Being an inquirer and a creative thinker, I always tried to decipher the scientific reason behind every observation in the world. This inquisitive nature has landed me to clear the concepts of physics and applications of physics and real-life situations. Though most of the curiosities are answered, however, few are left unanswered. The reason behind this exploration was one of such an inquiry. While travelling from my brother’s place to my hometown, due to unavailability of flight ticket, I travelled via train where this curiosity came to my mind. I observed that there are gaps at the junction of two consecutive rails. After reaching my place, I started research on it. Recently, in my curriculum, I learnt that when the temperature increases, the solid metal usually expands due to the phenomenon of thermal expansion of solid. Learning the above phenomenon, I tried to relate the observation with different instances where this case will be prevalent. Hence, one thing which come to my mind was electricity. What is the effect of extension of solid on current flow in conductor? Does it affect the other microscopic properties that controls the flow of charge?
To decipher the answer, I started my research on it. I read quite a few research papers and journals on thermal expansion of solid and understood that the temperature difference is the factor that affects the extension and not the exact value of temperature. However, after taking a few video lectures and tutorials, I was able to grasp confidence and knowledge on the domain of physics but couldn’t find any relationship between expansion of solid due to change in temperature and the factors controlling current flow. This inquiry resulted me landing in this research question stated below.
How does the coefficient of thermal expansion of a wire (expressed in ℃-1) due to heating by continuous flow of current through the wire depend on the resistivity of the wire, expressed in ohm. m (copper - 0.0171, aluminum - 2.8 × 10-8, nichrome - 1.1 × 10-6, iron - 1.0×10-7 and tungsten - 4.9 × 10-8), determined using length versus temperature scattered graph?
With an increase in temperature of any object, the length of solid increases because of an increment in the kinetic energy of the molecules of the object. The formula of increase in length is expressed as:
∆L = L0 × α × ∆T
= > α = \(\frac{∆L}{L_0×∆T}\)..........(5)
Here, ∆L indicates the extension in length of object, indicates the coefficient of thermal expansion, and ∆T indicates the difference in temperature.
Co-efficient of thermal expansion can be defined as the increase in length of a solid object of unit length for an increase in temperature of unity.
Resistivity of conductor (ρ) is the resistance offered by a unit length and a unit area of cross-section of a conductor. It is expressed in ohm. m.
Resistance offered by a conductor depends on certain factors which are shown below:
Length is directly proportional to the resistance offered by the conductor.
R∝L………(equation-1)
Cross-sectional area is inversely proportional to the resistance offered by the conductor.
R∝ \(\frac{1}{A}\)………(equation-2)
Combining equation (1) and (2), it has been experimentally found that:
R = ρ × \(\frac{L}{A}\) ………(equation-3)
R = Resistance offered by the conductor
L = Length of the conductor
A = Area of cross-section of the conductor
ρ = Resistivity of the conductor
A conductor dissipates heat current flows through it. It is stated that, amount of heat liberated is directly proportional to the square of current, resistance and time (in seconds) through which the current is flowing through the conductor.
