Heat of Fusion Lab
Autor: Mikki • September 13, 2017 • 655 Words (3 Pages) • 932 Views
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- My calculations represent the Heat of Fusion because at the beginning, the mass of the hot water in the calorimeter was 94 g and by having that, you have the mass portion of the heat equation. After that, you have to measure the temperature of the hot water, using a thermometer and that was 64oC. Next, me and my lab partner added ice to the hot water and ended with the temperature of 0oC. Now, we can find delta T since we already have our initial and final temperature. Lastly, I used the original joules of water since that is the specific energy. After that, I got the heat amount which is -25170.94 joules. Now that we have the heat, we have to find the mass of the ice water itself by subtracting the mass of hot water and the mass of water after ice (175-94=81). The equation for heat of fusion is q=mHf, and to find heat of fusion itself, you have to isolate Hf by dividing the mass (-25170.94/81=-310.75). Adding on, by looking at the heat of fusion diagram, we started out with water as the liquid, then the temperature of the water decreased, which lowered the temperature of the water to 0 Celsius. When the state of the phase ended in the ice and water phase, the intermolecular force was dominating (Hf).
Rebuttal: Show the calculations for the percent error AND explain how the design or flaws of the calorimeter would affect your results (In terms of Heat loss).
- The purpose of this lab was accomplished since the heat of fusion of ice was found with only a sdfdsd % error. The percent error was low considering the equipment used in this lab. One possible source of error for this lab was the Styrofoam cup that was used as a calorimeter. There was no lid on the cup and heat could have been lost to the surroundings or heat could have entered the cup from the top. Another possible source of error could have been in measuring the volume of water used in the calorimeter. Since the measurements of volume using the graduated cylinder have one estimated number, the estimated number could have been off resulting in an incorrect volume.
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