The mass of an empty coffee cup calorimeter was 5.25 g. To the calorimeter was a

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°C. To the calorimeter was added three ice cubes, each at 0.00°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°C, and the mass of calorimeter and water was 100.03 g. The specific heat of water is 4.18 J/g·°C, and the heat capacity of the calorimeter is 15.0 J/°C. Calculate the heat of fusion for water on a per mole of water basis. The heat of fusion for water is the energy change at 0.00°C for the following: H2O(s) H2O(l).

Explanation / Answer

1) The ice does three things:

b) melt at 0
c) heat up from 0 to 3.7

2) The two calculations are:

qa = (20 / 18.015 g/mol) (x kJ/mol)
qb = (20) (3.7 °C) (4.184 J / g °C)

The water and calorimeter do this:

d) water ---> cool down from 25 to 3.7
e) calorimeter ---> cool down from 25 to 3.7

3) The two calculations are:

qc = (75) (16.3 °C) (4.184 J / g °C)
qd = (16.3 °C) (15 J / g °C)

Heat lost by ice cube = heat gained by water + calorimeter constant

qa + qb = qc + qd

1.110 x + 309.62 = 5114.94 + 244.5

X = 4549 kj

Hence heat of fusion is 4.54 kj

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.