H = I2 Rt
H = Heat dissipated by the conductor
I = Current flowing through the conductor
R = Resistance of the conductor
t = Time through which current is flowing through the conductor
The molecular arrangement of solid or the properties of a solids are stated below:
It predicts the strength of a derived relationship between a dependent variable and the independent variable. The maximum value of the coefficient is ±1 which indicates maximum strength and the minimum value of the coefficient is 0 which signifies no correlation between the two variables. A negative value indicates that the relationship between the independent and the dependent variable is decreasing and, a positive value signifies that the relationship between the independent and the dependent variable is direct in nature. The formula of Pearson’s correlation coefficient for a linear trend is shown below:
R = \(\frac{\sum(x\ -\ \bar{x})(y\ -\ \bar{y})}{\sqrt{\sum(x\ -\ \bar{x})^2 \ ×\ \sum(y\ -\ \bar{y})^2}}\)
R = Pearson's Correlation Coefficient
x = each value of independent variable
\(\bar x\) = mean value of independent variable
y = value of dependent variable corresponding to the value of x
\(\bar y\) = mean value of dependent variable
In this process, a very simple circuit was formed with a voltage source and a limiting resistance connected in series in the circuit. Each type of wire was used as the connecting wire in the circuit for different trials. When the voltage source was turned on, due to flow of current through the circuit, some amount of heat was dissipated by the resistor as well as the wire. As the exploration is mainly about the resistivity and thermal expansion of wire, the temperature of the wire was monitored using a temperature probe. At an interval of 10.00 ℃ from the room temperature of 20.00℃, the battery was turned off and the expansion of length of wire was measured using a meter scale and a vernier calipers. The length of the wire was measured till a temperature of 60.00℃. Using the length of extension of wire at different temperature, the coefficient of thermal expansion of different materials were calculated and its relationship with the resistivity of wire was studied.
In a research article titled as – ‘Thermal expansion of solids’ in Moscow Izdatel Nauka (1974), by Novikova, So I, it was concluded that the coefficient of thermal expansion is a function of time and not a constant parameter. As a result, it could be assumed that for different materials, if the experimental duration varies, then the coefficient of thermal expansion would behave differently.
It was assumed that with an increase in resistivity of material, the coefficient of linear thermal expansion of the material would increase. The reason behind this hypothesis could be explained in two steps. Firstly, with an increase in temperature of a conductor, the resistance of the conductor increases. This was because, with an increase in temperature, the molecules of conductor (solid) vibrate with an intensified rhythm which, consequently, has increased the collision between the molecules and the free moving electrons. Due to increased collision, the flow of electrons was interrupted which resulted in an increase in resistance of the conductor. As the resistance of the conductor was increased due to microscopic property, it was evident from the explanation that the increase in resistance was due to an increase in resistivity of the material.
Secondly, with an increase in coefficient of thermal expansion, it could be assumed that the intensity of vibration of molecules of solid would increase due to which the length of the solid eventually increase. As the intensity of vibration of molecules would increase with an increase in coefficient of linear expansion, the resistivity would increase.
The resistivity of material, measured in ohm. m was chosen to be the independent variable of the exploration. Five materials were considered in this exploration to study the variation of the dependent variable with respect to the change in resistivity of materials. The five materials chosen for exploration are as follows: 1. Copper, 2. Aluminum, 3. Nichrome, 4. Iron and 5. Tungsten. These five materials are chosen because of the exploration methodology. The dependent variable is obtained from a length of metal wires (solid) of different resistivity was measured against temperature of the material using a scatter plot. In the process, as resistivity of the material is the independent variable, five materials are chosen in such a way that those are used as materials of wire and conduct electricity. Moreover, in the experimental procedure, the principle of Joule’s Law of Heating is used to raise the temperature of the wires. As a result, it is very important for the materials to conduct electricity through them. As the materials are chosen in accordance with the requirement of flow of current through it, the interval of resistivity between any two material is not constant. Rather, the intervals are quite irregular and invariably large.
2.8 × 10-8
1.1 × 10-6
1.0 × 10-7
4.9 × 10-8
The coefficient of linear thermal expansion was considered to be the dependent variable of the exploration. It was measured in the units of ℃-1. For each wire (material), the coefficient of linear thermal expansion was calculated using the formula of change in length of a solid material due to change in temperature (refer to section 3.1), from the experimentally observed value of the exploration.
There were no controlled variables in this exploration. This was because, both the dependent and the independent variable are intensive properties of matter. However, few physical parameters were considered to constant throughout the experimental procedure to maintain a uniformity in the exploration.
The wires were not heating by consuming fossil fuel which would result in emission of green house gases. Rather, the principles of Joule’s Law of heating have been used in this exploration to heat to wires to make the experimental procedure an environment friendly process.
