• Title Page
Title: Lab 1: Heat Capacity and Latent Heat of Fusion
Name: Mohamed Joumaa Zabadne
Student Number: 18827956
Course: Civil Engineering with Integrated Foundation Year
Module: Core Science – FY003
Date of experiment: Friday, 5th October 2018
Date of submitting report: Thursday, 18th October 2018
• Abstract
This is an investigation where adding a hot object to an object of less temperature will cause the cooler object’s temperature to increase. This also investigates how adding a cold substance to a substance of higher temperature causes the hotter object’s temperature to decrease.
• Table of Contents
Title Page 1
Abstract 1
Table of Contents 1
Introduction 1
Theory 1-2
Risk Assessment 2
Method 2-3
Results & Discussion 3-9
Conclusion 9
References 9
• Introduction
This is an investigation to see how adding a hot metal block effects the change in temperature of water and hence calculate the specific heat capacity of the metal to try and identify what it is. Also, an investigation to observe how adding ice to water causes a decrease in the temperature of the water and hence calculate the latent heat of fusion of the water.
• Theory
Heat is a form of energy. It can be transferred between two objects if they are of different temperatures. Heat is transferred from the higher temperature object to the object with lower temperature. Heat is ‘lost’ from the object of higher temperature and ‘gained’ by the object of lower temperature. The heat energy lost by the object of higher temperature is the same as the heat energy gained by the object of lower temperature.
The amount of heat gained or lost can be calculated using the equation E=mc (1), where E, is the Energy transferred, m is the mass of the object, c is the specific heat capacity of the object and is a characteristic value for a specific substance, is the change in temperature of the object.
Latent heat of fusion is the heat energy needed for a substance to change state from a solid to a liquid. The latent heat of fusion can be calculated by rearranging the equation E=mlf (2), where E is the energy required, m is the mass of the substance and lf is the specific heat of fusion.
• Risk Assessment
Metal will be very hot. Hold using a few napkins and handle carefully.
Glass beakers may break. Handle with care and in case a beaker breaks, collect the glass carefully using a dustpan and brush.
Mercury in the thermometer is toxic. Handle thermometer carefully
• Method
Materials:
1.
Get a 100 ml glass beaker and measure and record its weight using a balance
Pour approximately 60ml of water into the 100ml glass beaker and measure and record its weight using a balance
Measure the initial temperature of the water using a thermometer and record it
Add an ice cube to the beaker, start the stopwatch and stir the water and ice gently
Read and record the temperature of the water every 10 seconds until the temperature is constant.
Measure the weight of the beaker using a balance and record the reading.
Repeat 3 times and take an average
Repeat the above steps using 2 ice cubes instead of 1.
2.
Get a 100 ml glass beaker and measure and record its weight using a balance
Pour approximately 60ml of water into the 100ml glass beaker and measure and record its weight using a balance
Get a piece of metal and measure its mass using a balance
Heat a piece of metal in a kettle of boiling water
When the water in the kettle is boiling, remove the metal carefully using a string and hold with a few tissues and place it in the 100ml beaker.
Record the initial temperature of the water, start the stopwatch and begin to stir gently
Read and record the temperature every 10 seconds until the temperature is constant.
