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

Specific Heat Capacity Lab data and processing

Extracts from this document...

Introduction

Justin Germani        Specific Heat Capacity        January 24, 2012

Data:

Object 1

Mass

m/g

∆m = ±0.1 g

Temperature

T/ (°C)      ∆T= ±0.1°C

initial

final

change

Object

62.3

100.3

26.4

-73.9

Aluminum Calorimeter w/ water

156.5

23.1

26.4

+03.3

water

114.0

-

-

-

Object 2

Mass

m/g

∆m = ±0.1 g

Temperature

T/ (°C)      ∆T= ±0.1°C

initial

final

change

Object

34.5

100.6

29.2

-71.4

Aluminum Calorimeter w/ water

155.6

26.2

29.2

+03.0

water

113.1

-

-

-

Object 3

Mass

m/g

...read more.

Middle

34.4

+06.3

water

112.9

-

-

-


For each table above:
∆- represents uncertainty
-mass of Aluminum Calorimeter is 42.5g

Sample Calculations:

Calculate specific heat capacity (c):

Object 1

Q lost by block = Q gained by water + Q gained by Al

image00.png

0.0623(c)(-73.9) = 0.114(4200)(03.3) + 0.0425(910)(03.3)
c = 371 J/kg/°C  (3 significant figures)

Results:

Specific Heat Capacity

c / JKg-1°C-1

Object 1

370.9

Object 2

625.6

Object 3

318.9

Data:

Freezing Aluminum block

Mass

m/g

∆m = ±0.1 g

Temperature

T/ (°C)      ∆T= ±0.1°C

initial

final

change

Aluminum block

62.3

16.2

Aluminum calorimeter

46.5

-

-

-

Aluminum Calorimeter w/ water

156.5

19.7

16.2

-03.5

water

114.0

-

-

-

Calculate Initial Temperature of Aluminum block:

Q

...read more.

Conclusion

The % difference of object two was very large at 31.3% this could be due to random error but more likely due to the block not being fully submerged as I mentioned earlier.  

The value of the percentage difference for Object3 which was Copper was relatively high at 17.2%,

In order to improve upon the experiment a scale that measures up to a higher mass such as 500g. Another method to improve the experiment would be to add more water to the calorimeter in order to insure the block was fully submerged. Another improvement that could be made would be to not drop the blocks into the calorimeter and therefore minimize the risk of losing water after the mass was taken.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate 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

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. AIM: To find the specific heat capacity of a solid by method of mixtures.

    �C 68.5 � 1 �C Heat Lost = Heat Gained Mass of metal*SIC of metal*change in temperature of metal=(mass of water*SIC of water*change in temperature of water)+(mass of calorimeter*SIC of copper*change in temperature of calorimeter) 0.0246�0.0001 kg * C3 * 68.5�1 �C = (0.03�0.0001 kg * 4186 J kg-1 �C-1

  2. Specific heat capacity of an unknown metal

    - (30 � 0.05))] = [0.0775 x 4200 x (5 � 0.1)] = [0.0775 * 4200 * (5 � 2%)] = [ (1627.5 � 2%) c] = [ (1629.18 � 32.6)] = (1627. 5 � 32. 6) Joules is the amount of heat gained by the cold water.

  1. IB Latent Heat of Fusion of Ice Lab

    � 334.40) � 100 = 36.60% The random error is:- (4 � 212) � 100 = 1.89% Hence, the systematic error is:- Systematic error = Total error - Random error Systematic error = 36.60 % - 1.89 % Therefore, the systematic error is 34.71% The possible sources of errors in

  2. IB Specific Heat Capacity Lab

    Specific heat capacity of bob = (Heat gained by water + Heat gained by calorimeter) � (Mass of bob � Temperature change of bob). VARIABLE: * Independent (manipulated) variables i. Initial temperature of solid ii. Mass of metal bob and calorimeter * Dependent (responding)

  1. Specific latent heat of fusion of ice

    Therefore, using the above calculations, we can now calculate the specific latent heat of fusion of ice: [ (5117. 28 � 25.6) ] = [(19.782 � 1.98) ] + [ 0. 00942 Lf ] + [ (1186.92 � 3.92) ] + [ (768.614 � 2.46)

  2. Investigate the Size of Craters in Sand Due to Dropped Object.

    Uncertainties: � 0.05 2.00 0.50 4.00 0.60 6.00 0.70 8.00 0.90 10.00 1.00 12.00 1.10 * Fourth Reading Height, cm Uncertainties: � 0.05 Depth, cm (d) Uncertainties: � 0.05 2.00 0.20 4.00 0.40 6.00 0.60 8.00 0.80 10.00 0.90 12.00 1.10 Analysis of Data: The shape of slotted mass is cylinder.

  1. specific heat of a solid

    heat capacity, which was 10.7%, compared to the 20.8% of the aluminium block with the insulator, meaning that the aluminium block by itself had a closer value to the real specific heat capacity. But equally neither of them was very accurate, meaning that the system had a systematic error.

  2. Physics - Specific Heat Capacity of An Unknown Material Lab Report

    Controlling Variables Masses and Specific Heat Capacities will be controlled by using the same materials. will be monitored in a insolated Styrofoam cup until desired temperature is reached. Collecting Data Procedure 1) Setup data table for finding . Hang unknown material in the Styrofoam calorimeter so it doesn't touch the walls.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work