Sample Calculation
Mean length at 30.00℃ = \(\frac{100.01+100.02+100.02}{3}\) = 100.02 cm
SD at 30.00℃ = \(\sqrt{\frac{(100.01-100.02)^2+(100.02-100.02)^2+(100.02-100.02)^2}{3}}\) = 0.01
Sample Calculation
Change in temperature between 20.00℃ and 30.00℃ = 30.00 - 20.00 = 10.00℃ Change in length at 30.00℃ = 100.02-100.00 = 0.02cm Coefficient of Thermal Expansion of copper = \(\frac{0.02}{100.00×10.00}\) = 2.00×10-5 ℃-1 (refer to eq. 5) For calculation of mean and standard deviation, refer to the sample calculation shown in section 10.1 under Figure 5.
Calculation of mean and standard deviation of coefficient of thermal expansion is same as that of length. For calculation of mean and standard deviation of coefficient of thermal expansion, refer to the sample calculation shown in section 10.1 under Figure 5.
Coefficient of Thermal Expansion (℃-1)
1.79 × 10-5
2.8 × 10-8
2.67 × 10-5
1.1 × 10-6
1.33 × 10-5
1.0 × 10-7
1.15 × 10-5
4.9 × 10-8
0.33 × 10-5
Variation In Coefficient Of Thermal Expansion (℃-1) With Respect To Resistivity (ohm. m) Of Different Material.
Analysis using Scatter-plot:
Variation Of Coefficient Of Linear Expansion (℃-1) Versus Resistivity Of Material (ohm.m).
Data Interpretation of Figure 7:
Analysis using Bar Graph
Variation Of Coefficient Of Linear Expansion (℃-1) Of Five Materials
Data Interpretation of Figure 8 and Figure 9:
Calculation of Pearson’s Correlation Coefficient
In this process, a processed data table would be used which will have certain notations which were explained in the paragraph. x denoted the resistivity of the material in ohm.m, \(\bar x\) denoted the mean resistivity in same unit, y denoted the coefficient of thermal expansion in ℃-1, \(\bar y\) denoted the mean coefficient of thermal expansion in the same unit.
\(x-\bar x\)
\(y-\bar y\)
\((x-\bar x)(y-\bar y)\)
\((x-\bar x)^2\)
\((y-\bar y)^2\)
2.8 × 10-8
2.67 × 10-5
1.23 × 10-5
-4.21 × 10-8
1.17 × 10-5
1.52 × 10-10
4.9 × 10-8
0.33 × 10-5
-1.12 × 10-5
3.85 × 10-8
1.17 × 10-5
1.27 × 10-10
1.0 × 10-7
1.15 × 10-5
-0.31 × 10-5
1.05 × 10-8
1.17 × 10-5
0.09 × 10-10
1.1 × 10-6
1.33 × 10-5
-0.13 × 10-5
0.44 × 10-8
1.17 × 10-5
0.02 × 10-10
1.79 × 10-5
0.33 × 10-5
4.54 × 10-8
18.71 × 10-5
0.11 × 10-10
Calculation
\(\bar x=\frac{\sum x}{5}=\frac{0.0171013}{5}\) = 0.0034203
\(\bar y=\frac{\sum y}{5}=\frac{0.0000729}{5}\) 0.= 0.00001458
\(\sum(x-\bar x)(y-\bar y)\) = 5.68 × 10-8
\(\sum(x-\bar x)^2\) = 0.000233
\(\sum(y-\bar y)^2\) = 3.01 × 10-10
Let, the Pearson’s Correlation Coefficient be ℜ.
