Question

Suppose that 20.50 g of ice at -10.7°C is placed in 36.35 g of water at 91.0°C in a perfectly insulated vessel. Calculate the final temperature. (The molar heat capacity for ice is 37.5 J K^-1 mol^-1 and that for liquid water is 75.3 J K^-1 mol^-1. The molar enthalpy of fusion for ice is 6.01 kJ/mol. You must answer in Kelvin, not °C.)

Answer 1

Heat required to bring ice to 0°C

Qice = m*C*(Tf-Ti)

Qice = 20.5*2.08*(0 +10.7) = 456.248 J

Heat required to melt ice at 0°C

Qice = m*HL = 20.50 *334 = 6847J

now... calcualte how much temperature will drop:

Qwater = m*C*(Tf-Ti)

-6847 = 36.35*4.184*(Tf-91)

Tf = -6847 /(36.35*4.184) + 91 = 45.980 °C

meaning, we will have water at 0°C and water at 45.980°C, we then have another euqilibrium

Qcold water = m1*C*(Tf-Tfreezing)

Qhot water = m2*C*(Tf-Tactual)

-Qcold = Qhot

m1*C*(Tf-Tfreezing) = m2*C*(Tf-Tactual)

-20.5(Tf-0) = 36.35*(Tf-45.980 )

-20.5Tf = 36.35Tf - 36.35*45.980

(-20.5 - 36.35)*Tf = -1671.373

Tf = 1671.373 / 56.85

Tf = 29.399 °C change to K = 29.399+273 = 302.9 K