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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound electrons are electrons with negative charge in an orbit around the nucleus; the bound electrons cannot leave their orbit. The free electrons are not in orbit but free to go where ever in the metal. The free electrons are the charge carriers, the charge is a negative charge in electricity and in conduction of heat the free electrons carry heat energy, which mean they are responsible for electrical conduction and heat conduction. As the plastics have no free electrons there can be no conduction, but as metals have free electrons this allows conduction to happen in the metals. An atom of a metal: - - + - - - An atom of a plastic: - + - There are no free electrons in the plastic so there will not be any conduction, because the free electrons are the charge carriers, without charge carriers no conduction. The free electrons are sub-atomic particles and have a negative charge so when there is a potential difference voltage the free electrons will move to the positive in one direction this is because opposite charges attract. The free electrons are moving in one direction but randomly, the free electrons don't all travel in straight lines but they are going in the same direction. This is a crystal of a metal or a space lattice: - The atoms are fixed meaning that they have a regular arrangement and that they are rigid and orderly. There will be collisions between the free electrons and the atoms. The atoms may not move and are rigid but they do vibrate. The reason why the free electrons are called charge carriers is because current is the flow of charge and free electrons carry this flow of charge. Free electrons move at high speeds as they travel large distances between collisions with the metal atoms, they can transfer energy at very quickly. Electrical conduction happens best in metals like silver, copper, aluminium, iron. Poor conductors and good resistors of electrical energy are glass, plastics, wood. Good conductors have low resistance. Bad conductors have high resistance. Resistance: - Resistance is the opposing of the flow of current. Electrical resistance is the property where electric flow or electric energy is transformed into heat energy this heat energy will then be opposing the electrical energy. Resistance involves the collisions between the free electrons and the vibrating atoms, to be more precise the collisions between free electrons and the bound electrons which makes up the structure of the conductor. The collisions with the free electrons and the vibrating atoms will slow down the flow of free electrons therefore slow down the flow of electricity. The resistance of a wire is directly proportional to its length and inversely proportional to its cross sectional area. The temperature and the number of collisions are related, if the temperature rises then the atoms will vibrate more. The more vibrations of the atoms, would decrease the space so there will be more collisions and the more collisions will slow down the charge carriers free electrons so there will be more resistance. From this diagram we can see that the charge carriers the free electrons are having more collisions because the atoms are vibrating more because of the higher temperature. P2a SAFETY Safety will always be a vital factor when carrying out any experiment. This experiment will not involve the use of electricity through the mains. This amount of electricity will be around 240volts coming through the mains, this is too much because it is dangerous enough too kill you. So the voltage should be lowered in this experiment so that is why cells are used and not the mains, this also is could because a low voltage would keep the temperature of the wire the same. It is much safer to experiment with circuits powered by low voltage source. In the experiment to make the level of voltage to be safe I would use battery cells which have low voltages, the lowest voltage the cells could be 2 volts and the highest would be 5volts so this source would be much safer. Another safety precaution that I will take is will I will not work near any desktops with water taps. The reason for this is that if water goes into the circuit or more importantly into the electrical socket it could results to hazards like fire hazards, so I will for safety do the experiment on a side desktop away from gas taps and water taps. P4a FAIR TEST For this experiment to be a fair and get reliable results I will have to undertake a fair text. The fair test will have to undertake these following things: - The non-variables or the things that I will keep the same through out the experiment will be the temperature; at room temperature cross sectional area of wire which is 0.5mm or 0.05cm, the material of the wire will be kept the same which will be Nichrome. I will also keep the same amount of cells which will be three cells throughout the experiment; keeping the same amount of cells the same will make the experiment a fair test because if the amount of cells were changed each time then this will affect the results because if the cells were changed the starting voltage of the power supply would be different so you the results would not be reliable. When the temperature of the wire is increased then the atoms of the wire will vibrate more and this will decrease the space in the space lattice so there will be more collisions and there will be more resistance. We know that if the temperature is not kept the same throughout the experiment it will not be a fair test, because of the changing temperature. If you keep the non-variables the same then you can get correct and fair results as you change the variable which in this case is the length of the wire. Each length will be repeated twice and the average taken to get fair results. P4b LIST OF EQUIPMENT AND APPROPRIATENESS Equipment/Instruments: - "¢ Ammeter "¢ Voltmeter "¢ Crocodile clips "¢ Three cells "¢ Nichrome wire "¢ Meter ruler "¢ Board to put cells in. For the measurement instruments they will need a least count for appropriateness and consistency. The least count for the meter ruler will be 1mm or 0.1cm. The voltmeter will be in volts and the least count will be 0.1volts. For the ammeter the measurement will be amperes and the least count will be 0.1amperes. P6a THEORY AND PREDICTION The factors that affect the resistance of a wire include the following: - "¢ Material "¢ Temperature "¢ Cross sectional area of wire "¢ Length of a wire The material of a wire has three things that affect the resistance: 1. The type of metal: - each different types of metal are structured differently; their space lattice or crystals are different to one another. The interatomic distance was big then there would be fewer collisions so there will be less resistance. If the interatomic distance in the structure of the metal was small there would be more collisions therefore more resistance because the free electrons flow has been slowed. 2. The amount of free electrons: - this also affects the resistance, if the amount of free electrons in one metal is more than in another metal then the metal with more free electrons will have less resistance. The more the number of free electrons then less resistance, the less the number of free electrons then the more the resistance. The reason for this is that if there are more free electrons means that more electrical charge is been carried so there is less resistance. 