R = \(\frac{\sum(x\ - \bar{x})(y\ - \bar{y})}{\sqrt{\sum(x\ -\ \bar{x})^2\ ×\ \sum(y\ -\ \bar{y})^2}}\)
R = \(\frac{5.68\ ×\ 10^{-8}}{\sqrt{0.000233 \ ×\ 10^{-10}}} = \frac{5.68\ ×\ 10^{-8}}{\sqrt{7.04 \ ×\ 10^{-14}}} = \frac{5.68\ ×\ 10^{-8}}{\sqrt{2.654 \ ×\ 10^{-7}}} \)
R = 0.219
The resistivity of a material and the coefficient of thermal expansion of a solid material both are intensive microscopic properties of any material. Resistivity of a material is the physical parameter of any material which determines the excitation or the availability of free electrons in any material. For example, if the resistivity of any substance is more, it signifies that in room temperature, the substance does not release a high number of free electrons. Thus, it can be stated that the band gap energy between the conduction band and valence band of the substance is high which restrict the flow of charge and hence, current. Thus, resistivity of any substance deals with the energy content of electrons and availability of free electrons in the atoms or molecules of any substance. Substances like copper exhibit half-filled valence shell electronic configuration, which makes them more stable than other substances. As a result, the ionization potential of copper increases and hence the availability of free electrons in room temperature decreases. As a result, resistivity of copper is more than that of other elements or compounds. On the other hand, coefficient of thermal expansion determines the measure of vibration or excitation (internal kinetic energy) of the molecules or atoms of any substance. For example, if the coefficient of thermal expansion of any substance is more, it signifies that with a unit increase in temperature, the vibration of the particles (molecules or atoms) of the substance will increase resulting in an expansion in volume (length and surface area). The coefficient of thermal expansion does not affect the energy content of the electrons present at the valence shell of the atom of the substance. Hence, it could be stated that the there is no direct or indirect relationship between the coefficient of thermal expansion of a substance and the resistivity of the substance.
How does the coefficient of thermal expansion of a wire (expressed in ℃-1) due to heating by continuous flow of current through the wire depend on the resistivity of the wire, expressed in ohm.m (copper - 0.0171, aluminum - 2.8×10-8, nichrome - 1.1×10-6, iron - 1.0×10-7 and tungsten - 4.9×10-8), determined using length versus temperature scattered graph?
The coefficient of thermal expansion and the resistivity are two physical parameters of any material which are completely independent and has no relationship between each other.
Strength
Random Error
There were no traces of random error obtained in the experiment.
Methodological Limitation
Further Extension
As an extension to this exploration, the relationship between the resistance of the material and the coefficient of thermal expansion of the material could be studied. In this exploration, as it was found that there exists no relationship between coefficient of thermal expansion and resistivity, however, the resistance of any substance does not depend on the resistivity of the material only. This exploration would provide information about the effect of different intensive properties of any substance except resistivity on the resistance of the substance. The research question of the exploration could be framed as: “How does the resistance of any conductor depends on the coefficient of thermal expansion of the material of the conductor (expressed in ℃-1) determined using Ohm’s Law?”
The exploration can be designed by making a very simple circuit with a voltage source and a limiting resistance in series orientation. The connecting wire will be changed for different trials. Here, five different types of connecting wire can be taken to get a wider range of coefficient of thermal expansion. The five different wires chosen can be – copper, aluminum, nichrome, iron and tungsten. Switching on the battery, current will flow through the circuit, and that current should be measured using a multimeter. Now, using Ohm’s Law, the resistance of the circuit will be calculated for different trials. Once the resistance of the circuit is achieved, the value of resistance of the limiting resistance should be subtracted from the resistance of the circuit to get the resistance of the connecting wire. Then, a comparative analysis can be done between the resistance of different materials and coefficient of thermal expansion of the solid. In this exploration, the length of the wire, area of cross section of the wire, the voltage of the battery used in the circuit and the limiting resistance of the circuit should be controlled.
For calculation of mean and standard deviation in Figure no. 12 to 15, refer to the sample calculation shown in section 10.1 under Figure 4.
For calculation of mean and standard deviation in Figuure no. 16 to 19, refer to the sample calculation shown in section 10.1 under Figuure 4.