This is a standardized lab write up for an American Advanced Placement Chem class dealing with the use of a calorimeter. In specific, the mass of a volatile unknown liquid was discovered with the use of the calorimeter.Experiment Eleven - Heat Effects and CalorimetryIn this lab, molar mass of a volatile liquid was determined through the use of a calorimeter The lab validates the equation Q=S.H * M * T, by using the known or calculable values of QH20 and setting them equal in magnitude to Qmetal. This equation depends on the conservation of energy in the form of Delta T.We concluded that the molar mass was around 74.81, and that the specific heat of this metal was about .334, This ...view middle of the document...
Trial 1 Trial 2Mass of stoppered test tube plus metal (grams) 77.8 76.9Mass of test tube and stopper (grams) 47.3 46.8Mass of Calorimeter (grams) 48.1 47.7Mass of Calorimeter and water (grams) 84.6 87.3Mass of Water (grams) 36.5 39.6Mass of Metal (grams) 30.5 30.1Initial temperature of water in calorimeter *C 21.1 20.4Initial temperature of metal (assuming 100*C unless directed to do otherwise) 100 100Equilibrium temperature of metal and water in calorimeter 26 25Change in water temperature 4.9 4.6Change in metal temperature 74 75qH2O 747.59 761.43Specific heat of the metal (Eq. 3) .3312 J/g*C .3373 J/g*CApproximate molar mass of metal 75.5 74.12Unknown No. Unknown.Sample Calculations For Trial One.Calculation Answer ExplanationChange in water temperature 4.9 T initial - T final21.1 *C - 26 *CSince this is a magnitude it can be positive.Change in metal temperature 74 *C T initial - T final100 *C - 26 *CSince this is a magnitude it can also be positive.qH2O 747.59 Q = S.H*M*T(4.18)(36.5)(4.9)Specific heat of the metal (Eq. 3) .3312 QH2O = Qmetal747.59 =S.H(30.1)(75)S.H = .3312Approximate molar mass of metal 75.5 Molar Mass = 25/S.H25/.3312=75.5Error AnalysisAn avoidable error may have been committed by under-heating the metal This would have made the result...