Calometry Lab free essay sample

Volume of water in the calorimeter:| 26. 0 mL| 26. 0 mL| 26. 0 mL| 26. 0 mL| Initial temperature of water in calorimeter:| 25. 3 Â °C| 25. 3 Â °C| 25. 3 Â °C| 25. 3 Â °C| Temperature of hot water and metal in hot water bath:| 100. 5 Â °C| 100. 5 Â °C| 100. 5 Â °C| 100. 5 Â °C| Final temperature reached in the calorimeter:| 31. 6 Â °C| 34. 8 Â °C| 33. 1 Â °C| Â  34. 5 Â °C| Part I: Part II: Metal:| Metal A| Metal B| Metal C| Mass of metal:| 15. 262 g| 25. 605 g| 20. 484 g| Volume of water in the calorimeter:| 24. mL| 24. 0 mL| 24. 0 mL| Initial temperature of water in calorimeter:| 25. 2 Â °C| 25. 3 Â °C| 25. 2 Â °C| Temperature of hot water and metal in hot water bath:| 100. 3 Â °C| 100. 3 Â °C| 100. 3 Â °C| Final temperature reached in the calorimeter:| 27. 5 Â °C| 32. 2 Â °C| 28. 0 Â °C| Part 12: Part I: 1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation qwater = m ? c ? ?T. We can assume that the specific heat capacity of water is 4. 18 J / (g ? Â °C) and the density of water is 1. 00 g/mL. qwater = m ? c ? ?T m = mass of water = density x volume = 1 x 26 = 26 grams T = T(mix) T(water) = 38. We will write a custom essay sample on Calometry Lab or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page 9 25. 3 = 13. 6 q(water) = 26 x 13. 6 x 4. 18 q(water) = 1478 Joules SPECIFIC HEAT: qmetal = -205 J = 15. 363 g X c X (27. 2 100. 3 C) c = 0. 183 J/gC PART2. Using the formula qmetal = m ? c ? ?T, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation. q(water) = q(metal) q(metal) = 1478 Joules q(metal) = m ? c ? ?T m = 27. 776 g ?T = T(mix) T(metal) ?T = 38. 9 100. 5 = 61. 6 C = q(metal) / m x ? T C = -1478 / (-61. 6 x 27. 776 ) C = 0. 864 J / (g ? Â °C) Part 3: 12: For #1 theres a specific heat of 0. 864 J / (g ? Â °C) and that is closest to the specific heat of aluminum. So, for this experiment, lets call your metal aluminum. Now, the percent error formula is this: |experimental actual value divided by actual value| x 100 (|0. 864 0. 900| / 0. 900) * 100 = 4. 00 % For #2, you got 0. 183 J/gC. Comparing it to my list, I would recommend some sort of tin or cobalt meltal. 3. 9(. 39-. 39)x100%)/. 39 = 0% So there is a 0% error. It makes sense, given that the experimental results were THE SAME as the known value. Its the same. There is no error. 4. The easiest error reason is that the calorimeter wasnt a perfect insulator. This is because you must have opened the calorimeter when you added the cold water. Thus, heat was lost not only to the cold water but to the surrounding environment. Also, you might not have waited long enough for the thermometer to read, so the temperature of the hot water was lower than it really was, or the temperature of the cold water was warmer than it really was. Another possible source of error is the increase in heat by stirring due to increased kinetic energy.