You make some iced tea by dropping 161 g of ice into 500.0 mL of warm tea in an

Question

You make some iced tea by dropping 161 g of ice into 500.0 mL of warm tea in an insulated pitcher. If the tea is initially at 19.8°C and the ice cubes are initially at 0.00°C, how many grams of ice will still be present when the contents of the pitcher reach a final temperature? The tea is mostly water, so assume that it has the same density (1.00 g/mL), molar mass, heat capacity (75.3 J K-1 mol-1), and heat of fusion (6.00 kJ/mol) as pure water. The heat capacity of ice is 37.7 J K-1 mol-1. g ice remaining

Explanation / Answer

Heat capacity of the warm tea = 75.3 J K-1 mole-1

= 75.3 / 18 J K-1 gm-1 (since, molar mass of tea = 18 gm)

= 4.184 J K-1 gm-1

Now heat required to cool down 500 gm tea ( 500 mL = 500 gm as density = 1 g/mL) to 0°C

= 500 x 4.184 x (19.8-0) = 41421.6 J

Heat of Fusion = 6 KJ/mol = (6 x 1000) / 18 = 333.3 J/g

Heat required to completely melt 161 g of Ice at 0°C = 161 x 333.3 = 53661.3 J

Hence the ice would not melt completely.

Amount of ice that would melt = 41421.6 / 333.3 = 124.28 g

Amount of ice left in the pitcher = 161 - 124.28 = 36.72 g

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