...Physics Lab 4 (design) Design Aim To measure the effect on the heat of fusion of ice with varying percentage impurities of NaCl crystals in the water. Apparatus Ice cubes Distilled water Sodium Chloride crystals Styrofoam cup Calorimeter Digital balance with uncertainty (±0.01 g) Clamps Freezer Digital thermometer with uncertainty (±0.1°C) Glass rod Glass dish Pipette (±0.01 cm3) Variables Manipulated variables: Amount of impurity added to the water The aim of the experiment was to find the effect that impurities have on the heat capacity of the solution. Therefore the amount of impurities added to the water was varied. The amount added was 0%, 10%, 20%. Responding variable: Heat of fusion of the ice cube Since the composition of the solution has changed so will the heat of fusion. This is what is going to be evaluated in the experiment. Controlled variable: Temperature of the room, amount of water taken, impurity used (NaCl) Method of controlling variables The temperature of room was kept constant. The amount of water taken was kept constant at 50 cc. The same impurity was used so that the composition of the impurity would remain the same. Method to control variables Procedure Fill the styrofoam cup with 50 cm3 of water and place it in the calorimeter. Measure the temperature of the he room temperature by dipping the thermometer in water. Place the glass dish on the digital balance and zero the digital balance (this step...
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...HEAT 4.1 UNDERSTANDING THERMAL EQUILIBRIUM 1. Define: The measure of the degree of hotness of an object. (a) Temperature Measured in SI unit Kelvin, K A hot object is at a higher temperature than a cold object. Form of energy, measured in Joules, J (b) Heat Heat is transferred from hotter object (higher temperature) to colder object (lower temperature) When an object is heated, it will absorb heat energy and the temperature will increase. When an object is cooled, it will release heat energy and the temperature will decrease. (c) Thermal Two objects are in thermal contact when heat energy contact can be transferred between them. (d)Heat transfer When two objects with different degrees of hotness come into thermal contact, heat energy is transferred between the two objects. (e) Mechanism of Thermal Equilibrium Energy is transferred at a faster rate from the hotter object to the colder object. Energy is also transferred from the colder object to the hotter one, but at a slower rate. There is a net flow of energy from the hotter object to the colder object. (f) Thermal When two objects are in thermal equilibrium, there is Equilibrium no net flow of heat between them. Two objects in thermal equilibrium have the same temperature 60 The hotter object cools down while the colder object warms up . After some time, energy is transferred at the same rate between the two objects. There is no net heat transfer between the objects. The two objects are said to be in thermal equilibrium...
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...Chemistry Lab Chemistry Lab Aim: To investigate the specific heat capacity of water, copper and aluminium by using a caloriometer. Results: 1. Ohms Law: I=v/r, equation 3 = E=vlt hence you get E= v2t/r when ohms law is applied to equation 3. 2. De ionized water is used as it is the purest form of water, as it doesn’t contain ions from the soil like normal water which allow the conduction of electricity as normal water allows electrons to flow, thus preventing any extraneous variables to affect the experiment. 3. The value for Tf = 100o as water has a boiling point of 100o 4. SH = 1.00cal.g-1.c-1 m= density x volume (Tf – Ti) = 100 – 19.9 = 1 x 1,000ml = 80.1 = 1,000g EH2O = SH.m.(Tf – Ti) = 1 x 1,000 x 80.1 = 80,100cal Converting to Joules (1cal = 4.184j) 1000*4.184* 80.1= 335,138.4 joules g-1 deg-1 EKettle= V2.t / R V= 240 t= 219s R= 34.7 2402 * 219/34.7 =363,527.38 joules = E of kettle 5. The amount calories required to raise 1 gram of water by 1 degree Celsius is specific heat. Hence 80,100 calories are required to heat up the water, this shows the association between question 4. 6. In question 4, it can be observed that the energy produced by the kettle was 363,527.38 joules, however only = 335,138.4 joules g-1 deg-1 of energy was required to raise the water 80.1 degrees Celsius. Thus...
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...Energy Content of Foods Written by Chris Papadopoulos The energy content of foods is investigated. The energy released by a number of food samples and absorbed by water is determined using technology. Inferences about the energy content of foods with high fat content and foods with high carbohydrate content are then made. Hypothesis Foods, depending on their carbohydrate/fat composition, have different energy content that can be determined by measuring the heat release from their combustion. Primary Learning Outcomes At the end of this lesson, students will be able to: • Be familiarized with data collection using the Vernier LabPro and TI calculator • Be familiarized with the use of the Vernier LabPro temperature probe • Collect, graph, display and make inferences from data • Determine the amount of heat released by a substance given the specific heat capacity (Cp) of water, the mass (m) of the food sample and the change in temperature (∆t) Assessed GPS Characteristics of Science: Habits of Mind: SCSh2. Students will use standard safety practices for all classroom laboratory and field investigations. a. Follow correct procedures for use of scientific apparatus. b. Demonstrate appropriate techniques in all laboratory situations. SCSh3. Students will identify and investigate problems scientifically. a. Suggest reasonable hypotheses for identified problems. c. Collect, organize and record appropriate data. d. Graphically compare and analyze data points and/or...