Repeat 3 times and take an average
• Results & Discussion
o Specific Heat Capacity
Specific heat capacity of water = 4182 Jkg-1k-1
1. Mass of metal = 0.055 kg
Mass of beaker = 0.0185 kg
Mass of beaker + water = 0.0953 kg
Initial temperature = 23.0oC
Final temperature = 31.0oC
For water:
E= mc
E= (0.0953-0.0185)(4182)(31-23)
E= 2569.4208 J
For metal:
E= 2569.4208 J
c= E/m
c= 2569.4208/(0.055)(100-31)
c= 677.0542292 Jkg-1k-1
c= 677 Jkg-1k-1
2. Mass of metal = 0.055 kg
Mass of beaker = 0.0185 kg
Mass of beaker + water = 0.0933 kg
Initial temperature = 23.0oC
Final Temperature = 31.5oC
For water:
E= mc
E= (0.0933-0.0185)(4182)(31.5-23)
E= 2658.9156 J
For metal:
E= 2658.9156 J
c= E/m
c= 2658.9156/(0.055)(100-31.5)
c= 705.7506569 Jkg-1k-1
c= 706 Jkg-1k-1
Table 1: Change in temperature when adding a hot metal rod to a beaker of water
Time/s Temperature/ oC
Trial 1 Trial 2 mean
0 23.00 23.00 23.00
10 25.00 25.00 25.00
20 26.00 26.00 26.00
30 26.50 28.00 27.25
40 27.00 30.00 28.50
50 27.50 31.00 29.25
60 28.00 31.00 29.50
70 29.00 31.50 30.25
80 30.00 31.50 30.75
90 30.00 31.50 30.75
100 30.00 31.50 30.75
110 30.00 31.50 30.75
120 30.00 31.50 30.75
130 30.00 31.50 30.75
140 30.50 31.50 30.75
150 30.50 31.50 30.75
160 31.00 31.50 31.25
170 31.00 31.50 31.25
180 31.00 31.50 31.25
190 31.00 31.50 31.25
200 31.00 31.50 31.25
210 31.00 31.50 31.25
220 31.00 31.50 31.25
230 31.00 31.50 31.25
240 31.00 31.50 31.25
Analysis: Data obtained supports the theory as it show that when adding an object of higher temperature (hot metal cylinder) to a substance of lower temperature (water at room temperature) the temperature of the water increased as a result of heat energy being lost by the metal and gained by the water. This can clearly be shown in the trend of the results where temperature of the water continuously increases until it reaches a peak, where thermal equilibrium is reached and the metal and water have the same temperature.
o Latent Heat of Fusion
1. Mass of beaker = 0.0176 kg
Mass of beaker + water= 0.0793 kg
Mass of beaker + water +melted ice= 0.0964 kg
Initial Temperature = 23.0oC
Final Temperature = 4.5oC
E= mc
E= (0.0793-0.0176)(4182)(23-4.5)
E= 4773.5439 J
E= 4774 J
E=mlf
Lf= E/m
Lf= 4773.5439/(0.0964-0.0793)
Lf= 279154.614 Jkg-1
Lf= 279155 Jkg-1
lf = 279.155 kJkg-1
lf = 279 kJkg-1
Table 2: Change in temperature when adding 1 ice cube to a beaker of water
Time/s Temperature/oC
0 23.0
10 20.0
20 15.0
30 14.5
40 12.0
50 10.0
60 8.0
70 7.5
80 6.5
90 6.0
100 6.0
110 6.0
120 5.5
130 5.5
140 5.0
150 5.0
160 5.0
170 5.0
180 5.0
190 4.5
200 4.5
210 4.5
220 4.5
230 4.5
240 4.5
2. Mass of beaker = 0.0176 kg
Mass of beaker + water = 0.087 kg
Mass of beaker + water + melted ice = 0.103 kg
Initial Temperature = 23.0oC
Final Temperature = 1.0oC
E= mc
E= (0.087-0.0176)(4182)(23-1)
E= 6385.0776 J
E= 6385 J
E= mlf
Lf= E/m
Lf= 6385.0776/(0.103-0.087)
Lf= 399067.35 Jkg-1
Lf= 399067 Jkg-1
Lf = 399.067 kJkg-1
Lf = 399 kJkg-1
Table 3: Change in temperature when adding 2 ice cubes to a beaker of water
Time/s Temperature/oC
0 23.0
10 14.0
20 10.0
30 7.5
40 6.0
50 5.0
60 4.0
70 3.0
80 2.0
90 2.0
100 2.0
110 1.5
120 1.5
130 1.5
140 1.5
150 1.5
160 1.5
170 1.5
180 1.5
190 1.0
200 1.0
210 1.0
220 1.0
230 1.0
240 1.0
Analysis: Data obtained supports the theory as it can be clearly seen from the results and graphs that when adding a substance of lower to temperature (ice) to a substance of lower temperature (water at room temperature), the temperature of the water decreases.
• Conclusion
All in all, it has been shown that heat is transferred between substances which are in contact until they both reach the point of thermal equilibrium, where the temperature of each substance is the same. The heat energy lost by the hotter substance is equal to the heat energy gained by the cooler substance.
• References
(1) BBC Bitesize – GCSE Physics (Single Science) – Energy and heating – AQA – Revision 5, BBC Bitesize, https://www.bbc.com/bitesize/guides/z2gjtv4/revision/5, accessed on Tuesday 16/10/2018
(2) BBC Bitesize – National 5 Physics – Specific latent heat – Revision 2, BBC Bitesize, https://www.bbc.com/bitesize/guides/zprhjty/revision/2, accessed on Tuesday 16/10/2018