Data Table 2: Final Results Metal used: Trial 1 Trial 2 Trial 3 Mass of metal (g) 34.14g 28.76g 31.55g Mass of water (g) 29.03g 28.68g 27.33g Initial temperature of water (°C) 23.00C 22.00C 22.00C Initial temperature of metal (°C) 26.4C 25.3C 50.00C Final temperature of water (°C) Final temperature of metal (°C) 74.2C 90.4C 32.00C 94.7C 90.4C 96.00C Name: Specific Heat of a Metal Lab Date: Per. Analysis and Calculations: Show your work including significant figures and appropriate units in order to receive FULL credit. 1. Knowing your initial temperatures and final temperatures, calculate At for both the metal and water. Trial 1 Trial 2 Trial 3 Water: At tinal- tinitial 74.2-23.00 51.1 C 90.2-22.00 68.2 32.00-22.00 = 10.00 C Metal: At tal- tinitial 94.7 26.4 68.3 C 90.4 25.3 65.1C 96.00-50.00 46.00C 2. Knowing that the specific heat of water = 4.184 J/g °C, calculate the heat gained by the water using the equation q = mxs x At. Do this for all three trials. Show your work neatly. Trial 1 Q (29.03)(4.184)(51.1) = 6206.6J Trial 2 Trial 3 Q (28.68) (4.184) (68.2) = 8160.9J Q (27.33) (4.184) (10.00) = 1143.4 J 3. Based on the Law of Conservation of Energy, calculate the heat lost by the metal: (- Heat Lost by the Metal Heat Gained by the Water). -qmetal-qwater Do this for all three trials. Show your work neatly. Trial 1 Trial 2 Trial 3 100 g x 4.18 J/g°C x 5 °C = 2090 J -2090 J 150 g x 4.18 J/g°C x 4°C = 2508 J 2508 J 200 g x 4.18 J/g°C x 3°C = 2508 J -2508 J 2090J/50 gx20 °C = 2.09 J/g°C 2508 J/60 gx 18°C 2.32 J/g° C Specific Heat of a Metal Lab Name: Date: Per. 2508J/70 gx15 °C 2.39 J/g°C 2.09-2.15/2.15 x 100-2.79% 2.32-2.15/2.15x 100% 7.91% 239-2.15/ 2.15 x 100% 11.16% 4. Knowing the heat lost by the water, calculate the specific heat of the metal(s) using the equation: sq/ (mx At). Do this for all three trials. Show your work neatly. Trial 1 74.2 degrees Celsius - 23.0 degrees Celsius = 51.2 degrees Celsius q 29.03 grams * 4.18 J/g°C * 51.2 degrees Celsius = 6211.92 74.2 degrees Celsius - 23.0 degrees Celsius = 51.2 degrees Celsius q = 29.03 grams * 4.18 J/g°C * 51.2 degrees Celsius = 6211.92 Joules 94.7 degrees Celsius - 26.4 degrees Celsius = 68.3 degrees Celsius Trial 2 6211.92 Joules (34.14 grams * 68.3 degrees Celsius) = 2.65 J/g°C 28.68 grams Change in temperature of water = 90.4 degrees Celsius - 22.0 degrees Celsius = 68.4 degrees Celsius q = 28.68 grams * 4.18 J/g°C* 68.4 degrees Celsius 8183.95 Joules 90.4 degrees Celsius -25.3 degrees Celsius = 65.1 degrees Celsius 8183.95 Joules/(28.76 grams * 65.1 degrees Celsius) = 4.41 J/g°C 27.33 grams Trial 3 Change in temperature of water 32.0 degrees Celsius - 22.0 degrees Celsius 10.0 degrees Celsius q=27.33 grams * 4.18 J/g°C* 10.0 degrees Celsius = 1142.94 Joules 96.0 degrees Celsius - 50.0 degrees Celsius = 46.0 degrees Celsius 1142.94 Joules/(31.55 grams * 46.0 degrees Celsius) = 0.79 J/g°C 5. Using the accepted specific heat value for the metal(s) to calculate % error. Show your work. Use the internet to google the accepted value in J/g*°C. Trial 1 Trial 2 Trial 3

xMail - Mary x D2L Grades - Sp X G when does X AT HOMEWORX Id=5735112480241329813180832311&elSBN=9781305862883&id=1707785907&snapshot!... Back References Use the References to access important values if needed for this question. grams oxygen Submit Answer MindTap - C X 2. Conservation of Mass: Macroscopic: This is group attempt 2 of 10 Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. G If 27.8 grams of mercury(II) oxide react to form 25.7 grams of mercury, how many grams of oxygen must simultaneously be formed? Autosaved at 8:38 PM G Which of th X Q Search this D

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