Heat of fusion of ice lab answer key

Purpose: To estimate the heat of fusion of ice.

Materials:

·         Ice

·         Foam cup

·         100-mL graduated cylinder

·         Thermometer

·         Hot water

·         Temperature probe (optional)

Procedure

1.       Use a graduated cylinder to measure 70 mL of hot water

2.       Pour the water into the foam cup.

3.       Measure the temperature of the water

4.       Add an ice cube to the water.

5.       Measure the temperature of the water when ice is completely melted.

6.       Pour the water into the graduated cylinder and measure the volume.

Data and Observations

Analysis

1. Calculate the mass of the ice.

The moles of NaCl contained in one level teaspoon are calculated by this equation:

12 g H2O     | 1 mol    0.66 mol H2O

         | 18 g

2. Calculate ΔHfus of ice (kJ/mol)

Q = mcΔT

Q = (70 g)(4.18 J)(13.3°C)

           g •°C

Q = 3891.58 J

3891.58 J = 5896.33 J/mol

 0.66 mol

3. Compare experimental value of ΔHfus of ice with the accepted value

| 6.01 – 5.9 | = .018

     6.01

1.8 % error

Error Analysis

The data was not comparable to the accepted value mostly due to improper insulation.  To improve the results, the Styrofoam cups could be doubled, and the tops can be covered to prevent heat loss from the hot water, or heat from the palm of the hand to transfer through the foam cup.  The experiment could be performed more quickly with less interruption to prevent heat loss.

Conclusion

The heat fusion of ice is 6.01 J/mol, but the experimental value yielded 5.9 J/mol.  The experiment shows that the heat fusion of ice is equal to the amount of heat needed to melt ice into water.

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Extracts from this document...

Physics HL Lab Report:

Experiment to Measure the Heat of Fusion of Ice

Title:         Experiment to Measure the Heat of Fusion of Ice

Aim:

The aim of the experiment is to measure the heat of fusion of ice. This is the amount of heat energy needed to change one gram of ice at its melting point (0°C) into one gram of water at the same temperature.

Apparatus:

Calorimeter, balance, set of metric basses, thermometer, warm water, ice, paper towels

Introduction:

When a substance changes state, it absorbs or releases a large quantity of heat. During the change of state, the temperature remains constant. The quantity of heat required to change a

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

1. 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:

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 = (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.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

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