• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

...read more.

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)

...read more.

Conclusion

0.08 kg

V = 4.5 volts

I = 0.69 amperes

 0.0079 = image03.png

image04.png

c = 4912.874684

c = 4.9 × 103 Jkg-1K-1

Error Propagation

Calculating the Percentage Uncertainty in:

Volume of Water = Mass of Water = image05.png = 12.5%

Voltage = image06.png = 2.22%

Current = image07.png = 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

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Physics essays

  1. Specific heat capacity of an unknown metal

    c] Joules = [ (1627.5 � 32.6) ] Joules + [ (124.03 � 2.48) ] Joules [ (4.43 � 0.0077) c] Joules = [ (1751.53 � 35.08) ] Joules c = [ (1751.53 � 35.08) ] = [ (1751.53 � 2.0%) ] = [ (386.7 � 2.17%) ] [ (4.53 � 0.0077) ] [ (4.43 � 0.17%)

  2. IB Specific Heat Capacity Lab

    Mass of metal bob and calorimeter * Dependent (responding) variables i. Final temperature of metal bob, calorimeter and water ii. Mass of water * Controlled variables i. Initial temperature of water (22�C) and calorimeter (41.35�C) ii. Room temperature (air-condition was kept at a fixed temperature) * Unknown variable Specific heat capacity of the solid (metal bob)

  1. IB Latent Heat of Fusion of Ice Lab

    165. The value obtained experimentally is 212 J g-1 � 4 J g-1, and therefore, its percentage deviation from the literature value, or the total error in the experiment, is:- ((334.40 - 212)

  2. Specific latent heat of fusion of ice

    that the specific latent heat of fusion of ice is (334, 000 � 1837) JKg-1. Therefore, the specific latent heat of fusion of ice is 334, 000 JKg-1 with an error of 0. 55%. This means that the range of the specific latent heat of fusion of ice is from (334,000 - 1837)

  1. Physics - Specific Heat Capacity of An Unknown Material Lab Report

    Final temperature, () Mass of hot water, (kg) Temperature of cold water, () Mass of cold water, (kg) 68.2 40.8 0.094 18.5 0.112 82 47.1 0.099 20.5 0.126 81 47.6 0.094 20.1 0.110 Raw Data Table for Volume measurements Trial Initial Volume (ml)

  2. THermal Physics Lab

    + (696/41470) + ( 603/35815) + ( 510/ 30160) + ( 448/ 26390) + (385/22620) + ( 290/16965) + ( 194/11310) + ( 130/7540)) * 301600 = 51039 J Total Heat energy lost from hot milk: -301600 + 51039 J Total heat energy gained by the cooler: (-301600 + 51236)J + (-301600 + 51039)J = -603200 + 102275 J

  1. Physics lab on propagation of errors. In this experiment I investigated the propagation ...

    the sphere using the formula for volume of a sphere: V=4/3 ? r3 V = 4/3 x ?x (2.46/2)3 Volume = 12.9cm3 The error is equal to thrice the percentage error of the radius as it is cubed. The percentage error is hence calculated as the percentage of the error of the true value.

  2. HL Physics Revision Notes

    The hotter it becomes the greater rate at which it loses energy. Molecules are arranged in different ways depending on the phase of the substance, (i.e. solid, liquid or gas) Solids: Fixed volume and fixed shape. The molecules vibrate about a fixed position.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work