...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...
Words: 761 - Pages: 4
...The refrigerant that is typically used in an absorption chiller is distilled water and the absorbent is lithium bromide. These two fluids, the refrigerant An introduction to the absorption refrigeration cycle Absorption cycle versus vapour-compression cycle There are two fundamental differences between the absorption refrigeration cycle and the vapour-compression refrigeration cycle with which we are more familiar. The first difference is that the compressor is replaced by an absorber, a pump and a generator. The second is that, in addition to the refrigerant, the absorption refrigeration cycle uses a secondary fluid, called the absorbent. The condenser, expansion device, and evaporator sections, however, are the same. Refrigerant solutions and the absorbent, are mixed inside the chiller in various concentrations. The term dilute solution refers to a mixture that has a relatively high refrigerant content and low absorbent content. A concentrated solution has a relatively low refrigerant content and high absorbent content. An intermediate solution is a mixture of dilute and concentrated solutions. The absorption refrigeration cycle The four basic components of the absorption refrigeration cycle are the generator and condenser on the high-pressure side, and the evaporator and absorber on the low-pressure side. The pressure on the high-pressure side of the system is approximately ten times greater than that on the low-pressure side. [pic] Starting on...
Words: 950 - Pages: 4
...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...
Words: 412 - Pages: 2
...The purpose of this experiment was to determine the specific heat of an unknown metal we were give. With this specific heat of the unknown metal we can then determine what the unknown metal is. As for the acid-base reaction experiment we will also be conduction in this lab we will use this acid-base reaction to determine the enthalpy of neutralization of a strong acid-base reaction. By following these precise guidelines in this experiment we can then determine the energy released per mole of water formed in this acid-base reaction. 3. Introduction: For this lab experiment, I sought out an objective to prepare and to determine the specific heat of my unknown metal #36. As I sought out to find the specific heat of my metal I also tried to find the specific heat in another experiment of an acid-base reaction....
Words: 417 - Pages: 2
...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. ------------------------------------------------- ...
Words: 634 - Pages: 3
...Noise 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...
Words: 4665 - Pages: 19
...|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...
Words: 346 - Pages: 2
... 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...
Words: 441 - Pages: 2
...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...
Words: 1109 - Pages: 5
...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....
Words: 618 - Pages: 3
...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...
Words: 1026 - Pages: 5
...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...
Words: 497 - Pages: 2
...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...
Words: 2319 - Pages: 10
...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 | ...
Words: 1201 - Pages: 5
...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...
Words: 890 - Pages: 4
