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IB Specific Heat Capacity Lab

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Introduction

Candidate Name: Sahaj Miyani

Candidate no.: 001424-049

Session: May 2009

 Physics Lab Report -1To determine the specific heat capacity of solid

IT used:

Graphical Analysis 3.1

Date:

Class : XI B5y

Contents

INTRODUCTION:

RESEARCH QUESTION:

BACKGROUD INFORMATION:

HYPOTHESIS:

VARIABLE:

HOW TO CONTROL VARIABLES:

APPARATUS:

METHOD:

DATA COLLECTION:

DATA PROCESSING and PRESENTATION:

Trial 1

Trial 2

Trial 3

CONCLUSION

Graph: Temperature of Water Bath V/s Total heat gained by Water and Calorimeter

Average Specific Heat Capacity of the metal bob:

Calculating Errors:

EVALUATION

DESIGN (D)

INTRODUCTION:

This lab report will find the specific heat capacity of a metallic bob.

RESEARCH QUESTION:

How many joules of heat are required for the temperature to rise by 1 K of metallic bob of 1 Kg.?

BACKGROUD INFORMATION:

A solid of known mass is heated to a known temperature and placed in a calorimeter containing water at room temperature.  When all the three things reach the same temperature it is said to be at the state of equilibrium. Using the law of conservation of energy (heat gained = heat lost) we can find the specific heat capacity of the solid (Bob).

• Specific heat capacity of water = 4182 Jkg-1K-1
• Specific heat capacity of calorimeter = 385 Jkg-1K-1.

HYPOTHESIS:

Increase in the temperature of Water Bath will result in the increase of the total heat gained by water and calorimeter and vice versa.

Middle

Initial – TFinal)MB × CB (T3 – T2) = (Mw × Cw (T2 – T1)) + (MC × CC (T2 – T1))CB = [(Mw × Cw (T2 – T1)) + (MC × CC (T2 – T1))] ÷ MB (T3 – T2)CB = (HW + Hc) ÷ MB × δTB

Specific heat capacity of bob = (Heat gained by water + Heat gained by calorimeter) ÷ (Mass × Temperature change of bob).

Trial 1

 Explanation Formula Calculation Error % Error Mass of water(Mw) (2) – (1)(see data table) = 103.75 – 41.35= 62.40 g ± (0.01 + 0.01)= ± 0.02 g = (0.02 ÷ 62.40) × 100= 0.032 % Temperature change of water + Calorimeter (δT) (7) – (5)(see data table)(T2 – T1) = 25.50 – 22.00= 3.50 oC ± (0.25 + 0.25)= ± 0.50 oC = (0.50 ÷ 3.50) × 100= 14.286 % Temperature change of Bob (δTB) (6) – (7)(see data table)(T3 – T2) = 72.00 – 25.50= 46.50 oC ± (0.25 + 0.25)= ± 0.50 oC = (0.50 ÷ 46.50) × 100= 1.075 % Heat absorbed by water (Hw) MW × CW × δT = 62.40 × 4.18 × 3.50= 912.91 J = (0.03 + 14.30)% × 912.91= 14.33 % × 912.91= ± 130.80 J Heat absorbed by Calorimeter (HC) Mc × Cc × δT = 41.35 × 0.385 × 3.50= 55.72 J = [((0.01 ÷ 41.35) × 100) + 14.286 %] × 55.72= 14.31 % × 55.72= ± 7.974 J Heat lost by Bob (HB) = Heat gained by water & Calorimeter HS = Hw + HC = 912.91 + 55.72= 968.63 J = 14.33 % + 14.31 %= 28.640 % × 968.63= ± 277.42 J Specific Heat Capacity of Bob (CB) Heat Lost ÷ (mass of bob × temperature change of bob) [HS / MB × δTB] = 968.63 ÷ ( 65.90 × 46.5)= 0.3161 J g-1 K-1= 316.1 J kg-1 K-1 = [((0.01 ÷ 65.90) × 100) + 1.075 %] + 28.640 %= 1.090 % + 28.640 %= 29.730 %= ±93.98

