Specific Heat Capacity Lab data and processing

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Justin Germani        Specific Heat Capacity        January 24, 2012

Data:

Object 1

Object 2

Object 3


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

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

Results:

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Data:

Freezing Aluminum block

Calculate Initial Temperature of Aluminum block:

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

0.0623(910)(16.2-T)=.114(4200)(03.5)+.0465(910)(03.5)

918.427-56.963(T) =1675.8 + 148.1025

T = -15.9°C

Percentage difference:

Sample Calculations:

Conclusion:

The specific heat capacity of object one which was Iron was calculated to be 370.9 J/Kg/°C.

The specific heat capacity of object two which was Aluminum was calculated to be 625.6 J/Kg/°C.

The specific heat capacity of object three which was Copper was calculated to be 318.9 J/Kg/°C.

The specific heat capacity of the objects can be calculated ...

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