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Question

A 20 g sample of ice at -10 degrees C is mixed with 100 g water at 80 degrees C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O (s) and H2O (l) are 2.03 and 4.18 J/g degrees C, respectively and the enthalpy of fusion for ice is 6.02 kJ/mol.

Answer 1

Step 1:

Total energy removed from the hot water = Total energy required by the ice

For hot water:

QH = m c(Ti - Tf)
   = 100 * 4.18 * (80 – Tf)
   =418(80 – Tf)

Convert the enthalpy units
Hf = 6020( J/mol) ÷ 18 (g/mol)
    = 334.44 J/g

Find the amount of energy that is required to melt the ice.

Qf =mHf  = 20*334.44 = 6688.88 J

Find the amount of energy that is required to increase the temperature of the ice to 0˚.

Q 0˚ = m Hc ∆T = 20 * 2.03 * (0 - (-10)) = 406 J
Find the amount of energy that is required to increase the temperature of the ice from -10 to 0.

Q = 6688.88 + 406 =7094.88 J

Find the total energy required by the ice.

QI = 7094.88 + 20 * 2.03 * (Tf – Ti)
   = 7094.88 + 20 * 2.03 * (Tf – 0)
    =40.6Tf + 7094.88
QH = QI

418(80 – Tf) = 40.6Tf + 7094.88

456.6 Tf = 26345.12

Tf = 57.7˚ C

Solution:

Final temperature is 57.7˚ C.