• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Experiment to Measure the Heat of Fusion of Ice

Extracts from this document...

Introduction

Diana Herwono D 0861 006

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

...read more.

Middle

  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:

Trial 1

Trial 2

Mass of Calorimeter mc (g)

183.5

183.5

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

107

118.5

Specific Heat of Water cw (J/g°C)

4.19

4.19

...read more.

Conclusion

 = (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.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Heat of fusion of ICE

    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.

  2. IB Latent Heat of Fusion of Ice Lab

    + (0.10 � 13.00 � 100) %)) + (180.88 J � ((0.01 � 36.14 � 100) + (0.10 � 13.00 � 100)

  1. Specific latent heat of fusion of ice

    - (mass of calorimeter)). * Then I measured and recorded the temperature of ice as T2. * Then after adding the ice to the calorimeter, immediately I added the hot water to the calorimeter and straight away covered the calorimeter in order to avoid any heat loss to the atmosphere.

  2. IB Specific Heat Capacity Lab

    (2) - (1) (see data table) = 103.75 - 41.35 = 62.40 g � (0.01 + 0.01) = � 0.02 g = (0.02 � 62.40) � 100 = 0.032 % Temperature change of water + Calorimeter (?T) (7) - (5) (see data table)

  1. specific heat of a solid

    12 4 330 15840 12 4 360 17280 12 4 390 18720 12 4 420 20160 12 4 450 21600 12 4 480 23040 Q=E therefore Q= V�I�t Aluminium mass (�0.00001kg) temperature ( � 0.5 k) ?T (�0.5 k) m?T (�0.51kg�k)

  2. The Latent Heat of Fusion

    Some errors were encountered despite the fact that I tried to be as accurate as possible. First of all, the main weakness of the determination was that all the determination was done theoretically and I could not measure how much heat was transferred to air during the water cooling process.

  1. HL Physics Revision Notes

    path of the incoming rays a distance of 1.5x 1011m from the sun. -Varying solar constant -Earth?s elliptical orbit -Tilt of Earth?s axis -Weather -Altitude of the Sun in the sky -Season ?Albedo Hydroelectric Power: Distinguish between different hydroelectric schemes: 1.

  2. IB Specific Heat Lab

    (122.6 ± .2) (33.0 ± .1 – 31.3 ± .1) Q gained = 511.1 ± .3 Cadmium Q lost = 511.1 ± .3 Q lost (511.1 ± .3) = (c) (59.25° C ± .05) (100 – 33.0 ± .1) c of zinc = .130 ± .04 J g-1 K-1 Actual Specific Heat Capacities SUBSTANCE Specific Heat Capacity (J g-1 K-1)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work