The mass of an empty coffee cup calorimeter was 5.25 g. To the calorimeter was a
ID: 476699 • Letter: T
Question
The mass of an empty coffee cup calorimeter was 5.25 g. To the calorimeter was added 75.0 mL of liquid water; the mass of calorimeter and water was 80.03 g. The temperature of the water in the calorimeter was 25.00 Degree C. To the calorimeter was added three ice cubes, each at 0.00 Degree C. After the ice cubes had melted and the temperature of the water was no longer changing, the temperature of the water in the calorimeter was 3.70 Degree C, and the mass of calorimeter and water was 100.03 g. The specific heat of water is 4.18 J/g- Degree C. and the heat capacity of the calorimeter is 15.0 J/Degree C. Calculate the heal of fusion for water on a per mole of water basis. The heat of fusion for water is the energy change at 0.00 Degree C for the following: H_2O (s) rightarrow ?_2? (1).Explanation / Answer
Mass of ice cubes added = 20g
Temperature raised= 3.7
Heat required to temperature of the ice cube = 4.18 ×3.7×20
=309.32J
Heat released by the water = - 75× 4.18×21.3=6677.6 J
Heat realeased by calorimeter =-21.3×15= 319.5J
Total heat released = -6677.6-319.5=-699.71J
Heat consumed for melting =-6997.1 + 309.32 = -6687.8J
Therefore, per mole the heat consumed = (6687.8/20)×18 = 6019J = 6.01KJ
Therefore , heat of fusion of water = 6.01KJ/mol
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