Lab: Heat fusion of ice

mc = mass of the calorimeter (g)

cc = specific heat of the calorimeter (J/g?C)

mw = mass of the warm water (g)

cw = specific heat of water (J/g?C)

?T = T1 - T2 = change in temperature of the calorimeter and the warm water (?C)

?Qgained = miHf + micw?Ti

Where:

?Qgained = heat gained (J)

mi = mass of the ice (g)

Hf = heat of fusion of ice (J/g)

cw = specific heat of water (J/g?C)

?Ti = T2 = difference in temperature between that of the ice and the final temperature of the mixture (?C)

?Qlost = ?Qgained

By substituting values of mc, cc, mw, cw, mi, ?T, and ?Ti,, we can solve for Hf.

Procedure:

. Determine the mass of the empty calorimeter and its specific heat.

2. Fill the calorimeter about half-full with warm water. Obtain the mass of the water and the calorimeter. Take the initial temperature of the warm water (T1).

3. Then immediately add several cubes of ice that have been dried with paper towels.

4. Continue stirring slowly until all ice has melted and the mixture reaches a steady temperature. Record the temperature of the ice-water mixture (T2).

5. Obtain the mass of the calorimeter and the mixture (mc + w + i).

6. Use these data to calculate the heat of fusion of ice.

Data Collection:

Trial 1

Trial 2

Mass of Calorimeter mc (g)

83.5

83.5

Specific Heat of Calorimeter cc (J/g?C)

0.9

0.9

Mass of Calorimeter and Warm Water mc + w (g)

290.5

302

Mass of Warm Water mw (g)

mw = mc + w - mc

07

18.5

Specific Heat of Water cw (J/g?C)

4.19

4.19

Initial Temperature of Calorimeter and Warm Water T1 (?C)

70

73.5

Final Temperature of Calorimeter and Warm Water T2 (?C)

37

43

Change in Temperature ?T = T1 - T2 (?C)

33

30.5

Heat Lost ?Qlost (J)

20245

20181

Mass of Calorimeter and Mixture mc + w + i (g)

323.7

335.4

Mass of ice mi (g)

mi = mc + w + i - mc + w

33.2

33.4

Change in Temperature of water formed from ice

?Ti = T2 (?C)

37

43

Calculations:

Heat Lost

?Qlost = mccc?T + mwcw?T

First Trial ?Qlost = (183.5*0.9*33) + (107*4.19*33) = 20245 J

Second Trial ?Qlost = (183.5*0.9*30.5) + (118.5*4.19*30.5) = 20181 J

Heat of Fusion

?Qlost = ?Qgained

mccc?T + mwcw?T = miHf + micw?Ti

Hf = ?Qlost - micw?Ti

mi

First Trial Hf = (20245 - 33.2*4.19*37) / 33.2 = 454 J/g

Second Trial Hf = (20181 - 33.4*4.19*43) / 33.4 = 424 J/g

Average Value Hf = (454 + 424) / 2 = 439 J/g

Error Analysis:

The accepted value for the heat of fusion of ice is 334 J/g.

Percent Error = (439 - 334) x 100 / 334

= 31.4 %

Sources of error: Some heat is lost to the environment because the calorimeter is not a perfect insulator.

Conclusion:

According to this experiment, the heat of fusion of ice is 439 J/g. However, the accepted value for the heat of fusion of ice is 334 J/g. There is obviously some error in this experiment.

Future Improvements:

To limit the heat lost to the environment, better apparatus should be used (i.e. better insulating materials for the calorimeter). Using larger quantities of ice and water and conducting the experiment repeatedly would eventually cause errors to have a smaller effect on the result, such that the result of the experiment approaches the accepted value.

Diana Herwono D 0861 006

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