3. Arrangement of atoms in the metal: - if the structure of the atom or arrangement is different then this can affect the resistance. If the arrangement of the atoms has an atom in the middle there would be more collisions so there will be more resistance. If the structure had no atom in the middle there would be fewer collisions therefore less resistance. This diagram shows two differently arranged atoms one is the body-centred cubic bcc arrangement the other is faced-centred cubic FCC arrangement. The bcc arranged atom has less atoms and only one atom in the middle so it would have less collisions than the FCC arrangement because the FCC arrangement has more atoms and more in the middle, this will increase the amount of collisions and then there will be more resistance. The temperature will affect the resistance of a wire. A falling temperature will increase the conductivity. Most of the resistance to the motion of the free electrons therefore resistance of the wire is the thermal vibrations of the atoms. If the temperature is reduced to absolute zero then thermal vibrations will stop, then this will stop the resistance due to temperature. The cross-sectional area of a wire will affect the resistance. A larger cross sectional area will mean a larger current I and that makes smaller resistance. A smaller cross-sectional area has a smaller current which is a bigger resistance. The resistance of a wire is inversely proportional to its cross-sectional area. The length of a wire affects the resistance. If I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. This means that the resistance is directly proportional to the length of the wire. When the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire. L no: of collisions/sec. Resistance=R 2L 2N Resistance= 2R p8a Ohms law: - these are laws that relate to current, potential difference, and resistance. The resistance is measured in ohms, the potential difference is measured in volts and the current is measured in amperes amps. Ohms law states the amount of steady current through materials is directly proportional to the potential difference Voltage. An example that proves this is that if the potential difference voltage between two ends of a wire is tripled, the current is tripled. Ohms law also says that I=V/R, or the current in the conductor equals the potential difference voltage across the conductor divided by the resistance of the conductor. It also says that the potential difference across a conductor equals the current in the conductor times its resistance, V=IxR. Ohms law also explain that resistance of a material is its potential difference voltage divided by the current, R=V/R. The larger the resistance the greater the voltage is needed to push each ampere through it: there is a resistance of one ohm and a voltage p.d of one volt will drive a current of one ampere through it. Conductors that obey ohms law are called ohmic conductors; conductors that don't obey ohms law like semiconductors are called non-ohmic conductors. Resistivity: - is the electrical resistance of a conductor of unit cross-sectional area and unit length. A property of each material resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. The unit of resistivity is ohm-meter or ohm-centimetre. A high resistivity is in poor conductors, a poor conductor has high resistance, so high resistivity has high resistance and a conductor with low resistivity will have low resistance. Resistivity P is quantitatively equal to the resistance R of e.g. a wire times its cross-sectional area A and then divided by its length L. P =R A /L. Resistance = R= P L /A p8a Method: - 1. First I will connect the cells to the circuit board where the cells are placed. 2. Then I will then use the three pairs of connecting wire crocodile clips to connect the cells to the ammeter and the wire and back to the cells to make a complete circuit. I will also add a voltmeter opposite to the power pack/in parallel. 3. Thirdly the wire will be on a board and on that there will be a ruler so I can measure the different lengths. 4. Next, when I have measured one length I will break the circuit by taking off a connecting wire, this is so that the wire does not change temperature and affect results. 5. Then when I am ready to take the next length I will connect the circuit and move the crocodile clip to the new length. 6. I will do all the lengths three times to find reliable readings. I will also have a good range of lengths so I can find a full set of results to see how the length of a wire affects the resistance of the wire. Cells Voltmeter V Before the power is put on the ammeter is at zero. The point where the crocodile clip and the wire meet is at zero. The ammeter has two scales and the first one is 0.00amps to 1.00amps this is for a low voltage that I am using so this is the appropriate scale that I should use as I am using low voltage cells. The voltmeters least count is 0.01volts and before the experiment the voltmeter was at absolute zero. In my method I could have used digital ammeter but there was not one provided, a digital display ammeter would give me even more accurate readings. p6b NO: OF READINGS The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm 55.00cm 60.00cm 65.00cm 70.00cm 75.00cm 80.00cm 85.00cm 90.00cm 95.00cm 100.00cm This is my table where I am going to show all my results for my experiment. I will take the readings from the voltmeter to find the potential difference voltage, I will take the ampere readings to find the current I. The table has three sets of reading to put in because this means I am repeating the readings and trying to find consistent results and reliable results, any anonymous results will be marked with an asterisk and then repeated again. The average will be used to find average resistance for the three sets of results and see if they are reliable e.g. 3.1Ω, 3.2Ω, 3.1Ω average will be 3.1Ω+3.2Ω +3.1Ω=9.4Ω divided by 3= 3.1Ω. The average can let me set a graph showing how the average resistance against the length of a wire. The final column will let me work out the resistance once I have collected all the data. The table has a suitable range for the length of the wire, this is because this is my variable and I need to get a full range of results. The table has the least counts of scale units and the decimal places I am using. p8b PRELIMINARY WORK The dissipation of electric energy in the form of heat, even though small, affects the amount of electromotive force, or driving voltage, required to produce a given current through the circuit. In fact, the electromotive force V measured in volts across a circuit divided by the current I amperes through that circuit defines quantitatively the amount of electrical resistance R. Precisely, R = V/I. Thus, if a 12-volt battery steadily drives a 2-ampere current through a length of wire, the wire has a resistance of 6 volts per ampere, or 6 ohms. Ohm is the common unit of electrical resistance, equivalent to one volt per ampere and represented by the capital Greek letter omega Ω. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Resistance also depends on the material of the conductor. The Electrical resistance of a conductor is dependent on cross-sectional area and length. A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents. High resistivity designates poor conductors. Most resistance is down to the motion of free electrons comes from the thermal vibration of the atoms; if the temperature is reduced to almost absolute zero, where thermal motion essentially stops, conductivity can increase several thousand times over. The value of resistivity depends also on the temperature of the material; tabulations of Resistivities usually list values at 20 C. Resistivity of metallic conductors generally increases with a rise in temperature; but resistivity of semiconductors, such as carbon and silicon, generally decreases with temperature rise. The resistivity of an exceedingly good electrical conductor, such as hard-drawn copper, at 20 C 68 F is 1.77 10-8 ohm-metre, or 1.77 10-6 ohm-centimetre. At the other extreme, electrical insulators have Resistivities in the range 1012 to 1020 ohm-metres. The unit of resistance is the ohm. In the metre-kilogram-second mks system, the ratio of area in square metres to length in metres simplifies to just metres. Thus, in the metre-kilogram-second system, the unit of resistivity is ohm-metre. If lengths are measured in centimetres, resistivity may be expressed in units of ohm-centimetre. OBSERVATIONS o4b o4a o6b o6b table of results: - The cross-sectional area will remain at a constant of 0.05cm through out the experiment. READING NUMBER ONE READING NUMBER TWO READING NUMBER 3 AVERAGE RESISTANCE Average= 1.d.p Ω Length of Nichrome wire L+-0.01cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I=Current R=V/I 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p Resistance Ohms R=resistance V= voltage p.d I= current R=V/I 2.d.p 05.00cm 1.18v 2.22 0.53 Ω 1.18v 2.22 0.53 Ω *2.90v 1.18v *2.97 2.22 0.53 Ω 0.5 Ω 10.00cm 1.64v 1.91 0.86 Ω 1.64v 1.91 0.86 Ω 1.63v 1.90 0.86 Ω 0.9 Ω 15.00cm 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.99v 1.67 1.19 Ω 1.2 Ω 20.00cm 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 2.27v 1.49 1.52 Ω 1.5 Ω 25.00cm 2.49v 1.34 1.86 Ω 2.49v 1.34 1.86 Ω 2.51v 1.33 1.89 Ω 1.9 Ω 30.00cm 2.67v 1.22 2.19 Ω 2.67v 1.22 2.19 Ω 2.67v 1.21 2.21 Ω 2.2 Ω 35.00cm 2.82v 1.12 2.52 Ω 2.83v 1.11 2.55 Ω 2.82v 1.12 2.52 Ω 2.5 Ω 40.00cm 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.95v 1.03 2.86 Ω 2.9 Ω 45.00cm 3.06v 0.96 3.19 Ω 3.06v 0.95 3.22 Ω 3.06v 0.96 3.19 Ω 3.2 Ω 50.00cm 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.15v 0.90 3.50 Ω 3.5 Ω 55.00cm 3.24v 0.84 3.86 Ω 3.22v 0.83 3.88 Ω 3.22v 0.83 3.88 Ω 3.9 Ω 60.00cm 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 3.31v 0.75 4.41 Ω 4.4 Ω 65.00cm 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 3.38v 0.73 4.63 Ω 4.6 Ω 70.00cm 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 3.43v 0.71 4.83 Ω 4.8 Ω 75.00cm 3.49v 0.68 5.13 Ω *4.00v 3.49v *0.98 0.67 *4.08 Ω 5.21 Ω 3.49v 0.68 5.13 Ω 5.2 Ω 80.00cm 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 3.54v 0.64 5.53 Ω 5.5 Ω 85.00cm 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 3.58v 0.61 5.87 Ω 5.9 Ω 90.00cm 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 3.62v 0.59 6.14 Ω 6.1 Ω 95.00cm 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 3.66v 0.56 6.54 Ω 6.5 Ω 100.00cm 3.69v 0.54 6.83 Ω 3.69v 0.54 6.83 Ω *4.29v 3.70 *0.32 0.54 6.85 Ω 6.8 Ω o8a: - *note: the circled and crossed out results are anonymous so I repeated the reading to get a reliable reading and result, I did this because I thought that the reading was wrong and I needed to do it again, it was common sense because the anomalous results obviously didn't follow certain patterns in the table. I did the reading again and I got the same as I got in the other two sets of readings for that length. o6b: - the table is an accurate record of observations because each column of reading is to the same decimal place and is accurate. * In the planning I planned to use an analogue ammeter and voltmeter but I was able to use a digital ammeter and voltmeter as the teacher could get one in the last lesson of the practical. * I rounded the final average resistance to one decimal place because this is common sense and this will make it easier to plot the graph. ANALYSIS A.2a After completing the experiment and getting a full range of reliable results from the experiment, I have found out that in general that as the length of the wire goes up the resistance of the wire also goes up. This states that my theory and prediction is correct. I have found out that this general finding says that the resistance of the wire is proportional to the length of the wire and in many occasions the length is directly proportional to the resistance of the wire. A4.b PATTERN IN READINGS: - After collecting my results from the observations I carried out I can now analyse the results and make sense of the readings in the table of results. The table of results has a general pattern that suggests that as the length of the wire is increased the current decreases and therefore the resistance increases as my theory states through different rules. This general finding means that the resistance of the wire is proportional to the length of the wire. The readings from the table of result shows as the length of the Nichrome wire increases the current in the circuit decreases and as the theory explains the resistance should increase and this occurs in the table of readings. The readings in my results table show that the length is proportional to the resistance of a wire and in some cases nearly directly proportional and other cases exactly the length is directly proportional to the resistance of the wire. E.g. the result from 5cm to 10cm, the average resistance for 5cm is 0.5Ω and the average resistance for 10cm is 0.9Ω, this shows that the length of the wire is proportional to the resistance, if it was directly proportional then the resistance for 10cm will be 1.0Ω, this is because when the length is doubled then the resistance is doubled this then would be directly proportional, this result was not accurate by 0.1 because I could of made minor human error like not putting the exact length. The results from 30cm has the resistance of 2.2Ω and the and the doubled length that is 60cm has the resistance of 4.4Ω, this shows that as the length is doubled the resistance is doubled which means that the length is directly proportional to the resistance of the wire, these result from 30cm to 60cm proves that my theory that the length of wire is directly proportional to the resistance of a wire. The mistakes I made called experimental like not breaking the circuit when changing the length could of have affected the temperature of the wire therefore create more vibrations for the atoms in the wire and cause a bigger resistance then it should have been without the experimental error. A4a A6a A4b Graph of results: - A4b PATTERN ON GRAPH: - The pattern that appears on the graph of results tells us that the length of the wire is directly proportional to the resistance of the wire. The line of best fit provides the graph with the results. The results that prove that the resistance of a wire is directly proportional to its length include the result of the lengths 15cm to 30cm, 15cm has the resistance of 1.1Ω and the resistance of 30cm is 2.2Ω this shows as the length is doubled the resistance also is doubled which means