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...Toronto Collegiate Institute SPH 3U1 Practical Examination Specific Heat Capacity of TCI Tap Water Date: 14th July 2014 Title: Specific Heat Capacity of water Purpose: To find the specific heat capacity of TCI tap water Hypothesis: I hypothesize that the specific heat capacity of water will be 4200 J/kg°C because water is known to have a high heat capacity Apparatus: * Top Pan Balance * Kettle * Alcohol Thermometer * Pencil * Paper * Water Procedure: 1. Record the mass of the empty kettle 2. Fill kettle with water 3. Record the mass of kettle with water 4. Calculate mass of water 5. Turn on kettle and place thermometer in kettle 6. Record the temperature on the thermometer every 30 seconds until the temperature reaches 100°C Readings: PKettle = 1200W mKettle = .946kg mKettle and Water = 2.402kg mwater = 1.456kg Theory: EH=mc∆T The relationship between heat energy, the mass of the substance, the specific heat capacity of the substance and temperature can be expressed in the equation above. The heat energy in joules required to heat a given substance to a certain temperature is equal to the mass of the substance heated multiplied by the value of the specific heat capacity of the substance multiplied by the difference between the initial temperature and the temperature of the substance after heating In this lab, the energy used to heat the water is calculated from the power of the kettle. The mass of the water...
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...error. The mean ΔHrxn for the heat of neutralization was calculated to be -49 kj/mol with a standard deviation of ±2 and a % error of -12%. The mean ΔHrxn for the heat of solution was calculated to be -23 with a standard deviation of ±2 and a % error of -22%. 2. Looking at the % errors, comment on the accuracy of each experiment. For the heat of neutralization, the percent error was calculated to be -12%. For the heat of solution, the percent error was calculated to be -18%. Because both percent errors are greater than the accuracy threshold of ± 10%, neither measurement is accurate. 3. Looking at the standard deviations, comment on the precision of each experiment. The standard deviation for the heat of neutralization, which was ±2, was precise because the mean ΔHrxn was calculated to be -47kj/mol. Because a significant figure did not have to be dropped, the measurement was precise. The standard deviation for the heat of solution, which was also ±2, was precise because the mean ΔHrxn was calculated to be -49 kj/mol. Because a significant figure did not have to be dropped, the measurement was also precise. 4. Why is the sign of heat gained or lost by a reaction the opposite sign of the heat gained or lost by the solution? The sign of heat gained or lost by a reaction is the opposite sign of the heat gained or lost by the solution because as the reaction commences, the reactants will either release heat or absorb heat. When this happens, the products...
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...CHM130 Lab 4 Calorimetry Name: Data Table: (12 points) |ALUMINUM METAL | | |Pre-weighed Aluminum metal | | |sample mass (mmetal) |19.88 | |Temperature of boiling water and metal sample in | | |the pot (Ti(metal)) | | | | | | | | | | | |dsdfa(Ti | | |Temperature of cool water in | | |the calorimeter prior to adding hot metal sample | | |(Ti(water)) | | |Maximum Temperature of | ...
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...cold Water = mass of Hot Water Ex. From trial 1 124.35 g of mixed water - 74.898 g of cold water=49.452 g of hot water Next in order to calculate the initial temperature of the hot water without being able to measure it with a thermometer the Calorimetry equation 3 must be used. Equation 3 q=m×C ×∆T * q is the heat change of the system (joules, J) * m is the mass(grams, g) * ∆T is the Temperature change, Tfinal-TinitialC * C is the specific heat for that substance(Jg ° C) After calculating the Heat change of the system using equation one we can use equation two ; The law of conservation of energy which states that while energy can be converted or transferred, it is not created or destroyed. Equation 4 qsystem + qsurroundings=0 Or qsystem = -qsurroundings Therefore if the heat change of the surroundings are known then the system and all of its variables are also known except Tinitial which can be found. Here is an example of finding the Initial temperature for Trial 1 of the experiment. 1. First use equations 1, and 2 to find the mass of cold water. 2. Next input that value into equation 3 as mass 3. Input the Specific heat of water in liquid form which is 4.186Jg °C 4....
