The objective of this laboratory was to measure the specific heat capacity of two different metals using calorimetry.
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Introduction
Specific Heat Capacity of Metals
Objective:
The objective of this laboratory was to measure the specific heat capacity of two different metals
using calorimetry.
Equipment used:
- Cooking pot for boiling water
- Stovetop
- Fork
4- Washers
3- Styrofoam cups, 8 oz
- Cylinder, 25 mL
- Scale-Spring- 10 g
- Thermometer
- String, 4.0 m
2- Weights, ½ oz sinkers (I didn’t have a third one)
Data:
Data Table 1
Object Description | First Metal | Second Metal |
Room temperature, °C | 25 | 25 |
Mass of water in calorimeter, g | 25 | 25 |
Mass of metal object, g | 21 | 27.7 |
Starting temperature of water, °C | 25 | 25 |
Starting temperature of object, °C | 25 | 25 |
Highest final temperature, °C | 31 | 28 |
Data Table 2
Objects | Mass (g) | T initial (°C) | T final (°C) | ΔT (°C) | C (cal/g °C) |
Water in Calorimeter | 25 | 25 | 31 | 6 | 1 |
First Metal | 21 | 103 | 31 | 72 | 0.099 |
Middle
25
25
28
3
1
Second Metal
27.7
103
28
75
0.036
Calculations:
ΔT = Tfinal - Tinitial
An example of this for water ΔT in Data Table 3 is:
28 °C – 25 °C = 3 °C
ΔT = 3 °C
ΔT = Tfinal - Tinitial
An example for this for the metal ΔT in Data Table 3 is:
103 °C – 28 °C = 75 °C
ΔT = 75 °C
Calculations 1 and 2;
An example of how I found the specific heat for the metals from Data Table 2 and calculation 1part 1 is:
Q lost by object = Q gained by water
cm= (mm) (cm) (ΔTm) = (mw) (cw) (ΔTw)
(25g) (31 °C -6 °C) (1 cal/g °C) = (21g) (cm) (103 °C - 31 °C)
Specific Heat Capacity = cm= 0.099 cal/g C° = 0.429 J/g °C
= 0.099 cal/g °C
For Data Table 3 and calculation 1 part 2 it was:
cm= (mm) (cm) (ΔTm) = (mw) (cw) (ΔTw)
(25g) (28 °C -25 °C) (1 cal/g °C) = (27.7g) (cm) (103 °C - 28 °C)
Specific Heat Capacity = cm
= 0.0361 cal/g C°
In order to determine the percent error for calculation 2 part 1 I used the following equation:
%error = experimental value – theoretical value x 100
theoretical value
The percent error for the first metal, iron or steel, was determined in this way:
%error = 0.099 cal/g °C – 0.107 cal/g °C x 100
0.107 cal/g °C
% error = +- 7.477%
For calculation 2 part 2, I found it for the second metal, lead, in the same way:
Conclusion
Error Sources and Why:
Errors were caused by many things in this laboratory. First, the equipment was not laboratory grade. The use of Styrofoam cups instead of a traditional calorimeter caused heat to escape. It was also difficult to read the thermometer when it was inside the calorimeter hole. I would have to keep pulling it half way out to get a reading. I know this must have caused the temperature to drop. It also may have caused me not to see the highest temperature reached. It is also possible that the temperature of the calorimeter may have heated up more because I had it too close to the stove on the first run.
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