that the resistance of the wire is directly proportional to the length of the wire, this result shows the graph of results to state my theory and to be correct as the result agrees with my theory. Other results in my graph also state that the resistance of the wire is directly proportional to its length. They include: - 26cm that had the resistance of 1.9Ω and then when the length was doubled to 52cm the resistance was 3.8Ω, this result shows that as the length is doubled the resistance is also doubled which means that the length is directly proportional to the resistance of a wire, the result again proves my prediction correct because I predicted that the length of wire is directly proportional to resistance of the wire, the results from the graph state this with these results. The other results from the graph also tell us that the length is proportional to the resistance of the wire they include: - 5cm R=3.5Ω and 10cm R=7.5Ω, 20cm R=1.5 & 40cm R=2.9Ω. These results that are not directly proportional but it is only by a few 0.1 ohms this could be down to not drawing a 100% straight line and this cause these results to be a bit off. A6b A8a: - explanation using theory From my readings in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω, so the results from the table of readings state my theory and prediction, this also proven by the diagram theory I did: - From the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. The resistivity is quantitatively equal to the resistance, from the results this would be the resistance R of the wire times the cross-sectional area A divided by the length L of the wire. From this you can find out if the resistance is directly proportional to the length and inversely proportional to the cross-sectional area and see if your results are correct. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory. Results are correct because the resistivity and the resistance are quantitatively equal. This result also shows that my main prediction that is, the length of the wire is directly proportional to the resistance of the wire; this prediction was proven by this result as the theory of resistivity and the equations P =R A /L and R= P L /A were worked out using my result that showed that the resistance and resistivity were quantitatively equal this is proof that my results are reliable and follow the theory of rules in resistance and resistivity and states my prediction. A8b: - conclusion of results and analysis. Throughout the analysis of the results I have found many results from my readings in the table of results and the results from the graph, they show the relationship of the length of the wire and to the resistance of the wire. The general conclusion I have made after analysing these results is that the length of a wire is directly proportional to the resistance of a wire. The proof of this conclusion lies within the results in the readings and the graph which shows the relationship between the length of the wire and its resistance. The results from the table of readings and the graph do agree with my theory this shows that my results are proof of the theory. This is shown e.g. in the experiment the results from table which I've conducted, I have found results that show the length of the wire is proportional to the resistance of the wire. On occasions in the readings I have seen that the length of the wire was directly proportional to the resistance of the wire. The results 30cm R=2.2Ω, 60cm R=4.4Ω shows that as the length is doubled the resistance is doubled as well which means that the length is directly proportional to the length as I predicted. This result proves my theory that if the length is doubled the number of collisions will also double then so will the resistance be doubled; this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this is the case in the result 30cm R=2.2Ω, 60cm R=4.4Ω. Also shown in from the graph I have found out patterns that generally tell me that the length of the wire is directly proportional to the resistance of the wire. This was the case in the graph results: - 15cm to 30cm as my theory explains when the length is doubled the resistance should also doubled this was the case it went from 1.1Ω ohms to 2.2Ω, so this means that the length of the wire was directly proportional to the resistance of the wire, this states my theory which says 'I predict that if the length was doubled so will the resistance, I can also predict that if the length was tripled the resistance will also triple, this is because of the rule the resistance of a wire is directly proportional to the length of the wire.' The result 15cm=1.1 ohms and 30cm=2.2 ohms suggests that when the length is doubled the number of collisions will also double then so will the resistance be doubled, this shows that collisions increase the resistance will also increase. This result proves my theory which states that if I were to double the length of the wire this would double the collisions, so a longer wire will have a bigger resistance. Doubling the length would double the number of collisions and therefore will double the resistance. In detail my theory explains this result, when the length is more than before there are more number of free electrons and when there is more free electrons there will be more collisions, the more collisions will slow down the charge carriers free electrons and this would cause bigger resistance. If a length was doubled this would double the amount of free electrons and double the amount of collisions between the free electrons and the vibrating atoms, this would slow down the free electrons the charge carriers x 2 times by two and therefore double the resistance, this was proven by the results; 15cm R=1.1Ω and 30cm R= 2.2Ω, this is exactly what I predicted through my theory. This result I worked out about resistivity also can prove that the results are proof of the theory. P =R A /L. E.g. A=0.05cm × R=6.83Ω ÷ 100.00cm= 0.3415 ohmic-centimetre = resistivity. To prove the result is correct I will know find out the resistance R using this equation: - R= P L /A. P resistivity =0.3415 × 100.00cm ÷ A 0.05cm = 683 ohmic-centimetre, this is quantitatively equal to 6.83Ω. So, these results show my theory of resistivity is correct and that my results are correct and follow the theory of resistivity and show results are correct so the results are the proof of the theory. EVALUATION E.2a: - This coursework that I undertook on the factors that affect the resistance of a wire was quite challenging but at the same time enjoyable to carry out. The reason for which I found it enjoyable was that the physics topic the coursework came under is one of my favourite areas as I have an interest within physics. The coursework to me was easy in parts but other parts of the coursework were challenging as I had to implicate the theory to real experiments and I also enjoyed the independent research that was needed. The coursework made me understand the factors affecting the resistance of a wire, and it also allowed me to understand certain laws in physics like the ohms law, and the theory of conduction. E.4a: - The coursework I took on was accurate this could be proven by the results and the graph. The points that lie on my graph is quite adequate out of twenty points of results ten where on the graph line of best fit the other