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...paid with salt, hence the word “salary” for their income. Many other processes like leather tanning with mineral salts depended and still depend on salts. Acids (vinegar) and bases (soaps) also have played important roles in human culture. And in general, life on earth is not imaginable without electrolytes. Dissolves in water the crystal is sWhat is conductivity? Conductivity is the degree to which a specified material conducts electricity, calculated as the ratio of the current density in the material to the electric field that causes the flow of current. It is the reciprocal of the resistivity. The rate at which heat passes through a specified material, expressed as the amount of heat that flows per unit time through a unit area with a temperature gradient of one degree per unit distance. Conductivity is the measure of the ease at which an electric charge or heat can pass through a material. Without the classical and quantum mechanical views of conductivity have both described the movements of electrons in a metallic solid. Pure water does not conduct electricity. Dissolve in water or some other solvent and form ions. Sodium chloride is an electrolyte. Electrolytes are substances that form solutions that conduct electricity. Water molecules do not dissociate significantly to form charge carriers (ions). Ions are atoms with extra electrons or missing electrons. When you are missing an electron or two, you have a positive molecules. The clumps of solvent molecules and ions are...
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...Intensity Levels. (From Jewett, J.W. and Serway R.A., Physics for Scientists and Engineers (with PhysicsNow and InfoTrac) 6th edition, copyright 2004. Reprinted with permission of Brooks=Cole, a division of Thomas Learning.) Vapor-pressure (boiling point) curves of common refrigerants. (From King, G.R., Modern Refridgeration Practice, McGraw Hill Company, 53, 1971. With permission.) Vapor pressure of water vapor at varying temperatures. (From King, G.R, Modern Refrigeration Practice, McGraw Hill Company, 31, 1971. Vapor-pressure curves between triple points and critical points. (With kind permission of Linde BOC Process Plants.) Conversion Table for Oxygen, Nitrogen, and Argon. (With kind permission of Linde BOC Process Plants) Molar Heat Capacity of Some Gases at 300 K. (From Data Book, Air Liquide Process and Construction. With permission.) Conversion Table for Vacuum Pressures. (From Product and Reference Book, D00– 142, 2003–2004. With...
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...1. The total surface area of all eight cubes equals out to 192 due to the equation below. [(2∗2)∗6]∗8=192 the equation for the larger equals 128 as seen here. (4∗4)∗6=96 192/96= 1:2 2. Water's density is ≈ 1,000kg/m^3 The displaced water occupies about 600/1000=.6m^3 The barge's area is 15m^2. V = L×W×H = .6m^3 H (depth of sinking) = .6m^3/15m^2 = .04m = 40mm” 3. When a steadily flowing gas flows from a larger diameter pipe to a smaller diameter pipe the speed of gas is decreased and pressure become increased and the spacing between the streamlines less and the streamlines come very close to each other. 4. Iron specific heat c i is 9 times lower if compared with water specific heat c w ...it means that the heat necessary for getting the same over temperature on the same mass is 9 times higher on water. * ------------------------------------------------- 5. 450 J = (45 g)*c*(78 degrees C - 22 degrees Celsius) 450 J = (45 g)*c*(56 degrees Celsius) 450 J = c*(2520 g degrees Celsius) 6. Given: wavelength=600 nm; h=6.62×10^-34 Solution: Since, Energy=hc/wavelength (6.62A—10^-34A—3.0A—10^8)/600A—10^-9 =3.39A—10^-19J =3.39A—10^-19/1.6A—10^-19 eV =2.11875 eV 7. ------------------------------------------------- I’m sorry, I don’t know the answer to this question. 8. ------------------------------------------------- You hear the difference frequency. You want it to be zero. So loosen it more. 9. ------------------------------------------------- ...
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...Executive Summary The goal of this experiment was to measure the latent heat of ice. The value of the latent heat of ice was obtained by measuring the amount of boiling water needed to melt a given mass of ice. After repeating the procedure three times, the latent heat of ice was calculated for each test, and then averaged. The heat released by the boiling water is the product of specific heat, mass, and temperature change, while the heat required to melt the ice is the product of the mass of ice and the latent heat. (1) Einitial = Mwatercpwater (Tw) + Micecpice (Tice) (2) Efinal = Mwatercpwater (Tw) + Micecpice (Tice) + MiceLice, where M is defined as mass, cp is specific heat, and Lice is the latent heat of ice. Then, the First Law of Thermodynamics dictates that the change in Energy is equal to zero, and the final Energy is equal to the initial Energy. Thus, latent heat was calculated by substituting experimental values for mass and the literature values for specific heat. The averaged values of the latent heat of ice was computed to be 233.6 kJ/kg—a percent error of 30.1% from the literature value of 334 kJ/kg. These equations and relationships are based on several assumptions. The heat transfer from the boiling water is assumed to only melt the ice, and that no heat had been lost to the environment. It is also assumed that the temperature after melting the ice was 0 oC, and so the change in temperature is the full range between 0oC and 100oC. Lastly, there is no net work...
