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The Latent Heat of Fusion

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  1. To determine the latent heat of fusion of ice.


  1. Ice;
  2. Water;
  3. Thermometer;
  4. Calorimeter;
  5. Digital scale.

This is the table which I filled during my determination:



m of the calorimeter/g; ±0.2g

Initial t of water/°C; ±0.5 °C

t of ice/°C;

Final t of water/°C; ±0.5 °C

m calorimeter + m water/g; ±0.2g

m calorimeter + m water + m ice/g; ±0.2g

m of the water/g; ±0.4g

m of the ice/g; ±0.4g

Latent heat of fusion of ice/ J*kg-1; ± 3x104J*kg-1









2.77 x 105

Recording raw data:

First of all, I prepare my working place and start my determination. All my measurements are recorded to the table above.

 The smallest graduation of the thermometer is 1 °C. According to this, I take the absolute uncertainty of my temperature measurements as ±0.5 °C. I do not add additional uncertainty as I did not encounter any further difficulties in weight measurement.


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After having determined the raw data, I fill it into the table.

Further I provide constants that will be used during my determination. These are taken from Giancoli Physics 5th edition, page 421:

cwater = 4186 J*(kg*°C)-1

caliuminium = 900 J*(kg*°C)-1

Data processing:

Now I will provide the mass of the water. To do this I will subtract the mass of the calorimeter from the mass of calorimeter and water together:

mwater =  (163.3 ±0.2) – (35.0 ±0.2) = (128.3 ±0.4) g

To determine the mass of the ice I will subtract the mass of water and calorimeter from the mass of calorimeter, water and ice together:

mice =  (181.6 ±0.2) – (163.3 ±0.2) = (18.3 ±0.4) g

I fill the table with my findings.

Now I theoretically assume that there are no temperature losses while the water gets colder to the final temperature. I assume that Qlost = Qgained.

Now to determine the latent heat of fusion of ice I will fill that formula:

caliuminium*mcalorimeter*(tinitial – tfinal) + cwater*mwater

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Furthermore, some systematic errors have occurred as I had to do a lot of calculations and roundings during the data processing part. Also, the instruments may have been badly calibrated and this could have affected my determination. However, systematic errors are not important because even if they even were encountered, they were very small. Another thing is with random errors and heat loss as they were really significant because the percentage uncertainty shows relevantly high result.

I could provide several suggestions to improve the determination but obviously I am not able to make this determination not theoretically in school’s conditions. First of all, I would rather use more ice, bigger calorimeter and more water. Then, as I would still use the same equipment with same absolute uncertainties, the percentage uncertainty would be reduced significantly. The uncertainty would be less important and more accurate results would come. However, my suggestions would only lesser the uncertainties, but they would not totally cancel them and it would still be only a theoretical determination.

...read more.

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