eight were only a few decimal places on the graph below or above, they were slightly off the line of best fit, while only two were anomalous but by only small margins. The anomalous results were for 60cm and 10cm readings as the points were the furthest away from the graph or line of best fit. The accurateness was also proven by the graph passing through the origin which is the co-ordinates 0, 0, this level of accuracy suggest that my results and graph is reliable and accurate. E.4b: - evaluation of method For this coursework the method was very important because if the method was faulty and had a lot of gaps in it the results would be affected therefore most of the coursework. The method that I used was generally a good one to use as with anything it could have improved e.g. I could have used a power pack connected to the mains as the power supply, the power pack could easily change the voltage you want and keep it the same, this would also subsequently avoid complications of finding the voltage with a voltmeter and make the method quicker. A power pack would also allow me to switch of the circuit and not affect the temperature and then the results. In the method instead of taking the readings at every 5cm I could take it at every 2cm or even 1cm to give a better range and reliable results. Other improvement I could make to the method would be to make more accurate by putting the crocodile clips together without any part of the clip not connected. These changes and improvement would make my method much better and get rid of the faults that affected the results I got. E.6a: - evidence. A table to show the reliability of the results and the anomalous results present. LENGTH cm RESISTANCE From GRAPH Ω ohms RESISTANCE from EXPERIMENT Ω ohms DIFFERENCE 2.d.p % DIFFERENCE =diff÷ R graph × 100 1d.p * 5.00 0.35 0.50 0.15 *42 % 10.00 0.80 0.90 0.10 1 % 15.00 1.10 1.20 0.10 *9 % 20.00 1.50 1.50 0.00 0 % 25.00 1.82 1.90 0.08 4 % 30.00 2.20 2.20 0.00 0 % 35.00 2.52 2.50 0.02 1% 40.00 2.90 2.90 0.00 0 % 45.00 3.25 3.20 0.05 1 % 50.00 3.55 3.50 0.05 1 % 55.00 3.90 3.90 0.00 0 % 60.00 4.25 4.40 0.15 3 % 65.00 4.60 4.60 0.00 0 % 70.00 4.91 4.80 0.11 2 % 75.00 5.25 5.20 0.05 1% 80.00 5.60 5.50 0.10 1 % 85.00 5.90 5.90 0.00 0 % 90.00 6.20 6.10 0.10 1 % 95.00 6.50 6.50 0.00 0 % 100.00 6.80 6.80 0.00 0 % AVERAGE ERROR = 3.35 % Here you can see the average error in % is acceptable within experimental error, this is evidence to justify that my conclusion, because it is within the scope of experimental error so the results I got should be considered reliable because it is a low % of error, so this would prove that the length of the wire is directly proportional to the resistance of the Nichrome wire and also prove my analysis and conclusion to be reliable. In the table which shows the reliability of results also showed that my experiment had only two anomalous results, this is acceptable because within the experiment I could have made only one significant error to cause this anomalous result. The rest of the results were reliable as the table shows. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.0 Ω 0.20 ÷ 5.0 × 100% = 4 %. This is within experimental error. Length =30.00 cm, Resistance = 2.20 Ω Doubled = 60.00cm, Resistance = 4.40 Ω 0.00 Ω off 4.40 Ω 0.00 ÷ 4.40 × 100% = 0%. This is within experimental error, because there is no experimental error in this result. Length = 15.00cm, Resistance = 1.10 Ω Doubled = 30.00cm, Resistance = 2.20 Ω 0.00 Ω off 2.20 Ω 0.00 ÷ 2.20 × 100% = 0%. There is no experimental error here. Length = 35.00cm, Resistance = 2.50 Ω Doubled = 70.00cm, Resistance = 4.80 Ω 0.20 Ω off 5.00 Ω 0.20 ÷ 5.00 × 100% = 4%, again this is within experimental error. All these results I have showed show that the experimental error was low or there was no experimental error, so the results can be expected to be reliable and therefore the results are reliable enough to justify my conclusion. With the table showing the reliability of results with the difference and the individual evidence of results I have picked out it can be said my results are reliable. E6b: - further work. For the coursework if I were to do it again I would do further work to get more evidence. The further that I would carry out to get more evidence would consist of doing a wider range of lengths of wire and also to try other materials, like copper, constantan in the experiment, all this so I could get more evidence. Further work could also consists of choosing another factor and investigate how that factor affects the resistance of the wire. The factor could be the S.W.G or the thickness of the wire. The principle of the S.W.G gauge is that the electrical resistance of a wire changes when either stretched or compressed or dependent on the thickness. If I were to the experiment for the factor of the thickness I would need to have a new circuit, equipment and a new method for the experiment. The equipment that will be need is:- 1. Crocodile clips or connecting wire. 2. Power pack. 3. Ammeter. 4. Voltmeter. 5. Constantan or Nichrome wire depending on what you want to use. Method: - Power pack Voltmeter 1. First I would connect the power pack to the main power but not turning on the switch for safety purposes. 2. I then would set the power pack to A.C and the voltage that could be suitable a voltage between 4 volts to 10 volts could be used. 3. Then I would have to connect the circuit, using the connecting wire and crocodile clips. I will connect the voltmeter opposite to the power pack and the ammeter anywhere in the circuit. 4. then I would have to connect the circuit to the wire using the crocodile clips I will have to make sure that al of the crocodile clip is in contact with the wire at both sides. 5. I will take the readings for that specific thickness of the wire then remove the wire putting another thickness then taking the readings, I will take the readings to the appropriate range of thickness that would have to be chosen. 6. When changing the wires I will have to shut the power pack so the circuit doesn't over heat and also the wire doesn't heat up because this can affect the resistance. Table to show how I would collect the readings for the experiment: - READING NUMBER ONE READING NUMBER TWO READING NUMBER THREE AVERAGE RESISTANCE Average= 1.d.p Cross sectional area/thickness of wire A A+-0.1cm cm Voltage Volts v+- 0.01v 2d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p Voltage Volts v+- 0.01v 2.d.p Current I Amperes amps Least count 0.01amps 2.d.p R=V/I Resistance Ohms R=resistance V= voltage p.d I= current 2.d.p 05.00cm 10.00cm 15.00cm 20.00cm 25.00cm 30.00cm 35.00cm 40.00cm 45.00cm 50.00cm [7,657] WORDS
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INTRODUCTION This piece of physics coursework is an investigation, which will try to find out the factors affecting the resistance of a wire and reasons behind these. Mechanism of conduction: - In metals there is conduction most metals follow ohmic-law but in plastics there is no conduction. If a metal is good conductor of heat then it should also be a good conductor of electricity and visa-versa. In an atom of a metal there are bound electrons, nucleus and free electrons; a plastics atom has no free electrons, while metals have free electrons. The bound...
count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p Voltage

