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# Specific Latent Heat Of Fusion Of Ice

" width="500" height="300">. Conversely, when 1 kg of water at 0 ºC freezes to give 1 kg of ice at 0 ºC, 334 kJ of heat will be released to the surroundings. The initial mass of the ice in the container is equal toa)0. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. Answer: 1 📌📌📌 question a 20 of of ice at 0c is dropped into water at boiling point, specific heat capacity of water =4200 J/kg•c, sepesific latent heat of fusion (Lf) =3. The specific latent heat of a substance is the amount of heat requires to change the phase of 1 kg of substance at a constant temperature. The amount of heat required to convert one unit amount of substance from the solid phase to the liquid phase — leaving the temperature of the system unaltered — is known as the latent. In this experiment the heat of fusion of water/ice will be determined by adding ice to water in a calorimeter cup. The purpose of this lab experiment is to measure the specific heat capacity of unknown metal samples and also to determine the latent heat of fusion of water. In contrast to sensible heat, latent heat is the energy released or absorbed that changes the state of a body during a constant temperature process. edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement NNX16AC86A. latent heat of fusion of ice kj/kg. A 200–g sample of water at 80˚C is heated to steam at 120˚C. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. For ice, its value is 3. l is the specific latent heat of fusion for that material. This value, 334. Heat given off the steam Q s is equal to the heat taken in by the ice Q i. 32 × 10 4 J/kg,a specific heat of 128 J/kg ∙ K,and a melting point of 228°C. 4) This question only require a knowledge of graph interpretation and the concept of latent heat. Specific heat Q=mc ∆T Q energy Latent heat of fusion or vaporization: How much energy would be required to melt 47. (Specific heat of ice = 2090 J/kg °C, Specific heat of water =…. How to increase brand awareness through consistency; Dec. 636 x 10^3 kJ applied to 10. 3 × 105 J/kgQ. Conversely, when 1 kg of water at 0 ºC freezes to give 1 kg of ice at 0 ºC, 334 kJ of heat will be released to the surroundings. They are same things and it is defined as follows:- The specific latent heat of fusion (when solid change to liquid) of ice is the amount of heat required to change 1 kg of ice to water without a change in temperature. What is the latent heat of fusion of ice (the specific heat capacity of water is 4. Solution for Calculate the total heat required to convert 100 g ice at –15°C to steam at 120 °C. How much heat does it absorb? 3. This discussion on Ice at 0 degree celcuis is added to 200g of water initially at 70 degree celcuis in a vacuum flask. Energy absorbed = mass x specific latent heat Q = mL How much energy is needed to boil all the water in the bath and change its state to water vapour? The specific latent. to a final temperature of. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. If latent heat of fusion of ice is 80 cals per g at 0°C, calculate molal depression constant for water. Calculate the heat lost by the hot water using the specific heat equation. Dimaunahan Instructor March 22, 2019 SCORE Group Report (40) Analysis (25) Conclusion (25) Presentation (10) Safe Assign Score (30% max) TOTAL. Small lumps of ice at 0 degrees Celcius were rapidly dried on filter appaer and dropped into a beaker containing 85g of water at 18. 108kJ/kg-K * 10. The heat added, Q. The third experiment is about the latent heat of fusion. edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement NNX16AC86A. 3 × 105 J kg–1 A mass of 350 g of water at a temperature of 25 °C is placed in a refrigerator that extracts thermal energy from the water at a rate of 86 W. J097) as "The quantity of heat necessary to change one gram of solid to a liquid with no temperature change. We know: Specific Heat of water = 4186 J/Kg Degree Specific Heat of copper = 387 J/Kg Degree Latent heat of vaporisation of water = 2. How many J are needed to melt 30 grams of copper? Q = (30 g)(209 J/g) = 6270 J 5. What is the heat of. 1 × 103 J kg–1 K–1 Specific heat capacity of water = 4. 2 × 103 J kg–1 K–1 Specific latent heat of fusion of ice = 3. Energy absorbed = mass x specific latent heat Q = mL How much energy is needed to boil all the water in the bath and change its state to water vapour? The specific latent. If the liquid cools as resolidifies (freezes), the stored potential energy is given back out as heat energy. Latent Heat of Fusion of Ice Cream 210. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. Additionally, the high value of latent heat in ice and water is also responsible for the slow melting of snow!. In the case of the latent heat of fusion it is the heat required to change a substance from a solid (ice) to a liquid (water) or vice versa while the latent heat of vaporization from a liquid (water) to a gas (steam) or vice versa. 30 kg of ice at 0 °C is added to 1. Latent heat of Fusion : It is the heat required to change 1 kg of a solid substance into liquid state at the melting point of the substance. Water molecules can move in three ways. Specific Heat Capacity Latent Heat (Fusion) Latent Heat (Vaporization) Melting Points Boiling Points Density Air (room conditions) 1012 1. Scientists deal with this by defining a new constant called the heat of fusion: heat of fusion = q / m. They are same things and it is defined as follows:- The specific latent heat of fusion (when solid change to liquid) of ice is the amount of heat required to change 1 kg of ice to water without a change in temperature. Remember that the specific heat capacity of water is 4. The second step is to determine the amount of heat energy to melt the ice. Likewise, 1 gram of ice at 0°C uses 80 calories of heat energy to convert to 1 gram of liquid water at 0°C. It should be noted that an object must first reach a critical temperature before it will begin melting or vaporizing (i. latent heat of fusion of ice kj/kg. Heat energy required to melt \(0. It is called specific latent heat of vaporisation when the change of state is between liquid and gas. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. Phase diagram of water. 7 calories) per gram, and the heat of vaporization at 100 °C is about 2,230 joules (533 calories) per gram. 34 x 105 J kg-1 and. Temperature of the each compartment remain homogeneous during whole heat transfer process. l is the specific latent heat of fusion for that material. It turns out that the specific heat capacity of water 4186 coul section kilogram degrees Celsius. Using the formula for latent heat of fusion, the three trials yielded 3. The initial mass of the ice in the container is equal toa)0. WORDS 1,252. So the formula tells us how much heat energy is needed to go into a material to change it from a solid to a melted liquid. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. Dimaunahan Instructor March 22, 2019 SCORE Group Report (40) Analysis (25) Conclusion (25) Presentation (10) Safe Assign Score (30% max) TOTAL. Nevertheless, the latent heat of sea ice seems negligible to William Connolley and probably also most of his colleagues; they prefer the atmospheric CO2 as the object that everyone should look at. Specific latent heat of fusion of ice = 3. 34 x 106J/kg). melted ice. A Lab Experiment to Calculate the Specific Latent Heat of Fusion of Ice PAGES 9. Scientists deal with this by defining a new constant called the heat of fusion: heat of fusion = q / m. A lot of heat energy is required to free the solid molecules, so the latent heat of fusion is usually large. 00 C, what is L f? (the specific heat of water is 4186 J/(kg C)). 5 kJ/kg •Liquid nitrogen –At atmospheric pressure, boiling point is ‐195. This means that to convert 1 kg of ice at 0 ºC to 1 kg of water at 0 ºC, 334 kJ of heat must be absorbed by the ice. 97 x 105 J/kg, and 2. To measure the specific latent heat of ice by the method of mixtures The first step is to measure the specific heat capacities of copper and water by the methods described earlier in this chapter. The specific heats of ice over the temperature range 0° C. The accepted value for the specific latent heat of fusion of ice is 3. Please show your work. 694 W/m-K 6. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. The purpose of this lab experiment is to measure the specific heat capacity of unknown metal samples and also to determine the latent heat of fusion of water. (Specific heat of ice = 2090 J/kg °C, Specific heat of water =…. Simce the substance in question is a block of ice (solid), the heat needed to cause it to melt is Latent heat H = mL. Let M be. Constant : Specific Heat Capacity of ice & water. Example #4: Using the heat of fusion for water in J/g, calculate the energy needed to melt 50. 50-kg aluminum calorimeter filled with 1. ” For this paper we will focus on the latent heat of fusion and those high latent heats associated with the phase change between solids to liquids and vice versa. Heat lost to surroundings at the initial stages of the experiment is approximately equal to the heat gained Reasons why specific latent heat of fusion of ice when calculated was not exactly 3. 