Question

Name: Date: Section Group: Instructor Data GUAR Copper Mass of copper metal Mass of dry calorimeter Mass of calorimeter + water Mass of water Trial 1 Trial 2 a 0.26 9 34.559 39.80 9 39.05 g 73.82.9 82.19 34.56 9 40.91 . 25.1 °C 24.0°C 101.0°C 100.0 °C 28.9.0 29.5 °C 28.9°C 29.c Water temperature - initial Copper temperature (hot) - initial Water Temperature - final Copper Temperature - final Unknown Trial 1 Trial 2 Mass of unknown sample Mass of dry calorimeter Mass of calorimeter + water Mass of water 09.10 9 .49 42.449 43.109 87.609 90-70 g 45.16 9 47.609 23.6°C 6.0°C 100 °C 101.0 °C 1 26.4°C 28.2°C a6.4°C 28.2 °C Water temperature - initial Unknown temperature (hot) - initial Water Temperature - fina/ Unknown Temperature - final 2. Calculate the specific heat of copper. a. Trial 1: Specific heat of copper Dr. 20.26% (28.9 - .) = b. Trial 2: Specific heat of copper 9. C c. Average Specific heat of copper = 3. Calculate atomic weight of copper (g/mol) where: Atomic weight of copper (g/mol) = 25 molec specific heat of copper 4. Calculate percent error for copper where: 100 accented value copper-experimental value copper 04 Error mnnor =

Answer 1

Specific heat capacity of a substance is the heat capacity of unit mass substance. It is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in its temperature.

Unknown sample is Anall calculations given in image

specific heat of water is = 4.186 Joule / goc for trial I heat gained by water qNater = 4.186 83.88 34.56 = 551.92J acity of spec. heat capacity of Copper = Copper = 551.92 551.92 = 0.3775/9°C 72.1 x 20.26 I Note change in temprature of copper is 72.10 and mass of copper is 20.26 1 for Trial 2 q water = sp. heat Capacity of copper = 4.186x5.58 42.91=987.91J / 9°C 987.91 - 0.394 J / goc 72.5*34.55 to copper = 0.377 +0. 394 = 0+385 J/gre Average specific heat Atomic weight = 25x1 x t = 64.93 g/mol mole'c 0.385 - % Error = 63:55-64.93 = - 2.17% 63.55 for unknown Trial I sample - qwater = 4.186x2,8x45.16=429131 J/g.c specific heat = 529. 31 - ou 74.18 29.10 Τη 2 9 water = 4.186 X 2.2 x 47.60=438.35Jlg.c. specific heal = 438.35 - 0.227 - 72.88 26:48 Average specific heat = 0.245 +0.227 - 0.236 Ila 2 Atomic weight = 25 x 1 x 1 236 = 105.93 g/mol Thus % unknown sample = Sn froom in atomic weight of sn - 118171-105-93 x 100 = 118.71 10.76%