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# Physics lab - Calculating the Specific Heat Capacity of Water

Extracts from this document...

Introduction

PHYSICS LAB – DATA COLLECTION AND PROCESSING

Calculating the Specific Heat Capacity of Water

DATA COLLECTION

Variable

Volume of Water taken:

 Volume of Water (cm3) Uncertainty (cm3) 80 ± 10

Measured Readings of Voltage and Current:

 Voltage (V) / V Current (I) / A (± 0.1) V (± 0.01) A 4.5 0.69

Middle

313.8

314.0

314.2

314.4

314.6

314.8

315.1

315.3

315.5

315.7

315.9

316.1

316.3

316.5

316.7

316.9

317.2

317.5

317.7

318.0

318.2

318.5

318.8

319.0

319.3

319.5

319.8

320.1

320.3

320.6

320.8

321.1

321.4

321.6

321.9

322.1

322.4

Calculating values for Change in Temperature

Change in Temperature (∆T) = Temperature (T) - Initial Temperature (Ti)

 Change in Temperature (∆T ) / K (± 0.2) K 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.1 2.3 2.5 2.7 2.9 3.1 3.3 3.5 3.7 3.9 4.2 4.5 4.7 5.0 5.2 5.5 5.8 6.0 6.3 6.5 6.8 7.1 7.3 7.6 7.8 8.1 8.4 8.6 8.9 9.1 9.4

Plotting Change in Temperature v/s Time

 Time (t)

Conclusion

0.08 kg

V = 4.5 volts

I = 0.69 amperes

0.0079 =

c = 4912.874684

c = 4.9 × 103 Jkg-1K-1

Error Propagation

Calculating the Percentage Uncertainty in:

Volume of Water = Mass of Water =  = 12.5%

Voltage =  = 2.22%

Current =  = 1.45%

Total % Error

= Sum of % uncertainties

= 12.5 + 2.22 + 1.45 = 16.17% = 16.2%

= Percentage Error × Final Value

= 0.162 × 4.9 × 103

= 7.9 × 102 Jkg-1K-1

Results

 Specific Heat Capacity of Water Uncertainty (c) / (Jkg-1K-1) (∆c) / (Jkg-1K-1) 4.9 × 103 ± 7.9 × 102

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