SPECIFIC HEAT OF A SOLID:

TOPIC 3.2 IN THE SYLLABUS – 1.5 HOURS

When the heater is used as shown in the diagram, it provides energy E

E = V x I x t

V being the voltage provided by the power supply (should be constant), I the current (if the voltage is maintained constant, so will be the current) and t the time.

As the voltage (and hence the current) is kept constant, you will register temperature and time first using no thermal insulator and then using it.

Using the previous information about the heater together with the specific heat capacity of the blocks (Aluminium = 878 J K-1 Kg-1; Copper = 361 J K-1 Kg-1; Mild steel = 480 J K-1 Kg-1) do the appropriate graph to obtain the specific heat capacity of the block you worked with.

Criteria to be assessed in this practical are: Data Collection and Processing; Conclusion and Evaluation and Manipulative skills. This last one will not be included in the grade of the report itself but will count for the overall grade of manipulative skills.

VERSIÓN MAYO 2009

Data collection and processing:

Join now!

     

Q E= V·I· t where voltage (V) and current (I) were constant.

To maintain the experiment as fair as possible, the values for voltage and current were maintained constant. Voltage=12V (±0.01V), current= 4A (±0.01A)

Uncertainty (U) of E= U of V + U of I + U of t

U of Q= 0.01+0.01+0.01    U of Q= ±0.03J

Q=E therefore Q= V·I·t

Gradient= (24.03744-13.02028)/ (23040-14400) gradient= 1.28 ·10^-3

Gradient= 1/c  c= 1/gradient c=784.2

% error of c= (real value- ...

This is a preview of the whole essay