Question

10g of ice at 273K is added to 20 g of water at 90oC in an insulated flask. The heat of fusion of ice is 6 Kj/mol and the specific heat capacity of water is 4.2 J/K/g. ignoring the heat capacity of the flask;
Determine the final temperature of the system (3 marks)
Dtermine ∆S of the system (3 marks)
2. Calcul

Answer 1

heat to melt ice + heat to move ice T to final T + heat to lower T of 90 C water to final T = 0
mols ice = 10/18 = 0.555
(0.555 mols ice x 6010 J/mol) + [10 x 4.2 J/mol*K(Tf-0)] +[20g x 4.2 J/mol*K x (Tf-90)] = 0
Solve for Tf = final temperature. Post your work if you get stuck.

Answer 2

what about ∆s of the system?

Answer 3

Gives a simpler formula for calculating final temperature

Answer 4

Final temperature =306.5

Answer 5

Tf=306.5K