6. Stir the water in the calorimetergently and record the temperature (Tf) once a final steady state temperature is reached. 7. Calculate the heat gained by the calorimeter cup, water and stirrer. 8. Find the specific heat of the specimen by applying equation 2. Data A. Aluminum Specimen Mass of aluminum specimen M = 85.37g Mass of calorimeter inner cup and stirrer Mc = 83.9g Mass of calorimeter inner cup, stirrer and water Mc+Mw = 371.95g Mass of water Mw = 288.05g Initial temperature of water Tc = 21.1C Initial temperature of aluminum specimen Tm = 100C Final Temperature = 25.7C Calculations attached to lab
Discussion Questions 1. Attached to lab 2. For aluminum it would also make a 20% difference because the calorimeter cup is made of aluminum. 3. The temperature change (Tf-Tc) would be much lower which can cause a large margin of error, also the mass of the water (Mw) would be changed. 4. It would change the mass of water (Mw) because the water has spilled out, and condensation would also affect the mass because the water is being lost. 5. So there is no condensation that could result in the loss of water, and and less heat is lost through the transfer. 6. It would make it higher because there is metal on the thermometer 7. Published result: 0.215 (.215-.2224)/.2224*100=3.3% The majority of the error probably came from condensation of the outer cup, ignoring the specific heat of the thermometer, and some of the hot water being transferred.