General Phase Change question

Question

Two 20.0 g ice cubes at −20.0 ∘C−20.0⁢ ∘C are placed into 295 g295 g of water at 25.0 ∘C.25.0⁢ ∘C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, 𝑇f,�f, of the water after all the ice melts.

heat capacity of H2O(s)H2O(s)	37.7 J/(mol⋅K)37.7 J/(mol⋅K)
heat capacity of H2O(l)H2O(l)	75.3 J/(mol⋅K)75.3 J/(mol⋅K)
enthalpy of fusion of H2OH2O	6.01 kJ/mol

Answer 1

Conservation of energy states that heat GAINED by ice must equal heat LOST by warm water.

Equation for heat gained or lost is q = mC∆T

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

When there is no change in temperature (phase change), then q = ∆H_fusion x mass

Since we are given masses in grams, but thermodynamic constants are given in per mole, we must change one of them to be consistent. We will change grams to moles:

grams of ice = 20 g + 20 g = 40 g x 1 mol / 18 g = 2.22 moles ice

grams H2O = 295 g x 1 mol H2O / 18 g = 16.4 moles H2O

Heat gained by ice:

heat to raise temp from -20º to 0º = mC∆T = (2.22 mol)(37.7 J/molº)(20º) = 1674 J

heat to melt ice @ 0º = ∆Hf x m = (6.01 kJ/mol)(2.22 mol) = 13.3 kJ = 13,300 J

heat to raise temp from 0º to final temp = mC∆T = (2.22 mol)(75.3 J/molº)(Tf - 0º) = 167Tf - 0 = 167 JTf

TOTAL heat gained by ice = 1674 J + 13,300 J + 167 JTf

Heat lost by water:

q = mC∆T

q = (16.4 mols)(75.3 J / molº)(25º - Tf) = 30,873 J - 1235 JTf

Heat gained by ice must equal heat lost by water:

1674 J + 13,300 J + 167 JTf = 30,873 J - 1235 JTf

14974 J + 167 JTf = 30,873 J - 1235 JTf

Solving for Tf (final temperature):

1402 JTf = 15,889 J

Tf = 11.3º

Be sure to check all of the math.

Answer 2

The ice must gain heat to go from -20 C to 0 C, then gain more heat to melt at 0 C (fusion)k, then gain heat to the final temperature Tf. The water at 25 C would lose heat to the ice system until it reaches T.

The equation is ΔHice = - ΔHwater. Substitute known values and solve for Tf.

ΔHice = 37.7 J/moleK x 40g /18 g/mole x (0 - -20)C + 6.01 kj/mole x 1000 J/kj x 40g /18 g/mole

+ 75.3 J/moleK x 40g / 18g/mole x (Tf – 0)C.

ΔHwater = -75.3 j/moleK x 295g / 18g/mole x (Tf – 25). Equate the two ΔH equations and solve for Tf. I got 1.90 C as a result. I rounded as I went along, so you may get a slightly different result.