Gainless Steel Lab

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Gainless Steel Lab
Specific Heat Experiment of Stainless Steel using Water

Data Table I: Specific Heat Capacities of Common Substances | Substances | Specific Heat Capacity J/(gC) | Water | 4.184 | Ice | 2.06 | Steam | 1.87 | Stainless Steel | 0.0927 | Iron | 0.0449 | Aluminum | 0.897 | Copper | 0.385 | Brass | 0.376 | Gold | 0.129 | Lead | 0.129 | Carbon (graphite) | 0.709 | Density of Water | dH2O= 1.0 g/mL |

Procedures:
1. Measure the cylinder that your teacher gave you using a triple balance scale and round to the nearest tenths place.
2. Heat about 350mL of water in a 400mL beaker until it is boiling gently.
3. While the water is heating, measure the mass of the metal cylinder to the nearest 0.01 g and record the measurement.
While the cylinder is heating, measure approximately 100 mL of water in a graduated cylinder. Record the mass of the water based on the volume of the water you measured and related to the density of water to one decimal place. Then pour the water into a plastic-foam cup.
5. While the cylinder is heating, measure approximately 100 mL of water in a graduated cylinder. Record the mass of the water to one decimal place. Then pour the water into a plastic-foam cup.
6. Measure and record the temperature of the water in the plastic-foam cup and of the water in the boiling bath. This temperature should be recorded to one decimal place.
7. Create a hole on another foam cup on the center
Mass of cylinder in grams | 49.47 | 49.95 | Initial temperature of water in cup in degrees Celsius | 20.3 | 19.9 | Initial temperature of metal cylinder in degrees Celsius (temperature of boiling water) | 101.9 | 100.1 | Maximum temperature of cylinder and water | 26.9 | 25.7 | Mass of water in grams | 97.3 | 99.7 |

Calculations:
∆TH2O (trial 1) , ∆TH2O (trial 2) , ∆T Metal (trial 1) , and ∆T Metal (trial 2) q=mC∆T Results:
Changes in temperature of the water and of the cylinder of each trial

∆TH2O (trial 1) , ∆TH2O (trial 2) , ∆T Metal (trial 1) , and ∆T Metal (trial 2)
6.6 C 5.8C 75C 74.4C
Heat gained by water in each trial
T1 = qH2O = q=mC∆T = (97.3g) (4.4184 J/gC) (6.6 C) = 2686.9 J T2 = qH2O = q=mC∆T = (99.7g) (4.184 J/gC) (7.5) = 31285.6 J

The specific heat of the cylinder
2686.9 J = (49.47) C (5.8)
C = 9.4 J/(g C)
31285.9 J = (49.95) C (74.4) = 8.42 J/(g C)
C = 8.42 J/(g C)
The average value for the specific heat of the cylinder
T1 2686.9 + 9.4 / 2 = 1348.2 C
T2 31285.9 + 8.42 / 2 = 31294.3 C
1348.2 C + 31294.3 C = 32642.5 C
32642.5 C / 2 = 16321.3 C
Percent error in the specific heat value that was

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