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specific heat of a solid

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Introduction

SPECIFIC HEAT OF A SOLID:

TOPIC 3.2 IN THE SYLLABUS – 1.5 HOURS

image02.png

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.

...read more.

Middle

11520

12

4

270

12960

12

4

300

14400

12

4

330

15840

12

4

360

17280

12

4

390

18720

12

4

420

20160

12

4

450

21600

12

4

480

23040

Q=E therefore Q= V·I·t

image01.png

Aluminium

mass (±0.00001kg)

temperature ( ± 0.5 k)

ΔT (±0.5 k)

mΔT (±0.51kg·k)

Q (±0.03J)

1.00156

295

0

0

0

1.00156

295

0

0

1440

1.00156

295

0

0

2880

1.00156

295

0

0

4320

1.00156

296

1

1.00156

5760

1.00156

298

3

3.00468

7200

1.00156

299

4

4.00624

8640

1.00156

301

6

6.00936

10080

1.00156

303

8

8.01248

11520

1.00156

305

11

11.01716

12960

1.00156

307

13

13.02028

14400

1.00156

309

15

15.0234

15840

1.00156

311

17

17.02652

17280

1.00156

313

19

19.02964

18720

1.00156

315

21

21.03276

20160

1.00156

317

23

23.03588

21600

1.00156

319

24

24.03744

23040

image04.png

image01.png

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- experimental value)/ (real value) · 100

%error of c= ((878-784.2)/ 878) · 100 % error of c= 10.7%

% uncertainty of c= %u of Q+ %u of m+ %u of T

Average Q= 11520J

%u of Q= (0.03/11520) · 100  %u of Q= 2.60 ·10^-4%

Average T = 304.3 k

%u of T= (0.5/304.3) ·100  %u of T= 0.164%

...read more.

Conclusion

Improvements

The readings for the temperature weren’t very accurate as the thermometer did only show whole numbers; therefore the value had to be approximated to the nearest unit.

Use other thermometers, which show not only the whole numbers but also the decimals between them, to have more accurate readings. This will definitely reduce the high systematic error experienced.

When the aluminium block with the insulator was heated, it was still hot because of the heating experienced when the aluminium block was heated by itself. This affected the final results and that’s why it has a higher percentage error.

So that the aluminium block could be cooled down to normal temperature, more time has to be waited. Therefore the readings could be much closer to the ones of the real value.

The power supply didn’t provide a constant supply of voltage, affecting also the constant value for current, affecting the accuracy of the experiment.

Don’t have a power supply that varies much in the flow of power, as the one used in the experiment.

...read more.

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