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1.10g 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 m |
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Answer» 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 after solving Tf = 306 K |
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