• 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 Latent Heat of Fusion of Ice Lab

    the experiment are:- * Extra heat generated within the system due to friction when the stirrer was shook up and down. * The system wasn't a perfect insulator and therefore, heat might have been lost to the surroundings. * Not all the ice might have melted.

  1. 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)

  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. 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

  2. Specific Heat Capacity Lab data and processing

    The temperature of the water with the Aluminum block was recorded until the constant temperature was reached. With this we can calculate the initial temperature of the aluminum block which gives us the temperature of the freezer. Evaluation: There were several limitations in the equipment used to carry out the procedure.

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

    Controlling Variables Masses and Specific Heat Capacities will be controlled by using the same materials. will be monitored in a insolated Styrofoam cup until desired temperature is reached. Collecting Data Procedure 1) Setup data table for finding . Hang unknown material in the Styrofoam calorimeter so it doesn't touch the walls.

  2. HL Physics Revision Notes

    Avogadro?s constant is the number of atoms in 12 g of carbon-12. It is 6.02 x1023. 3.2 Thermal properties of matter: Thermal capacity (C) is the energy required to raise an object?s temperature by 1K. C=Q/?T Specific heat capacity is the energy required to raise a unit mass of substance by 1K.

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