T = change in temperature

(1) q = (2.22 mol)(37.7 J / molº)(20º) = 1676 J

(2) q = (2.22 mol)(6.01 kJ/mol) = 13.34 kJ = 13340 J

(3) q = (2.22 mol)(75.3 J / molº)(Tf - 0) where Tf is the final temperature

The sum of all of these will equal the heat gained by the ice and will be equal to the heat lost by the 25º water

which is q = mC∆T

Heat lost by the 25º water is ...

q = (215 g x 1 mol/18g)(75.3 J/molº)(25º - Tf)

Setting heat gained by the ice = heat lost by the warm water, we have ...

1676 J + 13340 J + (2.22 mol)(75.3 J / molº)(Tf - 0) = (215 g x 1 mol/18g)(75.3 J/molº)(25º - Tf)

15016 + 167Tf = 22485 - 899Tf

1066Tf = 7469

Tf = 7.00ºC = final temperature