There were no independent variables in the experiment as the parameters were not varied.
DEPENDENT VARIABLE:
The dependent variables were: the steady, final temperature measured after the hot water was added to the cold ice, the temperature of ice, the mass of water used and the mass of the ice used.
CONTROLLED VARIABLES:
- Room temperature and pressure remained constant.
- The purity of ice remained the same.
MATERIALS AND METHOD
Apparatus Used:
- Wire gauze
- 1 Tripod stand
- 1 Thermometer
- 1 Insulated Calorimeter with stirrer
- 1 250 ml beaker
- Electronic Balance
- Block of the unknown metal
- Water
- Cotton thread
DIAGRAM:
PROCEDURE:
- First, I measured and recorded the mass of an empty beaker and the calorimeter with the stirrer and noted down the readings.
- Then I filled the beaker with water up to a certain level and measured the mass of it with the water.
- Then I calculated the difference between the masses ((beaker + water) – empty beaker), which gave me the mass of the water used.
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Then I heated the water in the beaker to a certain temperature, and recorded the temperature as T1.
- Then I filled the calorimeter with some ice, and measured the new mass. Then I calculated the mass of the ice added by ((mass of calorimeter + ice) – (mass of calorimeter)).
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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.
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Then I left the ice and water in the calorimeter, until both the substances inside the calorimeter reached a constant temperature which was the final temperature of both the ice and the water. I measured and noted down this final temperature as T3.
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%) Joules ] = [ (19.782 ± 1.98) Joules ]
= [ (19. 782 ± 1. 98) Joules ] is the amount of heat gained by ice.
miLf = [ 0.00942 * Lf (Latent Heat of Fusion) ]
= [ 0. 00942 Lf ] is the amount of energy used when ice at 0oC changed to water
at 0oC.
mwcw ∆Tw = [ 0.00942 * 4200 * ((0 ± 0.05) – (30 ± 0.05)) ]
= [ 0.00942 * 4200 * (30 ± 0.1) ] = [ 0.00942 * 4200 * (30 ± 0.33%) ]
= [ (1186.92 ± 0. 33%) Joules ] = [ (1186.92 ± 3.92) Joules ]
= [ (1186. 92 ± 3. 92) Joules ] is the amount of heat gained by water
[which melted from ice].
mccc ∆Tc = [ 0.0644 * 385 * ((-1 ± 0.05) – (30 ± 0.05)) ]
= [ 0.0644 * 385 * (31 ± 0.1) ] = [ 0.0644 * 385 * (31 ± 0.32%) ]
= [ (768.614 ± 0.32%) Joules ] = [ (768.614 ± 2.46) Joules ]
= [ (768. 614 ± 2. 46) Joules ] is the amount of heat gained by the copper
calorimeter.
Therefore, using the above calculations, we can now calculate the specific latent heat of fusion of ice:
[ (5117. 28 ± 25.6) ] = [(19.782 ± 1.98) ] + [ 0. 00942 Lf ] + [ (1186.92 ± 3.92) ]
+ [ (768.614 ± 2.46) ]
[ (5117. 28 ± 25.6) ] = [ (1975. 32 ± 8.4) ] + [ 0. 00942 Lf ]
[ 0. 00942 Lf ] = [ (5117. 28 ± 25.6) ] – [ (1975. 32 ± 8.4) ]
[ 0. 00942 Lf ] = [ (3141.96 ± 17.2) ]
Lf = [(3141. 96 ± 17.2) ] = [ (3141. 96 ± 0.55%) ]
[ 0. 00942 ] [ 0. 00942 ]
Lf = [ (333541.4 ± 0.55%) ] ≈ [ (334,000 ± 0.55%) ] JKg-1
Lf = [ (334,000 ± 0.55%) ] JKg-1 = [ (334,000 ± 1837) ] JKg-1
Lf = [ (334, 000 ± 1837) ] JKg-1
Conclusion:
After conducting the above mentioned experiment and after carrying out all the calculations outlined above and in the previous pages, I hereby conclude that the specific latent heat of fusion of ice is (334, 000 ± 1837) JKg-1. Therefore, the specific latent heat of fusion of ice is 334, 000 JKg-1 with an error of 0. 55%.
This means that the range of the specific latent heat of fusion of ice is from (334,000 - 1837) to (334,000 + 1837). Hence, the range of the specific latent heat of fusion of ice from the readings gathered and calculations made in this experiment is 332, 163 JKg-1 to 335, 837 JKg-1.
According to the calculations made, I can also conclude that the hypothesis I made earlier is correct as the predicted value of the specific latent heat of fusion of ice was 334,000 JKg-1, and this value lies between the ranges calculated from the readings and results of the experiment.
Evaluation:
The method and materials used for this experiment was pretty good, but because an error has been experienced, some improvements could be made for a more accurate and correct result. The following are some of the errors which were experienced while conducting the experiment and improvements which could be made to overcome the errors:
- 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.
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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.
