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# Specific latent heat of fusion of ice

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Introduction

PHYSICS LAB REPORT

SPECIFIC LATENT HEAT OF FUSION OF ICE

AIM: To determine the specific latent heat of fusion of the ice provided by using a Calorimeter.

GENERAL BACKGROUND:

Latent heat of fusion:

 If you have a glass of a cool drink, well supplied with ice, you can expect its temperature to drop until it is close to 0 ºC. You also can expect (and can easily check with a thermometer) that it will remain cold, regardless of the outside temperature, as long as there remains some unmelted ice in the drink. Only after all the ice has melted will the temperature of the drink begin to rise. Why is this? When a solid substance changes from the solid phase to the liquid phase, energy must be supplied in order to overcome the molecular attractions between the constituent particles of the solid. This energy must be supplied externally, normally as heat, and does not bring about a change in temperature. We call this energy latent heat (the word "latent" means "invisible"). The latent heat is the energy released or absorbed during a change of state. With this in mind, we define the specific latent heat of fusion:

Middle

3.

RESULTS

Data Collection:

Mass of empty beaker = 92.37 g = 0. 09237 kg

Mass of beaker and water = 153.29 g = 0.15329 kg

Mass of water = 0.15329 – 0.09237 = 0. 06092 kg

Mass of empty calorimeter and stirrer = 64.4 g = 0. 0644 kg

Mass of calorimeter and stirrer and ice = 73.82 g = 0. 07382 kg

Mass of ice = 0.07382 – 0.0644 = 0. 00942 kg

Temperature of hot water – T1 = (50 ± 0. 05) oC

Temperature of ice – T2 = (-1 ± 0. 05) oC

Final Temperature – T3 = (30 ± 0. 05) oC

Specific Heat Capacity of water = 4200 JKg-1K-1

Specific Heat Capacity of copper (Copper Calorimeter) = 385 JKg-1K-1

Specific Heat Capacity of ice = 2100 JKg-1K-

Data Processing and Presentation:

• Heat lost by hot water = Heat gained by ice. Therefore:

Flow Charts:

Water → Water                                Ice → Ice → Water → Water

50oC         30oC                                -1oC    0oC      0oC         30oC

Total Heat lost by hot water = Total Heat gained by ice + Total Heat gained by

the copper calorimeter

mwcw ∆Tw = mici ∆Ti + miLf + mwcw ∆Tw + mccc ∆Tc

mwcw ∆Tw = [ 0.06092 * 4200 * ((50 ± 0.05) – (30 ± 0.05)) ]

= [ 0.06092 * 4200 * (20 ± 0.1) ] = [ 0.06092 * 4200 * (20 ± 0.5%) ]

= [ (5117. 28 ± 0. 5%) Joules ] = [ (5117. 28 ± 25.6) Joules ]

= [ (5117. 28 ± 25. 6) Joules ] is the amount of heat lost by the hot water.

mici ∆Ti = [ 0.00942 * 2100 * ((-1 ± 0.05) – (0 ± 0.05)) ]

= [ 0.00942 * 2100 * (1 ± 0.1) ] = [ 0.00942 * 2100 * (1 ± 10%) ]

= [ (19. 782 ± 10%)

Conclusion

• The experiment could have been repeated and conducted more than once, in order to obtain average readings which could lead to more appropriate results and more accurate values and it would help to reduce random errors which affected the results obtained.
• There might have been significant heat loss to the surrounding while transferring the hot water from the beaker into the calorimeter and this heat loss might have affected the values and results.
• There might have been significant heat gain from the surrounding while measuring the mass of the ice in the calorimeter and this heat gain might have affected the values and results.
• The use of an electric balance to measure the masses of the substances used in the experiment helped to avoid the uncertainties and errors related to the mass values.
• The main source of error in this experiment was due to the limitation in the choice of apparatus used, the use of better and more accurate equipment would lead to more accurate results (results without errors). For example, rather than using a normal laboratory analogue thermometer which has an error of ± 0.05 oC, it is better to use a digital thermometer which is more exact leading to more accurate temperature readings and hence, results.

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

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