Volts

v+- 0.01v

2.d.p Current I

Amperes amps Least count 0.01amps

2.d.p R=V/I

Resistance

Ohms

R=resistance

V= voltage p.d

I= current

2.d.p

05.00cm

10.00cm

15.00cm

20.00cm

25.00cm

30.00cm

35.00cm

40.00cm

45.00cm

50.00cm

[7,657] WORDS

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HYPOTHESIS We can recognize four... HYPOTHESIS We can recognize four types of substances due to their structure: ionic, metallic, covalent, and molecular. If a given substance has a metallic luster, is malleable and ductile, is a good conductor of heat and electricity, and has high melting and boiling points, than it is supposed to be a giant metallic structure. If a given substance has low melting and boiling points and does not conduct electricity, it surely is a molecular structure. If a given substance is soluble in water and in other polar solvents, if it conducts electricity after being melted or dissolved, and if it has high melting and boiling points, we can predict that it is a giant ionic structure. So we will have to check which of these properties does a given substance have. APARATUS spatula stirring rod open electric circuit batteries, light bulb, electric wire with two dismantled endings two metal plates the first made of copper, the second made of zinc plastic wash bottle test tubes in amount of 4 watch glasses in amount of 5 Bunsen burner test-tube rack porcelain crucible crucible tongs triangle tripod I decided to use separate watch glass for each substance to avoid possible laboratory errors resulting from contamination with the previous one. CHEMICALS substance A "“ white, granulated powder substance E "“ silver nodules, apparent metallic luster substance C "“ tiny, white crystals substance D "“ a bit larger white crystals substance B "“ black powder distilled water SAFETY RULES Be careful while burning substances in a flame! Don't put your hand into water when the electric current flows "“ you can have your skin seriously damaged! Do not touch hot crucible with bare hand, use crucible tongs! PROCEDURE I put a few grams of each substance except for substance E, which I put into a watch glass using a spatula into separate test tubes, placed in a test tube rack. I put a hint of each substance into separate watch glass. I use open electric circuit in order to investigate electric conductivity of each substance in solid state. I pour a few droplets of water into each watch glass using plastic wash bottle. Then I mix each substance with a stirring rod in order to make process of dissolving faster and more effective. I put two metal plates into each watch glass, so they are partly sunk in the water or solution, if it was formed in the manner one ending of the electric wire sticks to the first plate, and the second ending sticks to the second plate, and it is important that plates do not touch each other. Then I observe whether the light bulb is shining. I take a hint of each substance one by one, using a spatula. I put each substance into a porcelain crucible. I put crucible on a triangle placed on a tripod above the Bunsen burner. Then I turn the burner on and wait up to a minute in order to check whether the melting point is low or high. To handle the crucible I use crucible tongs. Note: I carefully clean spatula before using it again and again, I do the same with the stirring rod and porcelain crucible. DATA COLLECTION A B C D E Conductivity in solid state - + - - + Conductivity after being dissolved - - + - - Solubility in water + - + - - Melting point low high high high high CONCLUSION Substance A is soft and granulated. This substance has low melting point, what indicates that the intermolecular forces are weak. It does not conduct electricity, because molecules are not charged. So substance A has undoubtedly molecular covalent structure. However, on contrary to other substances with molecular covalent structure, it is quite soluble in water, what means that its' molecules can form hydrogen bonds to the water to compensate for the water-water hydrogen bonds broken. Example of such molecules are sugar molecules, so this substance is probably sucrose. In the case of the substance E there is an apparent metallic luster, so it has the giant metallic structure. This metal has high melting point, because it takes a lot of energy to break up a lattice of ions in a sea of electrons with strong forces of attraction, called metallic bonds, between them. Metals are good conductors of electricity because the delocalized, free electrons can move through the lattice carrying charge, when a voltage is applied across the metal structure. The substance C is the only substance aqua solution of which conducts electricity, so it has to have giant ionic structure. It's because the water molecules, which are dipoles, can attract the ions away from the lattice. The ions move freely, surrounded by water molecules. Dissolved or melted ionic compound conducts electricity, because the lattice breaks up and the ions are free to move as charged particles. It can be assumed that substance D is a giant covalent structure, because it is insoluble, it is very hard, but brittle, it forms crystal lattice, and it has high melting point. In addition, this substance does not conduct electricity at all. Substance B is soft and brittle in touch - the sheets can slide over each other easily. It may indicate that this substance has a molecular structure, like the first one. But it has much higher melting point than molecular substances. Besides that, it conducts electricity in solid state, and it does not dissolve in water. This set of properties is very specific "“ it is a combination of single properties of different types of structures. The fact that this substance could well be used as a lubricant layers are easily rubbed off could indicate that this substance can be graphite. EVALUATION After an experiment was finished, our chemistry teacher wrote the names of substances that we were to determine structures of, on the blackboard, so we could verify if our findings were correct and propose improvements to the method in case they were not. And so: substance A appeared to be glucose, substance B "“ graphite, substance C "“ sodium chloride, substance D "“ silicon dioxide, and substance E "“ chromium metal. My predictions according to substance A appeared to be correct. In case of substance E, which is chromium metal, I also obtained correct results. I think that this substance, like it is in case of all metallic substances, has a structure very easy to determine experimentally, even, to say, with bare eye, because we know that metals are the only type of substances that perform a property called metallic luster. Other properties I observed also form a set of properties typical for metal, which is chromium in this case. I was right in case of substance C, which, as it appeared later, is sodium chloride, and sodium chloride is the most characteristic representative of ionic substances. My assumptions relating to substance D are also proved to be correct, since I know now that this substance was silicon dioxide, commonly occurring as quartz, being a good exemplification of properties connected with a giant covalent structure. In case of substance B, I was again right, due to the fact that this substance appeared to be graphite, as I have predicted. Graphite is another example of giant covalent structure, but, on contrary to silicon dioxide, it conducts electricity "“ this property is specific only for this particular substance.   

HYPOTHESIS We can recognize four types of substances due to their structure: ionic, metallic, covalent, and molecular. If a given substance has a metallic luster, is malleable and ductile, is a good conductor of heat and electricity, and has high melting and boiling points, than it is supposed to...