Trial 2

 Explanation Formula Calculation Error % Error Mass of water (Mw) (2) – (1)(see data table) = 105.70 – 41.35= 64.35 g ± (0.01 + 0.01)= ± 0.02 g = (0.02 ÷ 64.35) × 100= 0.031 % Temperature change of water + Calorimeter (δT) (7) – (5)(see data table)(T2 – T1) = 26.00 – 22.00= 4.00 oC ± (0.25 + 0.25)= ± 0.50 oC = (0.50 ÷ 4.00) × 100= 12.5 % Temperature change of Bob (δTB) (6) – (7)(see data table)(T3 – T2) = 78.00 – 26.00= 52.00 oC ± (0.25 + 0.25)= ± 0.50 oC = (0.50 ÷ 52.00) × 100= 0.96 % Heat absorbed by water (Hw) MW × CW × δT = 64.35 × 4.18 × 4= 1075.93 J = (0.03 + 12.5)% × 1075.93= 12.53 % × 1075.93= ± 134.810 J Heat absorbed by Calorimeter (HC) Mc × Cc × δT = 41.35 × 0.385 × 4= 63.68 J = [((0.01 ÷ 41.35) × 100) + 12.5 %] × 63.68= 12.52 % × 63.68= ± 7.975 J Heat lost by Bob (HB) = Heat gained by water & Calorimeter HS = Hw + HC = 1075.93 + 63.68= 1139.61 J = 12.53 % + 12.52 %= 25.05 % × 1139.61= ± 285.47 J Specific Heat Capacity of Bob (CB) Heat Lost ÷ (mass of bob × temperature change of bob) [HS / MB × δTB] = 1139.61 ÷ ( 65.90 × 52.00)= 0.33256 J g-1 K-1= 332.56 J kg-1 K-1 = [((0.01 ÷ 65.90) × 100) + 0.96 %] + 25.050 %= 0.977 % + 25.050 %= 26.027 %= ±86.55

Conclusion

3
 DATA SYMBOLS UNCERTAINITIES TRIAL 1 TRIAL 2 TRIAL 3 1 Temperature of Water Bath ( oC ) T3 / TB ± 0.25 72.00 78.00 78.50 2 Heat gained by water & Calorimeter (J) HS ± 278.78 968.63 1139.61 1227.89

Graph: Temperature of Water Bath V/s Total heat gained by Water and Calorimeter Graph is made using software - Graphical Analysis 3.1.

As the slope is positive (+35.1), Temperature of Water Bath is directly proportional to Total heat gained by Water and Calorimeter.

Hence, I have proved my hypothesis.

Average Specific Heat Capacity of the metal bob:

 = (316.10 + 332.56 + 358.32) ÷ 3= 335.66 J kg-1 K-1 ± 26.34 %= 335.66 J kg-1 K-1 ± 88.4

Literary value of Specific heat capacity of Brass: 377 J kg-1 K-1

Experimental value of Specific Heat Capacity of Brass: 335.66 J kg-1 K-1 ± 88.4

Hence, the experimental value for the Specific Heat Capacity of the Brass Bob matches the literary value of the Specific Heat Capacity of the Brass, thus Equation of Thermal Equilibrium holds true:

M C δΤRISE= M C δΤFALL

Calculating Errors:

 Total % error:= (Literary value – Experimental value) ÷ Literary value × 100= [(377 – 336) ÷ 377] × 100= 10.88 % Random error:= (88.40 ÷ 335.66) × 100= 26.34 % Systematic error:= Total error – Random error= 10.88 – 26.34= 15.46 %

EVALUATION

Hence, according to my experiment the specific heat capacity of metal bob is 335.66 J kg-1 K-1. The value may contain some error due to following reasons:

1. Efficiency of the apparatus wasn’t up to the mark.
2. Unexpected interruptions delayed the experiment in between.
3. Heat loss in any spilling of water, may be while dropping in the solid.
4. Heat loss from apparatus to the atmosphere.
5. Non uniform heating of water may cause incorrect readings for temperature of water bath.

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