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The objective of this laboratory was to measure the specific heat capacity of two different metals using calorimetry.

Extracts from this document...

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:  

  1. Cooking pot for boiling water
  1. Stovetop
  1. Fork

4-        Washers

3-        Styrofoam cups, 8 oz

  1. Cylinder, 25 mL
  1. Scale-Spring- 10 g        
  1. Thermometer
  1. 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

...read more.

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:

...read more.

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.

...read more.

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