• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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

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.

This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related AS and A Level Electrical & Thermal Physics essays

1. ## An experiment to measure the specific heat capacity of water

We considered the risk of uneven heating of the water and decided to stir the water with the thermometer prior to taking the reading. The stirring was done prior to reaching the time that the reading was due to ensure accurate timing of readings.

2. ## The aim of my investigation is to determine the specific heat capacity of aluminium.

Within the spaces provided by the lattice there exists a sea of delocalised electrons, and these outer electrons are therefore able to move throughout the lattice. The mechanism responsible for heat flow depends on coupling between adjacent molecules. The atoms in a solid have vibrated kinetic energy.

1. ## I am going to investigate what the resistivity is of a pencil lead. ...

The second graph shows that the resistance of this particular pencil lead is 1.798? (1.80? to three significant figures). The first thing that is immediate is that the graph is a straight line that is the characteristic of an ohmic resistor.

2. ## Investigating the monitoring systems used on modern day large A/C for detection of specific ...

harmonics frequencies, as well as a comparison of measured and extreme limits of vibration, 'excessive vibration' and 'vibration limit' signals formed when limits are exceeded. Engine vibration is, of course, something that is unwanted, but unfortunately it cannot be entirely eliminated even with turbine engines, which have no reciprocated parts.

1. ## To investigate the effects of two different variables on a solar cell output.

For the distance variable, using the equation stated before, Light Intensity= Power/area, we can work out the light intensity as the distance between the source of light and the solar cell varies. Light spreads out in all directs, in the shape of a sphere, therefore it will occupy 4?r� amount

2. ## Investigate the effects of two different variables on a solar cell output.

When rearranging the first formula to display frequency as the subject of the formula, and then substituting the value for frequency given (wave speed/wavelength) into the second formula, we get: Planck's Constant x wavespeed/wavelength= energy Using this formula, we can find out what kinds of light give out the most energy.

1. ## To find which of the circuits, shown below, are most suitable to measure a ...

At the start the resistance I specified on the variable resistor was 50000?. The reading above that I calculated from the theoretical work is 25000? off this value. 25000? expressed as a percentage of the original 50000? is 50%. This is a vast reading off the labelled resistance, so I

2. ## The purpose of this experiment was to measure the specific heat capacity (Cb) of ...

always use the equation Heat lost by brass = heat gained by water+ heat gained by caloerimeter Or MbCb(?3- ?2)= MwCw(?2- ?1)+ McCc(?2- ?1) ......(2) Mb = mass of brass (Kg) Mw = mass of water (Kg) Mc = mass of copper caloerimeter (Kg) • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 