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...Reference: Mr. Smith’s Instructions B. Problem: To calculate the specific heat of Lead. C. Hypothesis: If the heat absorbed by water when lead is added is known then the specific heat of lead can be calculated. D. Theory: i. In this experiment it is assumed that the density of liquid water is 1.00 g/mL ii. The equation qlost = -qgained can be applied when the change in temperature of the water is calculated. The energy gained by the water is equal to the energy lost by the lead. iii. The equation q =m ×Cp ×∆T can be simplified to solve for lead’s specific heat (Cp) once the heat of lead (q) is solved for using the equation shown in (ii). E. Variables: i. Independent: type of metal (Pb) ii. Dependent: final temperature of water iii. Control: temperature and pressure of the lab environment, the volume of water II. Experimental Design A. Equipment: i. Hot plate ii. 200 mL tap water iii. Balance iv. Styrofoam calorimeter v. 250 mL beaker vi. Test tube vii. Lab Quest viii. Lab Quest temperature probe ix. Test tube clamp B. Chemicals: metal shot (Lead - Pb) C. Apparatus: D. Method: i. Heat 150 mL water in the 250 mL beaker on the hot plate at its highest setting ii. Add the metal (Pb) shots into the test tube and place the test tube into the heating water iii. When the water boils, turn down the heat setting so that the water is barely boiling iv. Place 9 mL of water...
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...PHYSICS SCIENCE Paper – 1 (One hour and a half) Answers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the Question Paper. The time given at the head of this Paper is the time allowed for writing the answers. ____________________________________________________________________ Attempt all the questions from Section I and any four questions from Section II. The intended marks for questions or parts of questions are given in brackets [ ]. SECTION I (40 Marks) Attempt all questions from this Section. Question 1 (a) (b) (c) (d) Derive the relation between C G S and S I unit of force. Classify the following into contact and non contact forces: 1. Tension 2. Friction 3. Gravitational force 4. Magnetic force. Classify the following into levers as class I, class II or class III: 1. A door 2. Catapult 3. Wheelbarrow 4. Fishing rod. [2] [2] [2] Where is the centre of gravity of the following objects situated? l. Ring 2. Rhombus 3. Scalene triangle 4. Cylinder. (e) Calculate the resultant torque from the following diagram: [2] [2] 30 N 4m 3m 5m C 80N 200 N D 50 N E B 20 N A O 1m 64 ICSE Specimen Question Paper www.examrace.com Question 2 (a) A uniform metre scale is in equilibrium position. Calculate the mass of the ruler. 0 40 gf 5 cm 30 cm 100 cm [2] (b) (c) (d) When a ray of light passes from air to glass, for what angle of incidence...
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...PRODUCTION OF FLOUR FROM YAM TUBER By FAKOKUNDE OLADOTUN SEGUN (051100) A project report In the DEPARTMENT OF AGRICULTURAL ENGINNEERING Submitted to Faculty of Engineering and Technology In partial fulfillment of the award of degree of BACHELOR OF TECHNOLOGY LADOKE AKINTOLA UNIVERSITY OF TECHNOLOGY, OGBOMOSO, OYO STATE. SEPTEMBER, 2010. DEDICATION This project is dedicated to the Almighty God the giver of wisdom, knowledge and Understanding, who has favoured me in all things ACKNOWLEGDEMENTS My appreciation goes to my project supervisor Dr. A.B Fashina for Initiating the Project and supervising it; and for his Fatherly love. Also to my late parents Mr. & Mrs. Olukunle Fakokunde (J.P.) for the values that they have in Education which gave them the desire to see me through school even to the point of their death. Appreciation also goes to some lecturers and staffs of the department. Dr. S.O. Jekayinfa, Mr. Idowu, Mr. Adegboyega, Mr. Abegunrin, Dr. Taiwo for their words of encouragements and for their great desire to make impact. CERTIFICATION I certify that this project work was carried out by FAKOKUNDE OLADOTUN SEGUN Matric no: 051100 in the department of Agricultural Engineering, Faculty of Engineering and Technology, Ladoke Akintola University of Technology, Ogbomoso, Oyo State. ________________________ ___________________ Dr. A.B...
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