Two 20.0 g ice cubes at −10.0 ∘C are placed into 265 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

Question

heat capacity of H2O(s)37.7 J/(mol⋅K)

heat capacity of H2O(l)75.3 J/(mol⋅K)

enthalpy of fusion of H2O6.01 kJ/mol

Answer 1

Heat gained by the ice must equal heat lost by water.

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperature

heat gained by ice to take it from -10º to 0º = q = mC∆T

q = ?

m = 2 x 20.0 g = 40.0 g x 1 mol / 18 g = 2.22 moles of ice(H2O)

C = 37.7 J/mol-degree (K or C doesn't matter because we will be using a CHANGE in temp)

∆T = 10º (to go from -10 to 10º)

q = (2.22)(37.7)(10) = 838 J

heat gained by ice to melt it @ 0º = q = m∆Hfusion

q = ?

m = 2.22 mol

∆Hfusion = 6.01 kJ/mol = 6010 J/mol

q = (2.22 mol)(6010 J/mol) = 13,342 J

Total heat gained by the ice = 838 J + 13,342 J = 14,180 J

This now can be set equal to the heat lost by the water

Heat lost by water = q = mC∆T

q = 14,180 J

m = 265 g x 1 mol / 18 g = 14.7 mols

C = 75.3 J/mol-degree

∆T = ?

14,180 J = (14.7 mol)(75.3 J/mol-deg)(∆T)

∆T = 12.8 degrees

Final temperature of the water = 25º - 12.8º = 12.2ºC