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

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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ONLY ANSWER NUMBER 5 the other questions are there in case you need information from them and the data table is there for the same reason

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
Transcribed Image Text: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
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