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Physics
specific heat1
specific heat1 The purpose of this laboratory experimentation is to calculate “c” and the specific heat of the metal slugs. (5) Slugs: tin, aluminum, copper, lead, and an unknown. 1. Construct a table or a chart in which data could be properly labeled and read before conducting any measurements. 2. Wash the tin cup, which is found in the calorimeter. 3. Dry the cup and locate the rod. Using a balance, measure the weight in grams of the tin cup and the rod. 4. Fill the tin cup with water and weigh the tin cup with the water. 5. Find the weight of the water by subtracting the weight of the tin cup. Denote the results in the proper location in a chart or table constructed. 6. Place tin cup, filled with water, onto the calorimeter and seal the tope with a wooden top. Two holes should be on the top, one for the stirring rod and the other for the thermometer. 7. Measure the initial temperature of the water and record the result in the proper location in a chart or table constructed. 8. The instructor should have the slugs in a hot medium ready for experimentation. Assign a member in your team to obtain a slug from the instructor. 9. The instructor should carefully set the slug in your tin cup. 10. Carefully place the tin cup in the calorimeter and seal it with the wooden top. 11. Stir the water with the rod and measure the initial temperature. Denote the results. 12. After the final temperature has been taken, carefully take out the slug from the tin cup and dry it with a towel or paper. 13. Weigh the experimented slug. Denote the result. 14. Continue steps 8 – 13 including the unknown. 15. Once the data is carefully calculated, place all laboratory materials in their corresponding locations. mslug Cslug (Tinitial-slug – Tfinal) = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) Cslug = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) mslug Ctin = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) mtin Ctin = (50 903 + 11 903 + 200 4180)(24 – 21) / (170 – 24) (65) Ctin = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) mlead Ctin = (50 903 + 11 903 + 200 4180)(26 – 24) / (160 – 26) (58.5) %-error = |Cactual – Cexperimental| / Cacutal = |130 – 113.7| / 130 = 12.6% Ctin = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) mAl Ctin = (50 903 + 11 903 + 200 4180)(33 – 26) / (155 – 33) (65) %-error = |Cactual – Cexperimental| / Cacutal = |903 – 786.7| / 903 = 13.0% Ctin = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) mCu Ctin = (50 903 + 11 903 + 200 4180)(36.5 – 33) / (160 – 36.5) (60) %-error = |Cactual – Cexperimental| / Cacutal = |385 – 420.9| / 385 = 9.3% C?? = (mcup CAl + mrod CAl + mh20 Ch20) (Tfinal – Tinitial-h20) / (Tinitial-slug – Tfinal) m?? Ctin = (50 903 + 11 903 + 200 4180)(39 – 36.5) / (155 – 39) (79.5) In this examination of physical reactions, observing, presenting, predicting, identifying, and recognizing were the essential elements in this laboratory experimentation. By correctly performing the correct procedures, it was relevant that many repercussions occurred. The great difference between the two calculations for the heat capacity of the calorimeter is evidence of an experimental problem. This problem probably lies in the fact that the hot water placed in the calorimeter was not de-ionized. This would mean that the specific heat of water, which was regular tap water, might also yield an inconsistent amount of additives per sample. This would also change the calculated heat capacity. The specific heat that was calculated for the aluminum was consistent. Both answers were within a tenth of a Joule per gram degree Celsius. The fact that an experiment that did not use the specific heat of water so any times, though wrong, and that was consistent, leads me to believe that the water was the source of error. Bibliography:
Word Count: 942
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