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Aim: To investigate the heat... Aim: To investigate the heat loss of the insulating materials, bubble wrap, cotton wool, blanket and foil Introduction: I will carry out this investigation by measuring the temperature of hot water in beakers insulated with these materials. Naturally the insulated beaker with the warmest water will have prevented the most heat loss. To understand how materials prevent heat loss I need to look at the three ways in which heat can be transferred or lost to the surroundings. One of the ways is conduction. This is when heat energy is transported along an object from the hotter region to the cooler region. The object itself does not move. So when an object is heated, its atoms start vibrating with the heat energy. The free electrons begin to diffuse through the object and collide into other electrons transferring kinetic energy. Substances that have particles close together, like metals are good conductors. Meanwhile gases have particles that further apart so they are poor conductors. Therefore things that are made of particles that are not close together, are poor conductors but good insulators such as gases. Another way is convection, which only happens in liquids and gases. This is when heated particles start moving faster so they move further apart. The heat also makes the particle expand so they become less dense than the unheated particles. As a result of this, the heated particles rise upwards taking the extra energy with them and are replaced by colder and denser regions. A form of convection is in evaporation where the particles in a liquid keep bumping into each other. Sometimes they collide with each other with kinetic energy. During these collisions some particles receive so much energy that they can break away from the liquid. The final way in which heat can be lost is through radiation. This is the transfer of heat energy by waves. Every object sends out infra red radiation but hotter objects send out more. This is completely different to conduction and convection as it does not require particles, so the so infra-red radiation can pass through a vacuum. Also dull and black surfaces emit more heat while shiny and white surfaces emit less. However this makes black and dull surfaces better absorbers Prediction: With the kinetic theory on mind, I predict that out of the materials of foil, bubble wrap, wool and blanket the bubble wrap will be the better insulator. This is because the bubble wrap has a lot of air trapped in its pockets. Air is a very good insulator as its particles are widely spread out therefore there is a less chance of them colliding and passing energy on. This means that there will be less energy transferred to the surroundings, as the bubble wrap would prevent heat loss mainly through conduction. Furthermore the material is made from plastic which in itself is a good insulator. As the aluminium foil is shiny it will reduce heats loss through radiation by reflecting heat back in, like a flask. However since it is a metal, some heat will be lost through conduction. Naturally the blanket material would be a good insulator as its purpose is to keep people warm. This because it has lots of layers and like the wool has air trapped between it fibres. So by reducing heat loss through conduction it should be a good insulator but not as good as the bubble wrap. This is because as it is a dull and black colour some heat will be lost through radiation. The cotton wool is also material that has a lot of trapped air in between its many fibres. Even though it is a good insulator, I feel that because of the thinness of the material the bubble wrap will still be the better insulator. All these materials except the foil are lagging which are insulating materials that have tiny air pockets that trap air to prevent heat from being conducted away. Safety: Since I am dealing with very hot water there are precautions I will need to take. Firstly I will abide to the general laboratory rules i.e. tidy workspace, loose clothing tucked away, and safety goggles. When I need the water for my experiment I will my teacher to pour the hot water. Preliminary investigation: For my investigation I need to know at what intervals I should record a temperature reading and for what period of time. Here are the results of my preliminary investigation: Time Secs Temperature °C 0 84.5 30 80.5 60 77.0 90 75.5 Time Mins Temperature °C 0 84 1 80 2 74 3 72 These results show that taking the temperature readings at 30 second intervals would make the results more accurate and to do this over a 5 minute period. I will also use 80 ml of boiling water and make sure all starting temperatures are the same. Range of results: My results will fit into a table like this, TIME SECS TEMPERATURE OF WATER IN BEAKER ºC CONTROL WOOL BUBBLE FOIL BLANKET Apparatus: In my experiment I will use the following equipments: Electric Kettle "“ To boil the water 5 Beakers "“ To hold the hot water for the experiments Aluminium Foil Bubble Wrap Cotton Wool Blanket Thermometer "“ To measure the temperature Stopwatch "“ To time the experiments Elastic Bands "“ To hold the materials in place Variables: In my experiment the volume of the water for each beaker will be 80 ml. Temperature readings will be taken at 30 second intervals for 5 minutes and I will make sure all beakers have the same starting temperature. Therefore there will 10 readings after the starting temperature. This will ensure my test is fair. The only thing that will be different is the material around the beakers. Method: First of all I will wrap four of the beakers in the materials with the elastic bands. The other beaker will stay uncovered as it is the controlled experiment. After placing the thermometer into the beaker, the 80 ml of water will be poured in and the stopwatch will be started straight away. Also the starting temperature recorded. After that the temperature reading will be recorded at 30 second intervals shown by the stopwatch. This will be done for 5 minutes and repeated for all the beakers. Results: TIME SECS TEMPERATURE OF WATER IN BEAKER ºC CONTROL WOOL BUBBLE FOIL BLANKET 0 83.0 83.5 83.0 83.5 83.5 30 80.5 80.0 81.5 79.5 80.5 60 77.0 78.0 80.0 77.5 78.5 90 75.0 76.0 78.5 76.0 76.5 120 73.5 74.5 76.5 74.0 75.5 150 72.0 73.5 75.5 73.5 74.5 180 70.0 72.5 74.5 72.5 73.5 210 69.0 71.5 74.0 71.5 72.5 240 68.0 70.5 73.0 70.5 71.5 270 66.5 70.0 72.5 69.5 71.0 300 65.5 69.5 72.0 68.5 70.5 Observations: For all the materials there is a sharp drop in temperature for the first minute and after that it gradually slows down. The starting temperatures for the insulated beakers are very close but at 5 minutes they the end readings are more spread out. As you can see on the graph the bubble wrap has the most gradual curve and consequently the slowest rate of heat loss. From the starting temperature of 83.0 °C, it dropped by 12.0 °C after 5 minutes. The blanket is the second best insulator with the temperature drop of 13 °C from a starting temperature of 83.5 °C. The cotton wool lost 14.0 °C from a starting temperature of 83.5°C The aluminium foil is the worst insulator from the materials as from a starting temperature of 83.5 °C it lost 15 °C The controlled experiment the beaker with no insulation has the steepest curve and therefore the fastest heat loss. From the starting temperature of 83.0 °C, it lost 17.5 °C after 5 minutes. Conclusion My prediction was correct as the bubble wrap material was the best insulator because it prevented the most heat loss being the best insulator. The bubble wrap was the best insulator due to its ability to reduce heat loss through conduction. This can be understood by looking behind theory of conduction. Since heat is energy associated with the motions of the particles making up a substance, it is transferred by these motions, shifting from regions of temperature, where particles are more energetic to regions of lower temperature. The rate of heat flow between the two regions is relative to the temperature difference and the heat conductivity of the substance. In solids the molecules are bound and add to the conduction of heat mainly by vibrating against neighboring molecules. However a more important mechanism is the migration of free electrons. Materials such as metals have a high free-electron density therefore they are good conductors of heat while non-metals do not conduct as well. Since gases and liquids have their molecules even further apart they are generally, very poor conductors which makes them good insulators. Therefore the bubble wrap which is made up of lots of air pockets has not only less free electrons to transfer heat but it has molecules the trapped air that are quite far apart. This makes harder for heat from the boiling water in the beaker to be conducted away through the bubble wrap and to the surroundings The blanket and cotton wool were also quite good insulators as they too have a lot of air trapped between the fibres. The foil was not as good because being a metal it had more free electrons with particles being closer together, for heat to be conducted away to other regions. However, since it was quite shiny it reflected some heat back in, to the beakers, such as the effect of a flask so it prevented heat loss better than the controlled beaker. The controlled beaker naturally let out the most heat since it had no form of insulation except the fact that it the beaker was made out of glass which in itself is a fairly good insulator. Evaluation The method that I used to carry out my investigation was fairy simple to carry out. It required me to pour out 80 mls of hot water into beakers insulated with different materials. The temperature of the water was then recorded over a 5 minute period at 30 second intervals. The results of my investigation were accurate as I took my readings at eye-level to the nearest half a degree centigrade. I had no anomalous results and the accuracy can be seen on my results graph where the readings are close to the best fit curve. However I could have been even more accurate as if the time h d been available tome, I would have repeated my experiments to get an average of the results. Another way in which I could have improved my experiment was if was available to me, I would have used a data logger with temperature probe. This would have given me very accurate readings every second. I could have also used polystyrene lids to reduce heat loss by convection in the beakers. To extend by investigation I could have used different beakers like copper cans to see how the material of the beaker affects the rate of heat loss. In addition I could look at the effect of the amount of layers of material insulating the beaker as well as looking at different volumes of boiling water.   

