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Lab report on specific heat capacity
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Experiment to Find the Specific Heat Capacity of an Aluminium Block
481 Words1 Page
Recommended: Lab report on specific heat capacity
Experiment to Find the Specific Heat Capacity of an Aluminium Block
DATA COLLECTION:
Mass(m) (kg) (5dp)
Current(I) (A) (2dp)
Voltage (V) (2dp)
Time(t) (s) (0dp)
Initial temp(T1) (ºC) (2sf)
Final temp(T2) (ºC) (2sf)
+/-( 5*10^ -4) %
+/- (1*10^ -1) %
+/- (4*10^ -2) %
+/- (5*10^-1)s
+/- 0.5 ºC
+/- (5*10^ -1) ºC
0.99577
3.70
12.07
0
16.0
16.0
0.99577
3.70
12.08
60
16.0
18.0
0.99577
3.68
12.10
120
16.0
20.0
0.99577
3.66
12.11
180
16.0
22.0
0.99577
3.68
12.10
240
16.0
25.0
0.99577
3.64
12.08
300
16.0
27.0
0.99577
3.64
12.08
360
16.0
30.0
0.99577
3.65
12.03
420
16.0
33.0
DATA PROCESSING AND PRESENTATION:
As Q=m.c.(T2-T1)
Where Q= energy transfer
C= specific heat capacity
We can rearrange this to give:
C= Q/(m(T2-T1))
And as power = energy/time
Therefore E= Pt = Q
And P = IV therefore Q = IVt
Hence C= IVt/(m(T2-T1))
Which is rearranged to the form y=px + c to give:
T2= (IVt / (m.C)) + T1.
Where p is the gradient, and equals 1/C, therefore x = IVt/m = Q/m,
and y = T2 the y intercept is equal to T1
Therefore I have calculated this table:
Energy transfer
Errors (J)
Q/m (j/kg)
Errors(J/kg)
Final temp(T2) ( ºC) (2sf)
(Q) (J) (0dp)
(0dp)
(0dp)
(0dp)
+/- (5*10^ -1) ºC
0
+/- 0
0
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