...Experiment Aim: To find the specific heat capacity of three different metal using the method of mixtures. Specific Heat Capacity – It is the amount of heat (thermal) energy required to cause a rise in temperature by 1°C in a solid of 1kg. Apparatus: 1. Metal cubes (3) 2. Beaker 3. Water bath 4. Thermometer (L.C. 1°C) 5. String 6. Digital Balance Risk Assessment: 1. Wear a lab coat to prevent any substance from falling you and potentially harming you. 2. Be careful when transferring metal from water bath to beaker as the water may fall and could burn you. If this scenario arises then put affected part under running cold water. Method: 1. Gather all required apparatus 2. Suspend the metal using the string into the water bath 3. Use the digital balance to measure the mass of the empty beaker 4. Pour water into the beaker 5. Measure the mass of the beaker containing water and find the difference between the two which will be the mass of the water and record the values 6. Measure the initial temperature of the water 7. Measure the initial temperature of the metal 8. Transfer the hot metal into the beaker 9. Monitor the change in temperature of the mixture and record the initial increase of the water when it stabilizes 10. Remove the metal from the beaker and find its mass 11. Using the formula E=MC (change in temperature) to find the total energy transferred in water 12. Since...
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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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... Experimental Determination of Heat Capacity Ratios of Gases, N2, CO2 and Ar, by the method of Adiabatic Expansion ABSTRACT The heat capacity ratios [Cp/Cv] of Nitrogen, Carbon Dioxide and Argon were obtained by the method of adiabatic reversible expansion. Four studies were conducted for each gas using an 18.0 L carboy and phthalate manometer at 297K. The initial and final pressures were obtained and used to calculate the experimental heat capacity ratios. Based on the results, Argon has an experimental heat capacity ratio of 1.66 and a Cv value of 1.56, which agrees with the theoretical Cp/Cv and Cv values obtained from the classical equipartition theorem. Based on the results, the diatomic molecule, N2 and triatomic molecule, CO2 have heat capacity ratios of 1.45 and...
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...|Length (in centimeters) = | |Weight (in pounds) = 20 |Weight (in kilograms) = | 2. State the law of mass and energy conservation. Describe how you might demonstrate this law. 3. Adapted from Exercise 116 in Ch. 3 of Introductory Chemistry: Global warming refers to the rise in average global temperature due to the increased concentration of certain gases, called greenhouse gases, in our atmosphere. Earth’s oceans, because of their high heat capacity, can absorb heat and therefore act to slow down global warming. Imagine that you have a container of seawater. Assume that the volume of the container is 50 cm3 and that the density of the seawater is 1.03 g/cm3. Also, assume that the heat capacity of seawater is the same as that of water. 1. What is the volume of the container in cm3? 2. What is the weight of 1 cm3 of seawater in grams? 3. What is the weight of the seawater in the container in grams? 4. What is the formula for specific heat...
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... 2. State the law of mass and energy conservation. Describe how you might demonstrate this law. The law of conservation of energy, is when the total amount of energy remains constant, but can change forms from one to another. The law of conservation of mass is when the total amount of mass remains constant, even if you change the shape or the chemical make up. I would demonstrate this law by proving how the energy and mass stays constant. 3. Adapted from Exercise 116 in Ch. 3 of Introductory Chemistry: Global warming refers to the rise in average global temperature due to the increased concentration of certain gases, called greenhouse gases, in our atmosphere. Earth’s oceans, because of their high heat capacity, can absorb heat and therefore act to slow down global warming. Imagine that you have a container of...
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...Data/Results Specific Heat Capacity of a Metal (Copper) | Trial 1 | Trial 2 | Mass of metal (g) | 11.6813 g | 12.1214 g | Mass of empty calorimeter (g) | 4.7507g | 4.7507 g | Mass of calorimeter & water (g) | 24.2425 g | 24.2422 g | Mass of water (g) | 24.2425-4.7507=19.4918 g | 24.2422-4.7507=19.4915 g | Volume added to calorimeter (mL) | 20.0 mL | 20.0 mL | 1 | Time (s) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 | 390 | 420 | 1 | Temp(°C) | 26.1 | 26.2 | 26.3 | 26.3 | 26.2 | 26.2 | 26.2 | 26.9 | 27.1 | 27.4 | 27.2 | 27.2 | 27.1 | 27.1 | 27.1 | | | | | | | | | | | | | | | | | | 2 | Time (s) | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 | 390 | 420 | 2 | Temp(°C) | 24.0 | 24.1 | 24.2 | 24.2 | 24.2 | 24.2 | 24.2 | 28.0 | 27.9 | 27.8 | 27.7 | 27.7 | 27.7 | 27.6 | 27.6 | Change in temperature of the water (°C) | 1.0°C | 3.6°C | Energy gained by the water (J) | 81.552J | 293.59J | Change in temperature of the metal (°C) | -72.9°C | -72.4°C | Calculated specific heat capacity (J/g°C) | 0.0958 J/g°C | 0.335 J/g°C | Molar mass approximation (g/mol) | 261.02g/mol | 74.73g/mol | Percent error of specific heat capacity (%) | 75.1% | 13.0% | Percent error of molar mass (%) | -310.76% | -17.60% | Enthalpy of Neutralization (Nitric Acid) | Trial 1 | Trial 2 | Volume of acid (mL) | 50.0 mL | 50.0 mL | Concentration of acid (M) | 1.1 M | 1.1 M | Volume of NaOH...
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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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...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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...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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...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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...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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...Specific Heat and Heat Exchange Problems 1. How much heat is required to raise 80.0 g of aluminum from 12°C to 46°? [2 448 J] 2. How much heat is required to raise 30.0 kg of iron from 40°C to 180°C ? [1 620 kJ] 3. How much heat must mercury lose so 290 g cools from 116°C to 31°C? [-3 451 J] 4. What temperature results when 500.0 J of heat are added to 50.0 g of silver at an initial temperature of 25 °C? [67°C] 5. What mass of copper at a temperature of 15°C must be used so 300.0 J of heat will cause it to warm to 35 °C? [38 g] 6. To warm 250 g of glycerine from 21°C to 46°Crequuired 3625 J of heat. Calculate the specific heat of glycerine. [0.58 J/g°C] 7. Two hundred grams of turpentine was heated from 20°C to 55°C by the cooling of 147 g of water from 75°C to 55°C. [1.76 J/g°C] 8. How much water at 25°C must be added to 15 kg of water at 10 °C so that the mixture becomes 20°C? [30 kg] 9. If 70.0 g of water at 95°C was mixed with 59 g of water at 25°C, what is the final temperature of the mixture? [63 °C] 10. What mass of copper at 90°C must be added to 200 g of water at 10°C so the final temperature of the mixture is 20°C? [300 g] 11. What mass of aluminum at 150°C must be added to 7.0 kg of water initially at 15°C if both materials end up at 30°C? [3.6kg] 12. When 200 g of mercury at 100°C was mixed with some water at 20°C, the final temperature reached 25°C. Calculate the mass of the water. [100 g] ...
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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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...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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