Two 20.0 g ice cubes at −19.0 ∘C are placed into 215 g of water at 25.0 ∘C.

Question

Two 20.0 g ice cubes at −19.0 ∘C are placed into 215 g of water at 25.0 ∘C. 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 the ice must equal heat lost by the water

Looking at heat gained by ice, we have 3 steps.

1). heat gained going from -19º to 0º = q = mC∆T = (40.0 g x 1 mol/18g)(37.7 J/molº)(19º) = 1592 J

2). heat gained melting the ice @ 0º = q = m∆Hf = (40.0 g x 1 mol/18 g)(6010 J/mol) = 13356 J

3). heat gained going from 0º to final temp = q = mC∆T = (40 g x 1 mol/18g)(75.3 J/molº)(Tf - 0) = 167 Tf

Sum of all the heat gained by the ice = 1592 + 13356 + 167Tf

Looking at heat lost by water, we have

q = mC∆T = (215 g x 1 mol/18g)(75.3 J/molº)(25 - Tf) = 22485 - 899 Tf

Setting heat gained by ice equal to heat lost by water, and solving for Tf, we have...

1592 + 13356 + 167Tf = 22485 - 899 Tf

1066Tf = 7537

Tf = 7.07º

(be sure to check the math)