Question

Part A. 1. Calculate the specific heat for your metal for each trial. Remember that the heat lost by the metal is equal to the heat gained by the water in the calorimeter and by the calorimeter itself. This can be expressed as metal = (qwater + (calorimeter). The specific heats are positive numbers. 2. Calculate the average specific heat for your metal sample. 3. Calculate the standard deviation of your average specific heat capacity. 4. Using the physical properties of your metal, i.e. density & color, identify your metal. 5. Calculate the percent error of your average specific heat as compared to the accepted value. Common values of specific heat capacities can be found online. Be sure to cite the website or reference text.

Answer 1

Room temperature of H2O calorimeter is not given for trial-2, here calculation done for trial-2 taking 22.1oC for room temp.

For en-metal, Ti=93.2°C Tf= 34.1°C AT = Tf - Ti = 34.1-93.2 -- 59.1°C Sm . In " mxat - 2 metal= Iwater. S. Sm = - 3494.64 J 163.35819) * - 59.1°C Sm = 0.362 J/g°C. Trial- 2. Mass of Calorimeter 38.745g Mass of Calorimeter & Werter- 112.95g. & Mass of water. 112.95-38.7450 = 74.2059 Ti= 22.1°C ; If = 34, 1°C AT = Tf - Ti = 34.1-22.1 = 12.0°C and Uw - Mw.sw. AT ! = 74.205 x 4.186 X 12 qu = 3727.46 Ives For en : Tio 93.9°C Tf = 34.4°C. AT = Tf - Ti - 3401 - 93,9 = -59.8°C -Zw wh - 3727.46 oms MYAT 163.358 X - 59.8°C Sm a 0.3816 5/9°C.

Trial-3 Mass of Calorimeter = 38.7459 plass of Calorimeter & Water - 108.2459 ir plass of water a 108.245-38.745=69.59. Tir 23.0°C, If = 35.2°C ATE = Tf-Ti Uw = mwx SwNAF = 35.2-23.0 - 12.2°C = 69.58 4.186 x 12.2% Uw = 3549.31 J for metal m = 163.358 g. Ti = 93.2°C , It = 35.2°C AT = Tf-Ti= 35.2 - 93.2 = -58.0°C - Uw - 3549.31 Sma MXAT = 163.358 * - 58.0 Sm = 0.3746 5/9°C NOM 2. The average specific heat of metal s = 0.362 + 0.3816+ 0.3746 S = 0.3727 J/gcc.