- Level: International Baccalaureate
- Subject: Chemistry
- Word count: 1406
Lab to determine The Specific Heat of a Metal
Extracts from this document...
Introduction
The Specific Heat of a Metal Data Collection and Processing Raw Data Table Initial Raw Data Volume of Distilled Water (mL) Temperature of Distilled Water (C�) Mass of Copper (g) Temperature of Copper (C�) Final Temperature of Distilled Water (C�) Trial 1 100.0 + .5 24.2 + .5 43.027 + .001 95.1 + .5 28.9 + .5 Trial 2 100.0 + .5 24.2 + .5 43.078 + .001 94.8 + .5 28.9 + .5 Trial 3 100.0 + .5 24.2 + .5 42.560 + .001 94.1 + .5 28.2 + .5 Some of the qualitative data observed in this experiment were: > The duration of time to stir, how hard to stir and the overall amount of time to stir were not easy to keep the same in all of the three trials. > It was hard to collect the temperature without touching the copper when deducing the final temperature. > Heat was severely lost in the transfer of the copper from the boiling water to distilled water which flawed the data. > Copper mass varied after each trial do to water partials left over from the boiling water soaking and the distilled water soaking. Data Processing Overview The data table helps delineate a unified order of numbers that correspond to the experiment which is used for calculations all calculations are portrayed in analyzed steps which also includes statements to explain how the calculations were conducted. ...read more.
Middle
-1964.6 + 21.8% J = (-2848.3874 + 1.5023% g? C�) (c) c = .690 + 23.3023% g? C� (not correct sig figs) c = .690 + 20% g? C� (correct sig figs) The formula q = mc?t was used here to find the specific heat of copper. The qcopper is just the inverse of the qH2O so it was -1964.6 + 21.8% J. Then the mass of copper was plugged in and multiplied times the change in temperature (final temperature - initial temperature). The product of these two things were then divided from the qcopper to get the specific heat, "c", of copper. Percent error (| Literature value - experimental value| / Literature Value) ? 100 (|.385 J/ g ? C� - .690 J/ g ? C�| / .385 J/ g ? C�) ? 100 = 79.2% To find the percent error, it is the absolute value of the literature value (.385 J/ g ? C�) - the experimental value (.690 J/ g ? C�). Then divide that by the literature value (.385 J/ g ? C�) and multiply that by 100 to change it into a percent. Overall Experiment: Average of Specific Heat in All Three Trials .690 + 20% J/ g ? C� .692 + 20% J/ g ? C� + .596 + 30% J/ g ? C� .659 + 23.33% J/ g ? ...read more.
Conclusion
The design the transferring was horrible because it allowed massive heat loss which flawed the outcome and accuracy of the data. When stirring the distilled water the concept of having the consistency during the entire experiment was not present because it was not measured which caused error in the data. Since the percent error was so substantially high for this experiment (2344) this lead to not only systematic error but also random error as well. 100mL of distilled water was to be measured by the graduated cylinder which has been shown not to be the most accurate way to measure a substance. Time was not properly collaborated when timing the duration of how long the copper bathed inside the boiling water. It was not measured to seconds. Also the stirring of the distilled water within the cup with copper was not timed. The digital scale already displayed an uncertainty which varied the data. The specific heat of copper was not accurately derived because the transfer of copper lost heat. Suggestions for Improvement Suggestions of improvement include the use of a different piece of equipment to mea use out the volume since the graduated cylinders have an uncertainty of + .5 which was rather unreliable. Another suggestion of improvement is to calculate the duration of time the copper is to ay within the boiling water down to a second. Be sure to stay consistent between trials. Lastly, the transfer of copper from coiling water to distilled water could be measured in ten second intervals to allow all trials the same amount of heat to escape. ...read more.
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