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

# IB Specific Heat Lab

Extracts from this document...

Introduction

Lab #3

Specific Heat

By: Rendol Williams

Mr. Smith

Irmo High School

4A Physics SL

Wednesday, November 16, 2011

## Design

ASPECT 1:

The purpose of this lab is to prove an understanding of the concept of calorimetry, and establish a basic knowledge of applying those skills in the category of specific heat capacity. Through this experiment we will outline a method for determining specific heat capacities experimentally in lab. Specific heat represents the amount of thermal energy required to change the mass of one kilogram of a substance by one Kelvin. We will manipulate the equation Q=mcT to calculate various quantitative data that will inevitably lead to us an experimental specific heat capacity that we could compare to the actual or accepted value. We will also exercise the use of thermal equilibrium as we experiment in moving heat from a warmer body to a colder body in the form of thermal energy. The amount of heat lost must

Middle

T

Q= Heat (Joules)                m= mass        c= specific heat capacity (J g-1 K-1)

T= Change in Temperature

Q lost= Heat Lost (Metal)                Q gained= Heat Gained (Water)                Q lost = Qgained

Q gained = (4.183 ± 0) (122.6 ± .2) (25.2 ± .1 – 22.5 ± .1)        4.183= c of water

Q gained = 1384.7 ± .3

Zinc                Q lost =1384.7 ± .3

Q lost (1384.7 ± .3) = (c) (56.25° C ± .05) (100 - 25.2 ± .1)

c of zinc = .329 ± .04 J g-1 K-1

Q gained = (4.183 ± 0) (122.6 ± .2) (29.5 ± .1 – 24.5 ± .1)

Q gained = 2564.2 ± .3

Tin                Q lost =2564.2 ± .3

Q lost (2564.2 ± .3) = (c) (70.5° C ± .05) (100 – 29.5± .1)

c of zinc = .510 ± .04 J g-1 K-1

Q gained = (4.183 ± 0) (122.6 ± .2) (32.0 ± .1 – 29.0 ± .1)

Q gained = 1538.5 ± .3

Copper                Q lost =1538.5 ± .3

Q lost (1538.5 ± .3) = (c) (70.25° C ± .05) (100 – 32.0 ± .1)

c of zinc = .320 ± .04 J g-1 K-1

Q

Conclusion

## Evaluation

As the percent error expresses the experiment overall could have been more accurate. Many simple adjustments could be made to result in better quantitative data being recorded. One source of error could be the error of validity of the scale. The triple beam balance is a good tool but if you desire more accurate results the electronic scale with zeroing would be a better alternative. Another source of error was the heat lost from the cup and the air flow coming into the cup. Heat from the cup would have dispersed into the open air since the top was open. A simple way of solving this problem would just be to put a lid over the top of the cup to prevent major heat loss. Some still would be loss but the ratio would be much smaller.

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

# Related International Baccalaureate Physics essays

1. ## Specific heat capacity of an unknown metal

mccc ?Tc = [0.06443 x 385 x ((35 � 0.05) - (30 � 0.05))] = [0.06443 x 385 x (5 � 0.1)] = [0.06443 x 385 x (5 � 2%) ] = (124.03 � 2%) c = [ (124.03 � 2.48)] = (124.

2. ## IB Specific Heat Capacity Lab

- 1 METHOD: i. I kept the metallic bob into the hot water bath to heat it. ii. During the bob was being heated by hot water bath I took an empty calorimeter and weighed it on the Top-pan balance. iii. I half filled the calorimeter with tap water and weighed it again. iv.

1. ## IB Latent Heat of Fusion of Ice Lab

Latent heat of fusion of ice based on trial 1 data = 185 J g-1 � 3 J g-1 Latent heat of fusion of ice based on Trial 2 data Using the method used to calculate the value for the latent heat of fusion of ice for trial 1 data,

2. ## Specific latent heat of fusion of ice

- (30 � 0.05)) ] = [ 0.00942 * 4200 * (30 � 0.1) ] = [ 0.00942 * 4200 * (30 � 0.33%) ] = [ (1186.92 � 0. 33%) Joules ] = [ (1186.92 � 3.92) Joules ] = [ (1186.

1. ## Ohm's Law lab

Uncertainty Propagation Systematic Errors: There is no literature value to compare our calculated Resistance to, for a better understanding of the accuracy of our experimentally calculated resistance. The procedure involves using a light bulb, which when an electric current passes through it begins to heat up.

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

Final Volume (ml) Displaced Volume (ml) 1 88 80 8 2 58 50 8 3 73 65 8 4 90 82 8 5 97 89 8 Mass of unknown material: . Processed Data Calculation: Calculation: Density Calculations: , , , Conclusion Based on the specific heat capacity obtained , the

1. ## Physics lab - Cantilever Beam

Hence, is equal to 3.3 � 0.3 Y-intercept will be equal to ln() [so =] Hence, = ] The y- intercept is -10.36, Graph of Beam 2: Gradient of the best-fit line: = = = = Measurement: 0.95 � 0.28 = 3.392857 Relative error of a: 0.02 � 0.95 =

2. ## HL Physics Revision Notes

the element absorbs light Evidence for the existence of atomic energy levels: The emission spectra of each element is unique as electrons can only occupy specific energy levels. Movement between energy levels requires electron to emit or absorb energy. Energy emitted or absorbed is in the form of packets of light called photons. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 