Aim: To investigate the heat loss of the insulating materials, bubble wrap, cotton wool, blanket and foil Introduction: I will carry out this investigation by measuring the temperature of hot water in beakers insulated with these materials. Naturally the insulated beaker with the warmest water will have prevented...

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Aim:- We will investigate... Aim:- We will investigate the length of a wire in a series circuit, and if it will affect its resistance. Prediction:- Resistance is the force of which opposes the flow of an electric current around a circuit so that energy is required to push the charged particles around the circuit. I predict the resistance will vary with the length. I also predict the longer the wire the less current will flow which increases the resistance. This is because electric current is the movement of electrons through a conductor, so when resistance is high, conductivity is low. Therefore, the electrons will have to push their way through a shorter path of atoms in the wire, reducing their resistance. Whereas, if the length was longer, then the number of atoms in the wire increase. Electrons are negatively charged particles, and protons are positively charged atoms. Electrons move around, but protons don't move, they stay in the same place. Current is a flow of electrons, and is measured in amperes A. When a current flows through a resistance, energy is given off as heat. I think the thicker and shorter the wire, the lower the resistance. I think this because, for example, if you had a road with cars parked to the side and only one car at a time can pass the cars parked on the side of the road as the road is so narrow that allows two cars to go at a time, but as it seems that there are cars parked, that only one car can move past the parked cars; in this case it will be slower for the cars to pass, because the road is long and narrow. Whereas, if the road was wider thinker and shorter it would be quicker. DIAGRAM OF THE THICKNESS AND LENGTH Planning:- Before I do start my investigation I will need to set up my circuit. I will need a variable resistor connected to a power supply, an ammeter and a voltmeter voltmeter parallel to the nichrome wire. I will move the knob on the variable resistor into five different positions for each one length e.g:- 10cm, 20cm, 30cm "¦"¦.. I will get five different readings for each length, and I will be doing five different lengths, which makes twenty-five readings all together, on the voltmeter and ammeter. I will calculate the resistance with this equation:- V = R x I OR Potential difference volts, V = Current amps, A x Resistanceohm, This is how my circuit will look like when I've finished setting it up:- DIAGRAM OF CIRCUIT I will link all the components together with the wire connected to the circuit with crocodile clips at the length of 10cm. I will use to measure the voltage using a voltmeter and recording the results on a table. I will also need to measure the current using an ammeter and recording the results for them too. When I have the results I require, I will use the calculator and divide the voltage by the current to get the resistance. I know that I will need to turn off and on the power supply every time I investigate another length of the wire. This is because the wire intends to warm up and this may have an effect on my other readings and also the wire can snap in half by melting. To keep my investigation fair, I will keep the voltage on the power supply the same, the type of wire and the thickness, and also do the investigation in the same surrounding temperature. Analysing:- I have calculated the resistance of each length on the nichrome wire. I have used these results of values to plot a graph of resistance against length. Length goes along the bottom axis because it is the dependent variable. Its value depends on the length of the wire chosen The points on my graph are a little scattered, none of the points touch the line of best fit, but they are quite close together.. On my graph of the length against gradient, I have rejected one point. I would of rejected two, but I have noticed that the 10cm point was very high, I was going to also reject the 40cm point too, but I was more curious on the 10cm. my table of results suggests that the voltage reading for one point in the 10cm trial was very high compared to the other results of 20cm, 30cm, 40cm and 50cm. but I reckon that I must of miss read the meters whilst investigating. I have noted my working out on the graph of current against voltage. On my graph of current against voltage, there is an anomalies point which I have circled. It is the 10cm point of 0.90V and 0.18A which I must have rejected on the graph of length against gradient. So this is the reason of my rejection on the graph of length against gradient. You can see that this one point has affected the gradient. And as I mentioned that I must of miss read the meters.   

Aim:- We will investigate the length of a wire in a series circuit, and if it will affect its resistance. Prediction:- Resistance is the force of which opposes the flow of an electric current around a circuit so that energy is required to push the charged...

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