3 In our calorimetry experiment, we assumed our calorimeter was a perfectly clos
ID: 304037 • Letter: 3
Question
3 In our calorimetry experiment, we assumed our calorimeter was a perfectly closed system (no heat was lost to the cup). We could have corrected our results with a calibration procedure. a. If we found that 24.39 g of water at 99.7 °C when added to 35.46 g of water at 23.6 °C resulting in a final temp of 40.3 °C, what would be the heat capacity of the cup in J/°C? b. Using the same cup, if we determined that 4.29 g of an unknown metal at 100.1 °C raised 20.00 g of water from 22.9 °C to 25.0 °C what would be the specific heat of the unknown metal with correction for the cup in J/g°C? *Or without correction if your cannot solve a. c. What do you think the identity of the metal is? What would be the experimental percent error of your choice?
Explanation / Answer
a) Heat lost by water = 24.39 x 4.184 x (99.7-40.3)
Heat lost by water = 6061.63 J
Heat gained by water = 35.46 x 4.184 x (40.3-23.6)
Heat gained by water = 2477.68 J
Heat gained by cup = 6061.63 - 2477.68 = 3583.95 J
Heat capacity of Cup = 3583.95 / 40.3 = 88.93 J/oC
b) Heat gained by water = 20 x 4.184 x (25-22.9)
Heat gained by water = 175.728 J
Heat lost by metal = 175.728 J
Specific heat of metal = 175.728 / 4.29 x 100.1 = 0.41 J/goC
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