8X10^5j/kg, it is found that the data from the experiment is relatively accurate. com/heat/latent-heat. From the latent heat of fusion calculated from the experiment it was observed that the avg. On freezing, specific heat of foods decreases. By weighing it was found the beaker contained 91. Ice absorbs a large quantity of latent heat during the process of melting. Latent Heat of Fusion Thermodynamics: phase change, latent heat of fusion, melting Qty 1 1 1 1 1L 0. It should be noted that an object must first reach a critical temperature before it will begin melting or vaporizing (i. View Full Essay. A sample of ice at 0°C melts after absorbing 300 kJ of heat. 34 + 10 J/Kq. The amount of latent heat energy required to boil water is 2260 J/g. 6 kgCorrect answer is option 'C'. k) and the S 75°C b) how and com la) if so g of ice at -20°C is added to the water Meits, what is the final temp. The intensive properties cv and cp are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T. Specific heat capacity of ice = 2. of vaporization. The initial mass of the ice in the container is equal toa)0. 8) Dry the ice on filter paper and add to the water in the calorimeter 9)Put on lid and stir continously 10)Record lowest temperature θ₂, using a digital thermometer 11)Find mass of calorimeter with water and melted ice 12)Calculate latent heat of fusion of the ice. Water has a known specific heat capacity of 4. Specific latent heat of fusion of ice = 330 kJ kg Specific heat capacity of ice = 2. What is the Latent Heat of Fusion of a substance? Energy needed to melt or freeze 3. The sunlight and the hot air around the cube heat it at a rate of 10 W. And these units are special creates an idea about heat capacity. 5 kJ/kg •Liquid nitrogen –At atmospheric pressure, boiling point is ‐195. The value for specific latent heat of fusion of ice is 0. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. They can move by vibration, rotation, and translation. Specific Heat of Ice Cream at 4°C 2. MAPÚA UNIVERSITY Department of Physics E107: LATENT HEAT OF FUSION DIOMAMPO, Maria Carmela G. Prezi’s Big Ideas 2021: Expert advice for the new year; Dec. Water has a high latent heat of fusion, so turning water into ice. Let the symbol Lf represent the latent heat of fusion for water and the symbols cw, ca, and cs represent the specific heat capacities of water, aluminum, and stainless steel, respectively. Assume that the stirrer was fully immersed. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. In addition, we will study the effectiveness of different calorimeters. D: Water boils and absorbs latent heat of vaporization. Latent heat means ‘hidden’ or concealed energy, in the sense that it does not show up on a thermometer. 20 J g–1 °C–1. Temperature of the each compartment remain homogeneous during whole heat transfer process. 0kg of ice at -10°C to steam at 120°C. Its specific gravity is 3. The temperature rise in each case is shown in the table. From this equation you can calculate the specific latent heat of fusion of ice. Heat of Fusion (H f) of IceObj: Determine H f of Ice Materials: ice at 0 o C, insulated cups, balance, thermometer Methods 1. 4 × 10 5 J kg -1. Solution for Calculate the total heat required to convert 100 g ice at –15°C to steam at 120 °C. Using the formula for latent heat of fusion, the three trials yielded 3. 947 kJ of heat. PRECAUTIONS 1. Starting below 0C as heat is added the line rises steeply to 0C where it temporarily levels out. Latent heat of fusion = 3. 19c Latent Heat of Fusion. 100 J of heat energy can melt a large amount of wax, but the same amount of heat can only. The accepted value is 334,000 J/kg. Specific heat Q=mc ∆T Q energy Latent heat of fusion or vaporization: How much energy would be required to melt 47. The accepted value for the latent heat of fusion of ice is 335 J/g. 9 Formula 10 Example 11 The specific. 2 Latent Heat, Specific Heat, and Work 1. Latent Heat of Ice We will first explore the latent heat of ice. The heat supplied is known as latent heat of fusion (melting) and is stored in the liquid as a kind of potential energy. How much heat must be added to 500 g of aluminum at 27° C to completely melt it?. C into 100% liquid water at 0. From this equation you can calculate the specific latent heat of fusion of ice. 000 C and it all reaches an equilibrium temperature of 19. (Specific heat of ice = 2090 J/kg °C, Specific heat of water =…. Synonyms for heat of fusion in Free Thesaurus. Thermal Conductivity of Ice Cream at -26°C. It takes 5. If the ice is melting, we would need to account for increase in temperature to 0 °C with specific heat capacity of ice as heat is absorbed by the ice. How many moles of H 2 O are contained in the sample? Assume that the molar heat of fusion of ice is 6 kJ/mol. The heat of fusion of a substance is the heat exchange required to melt one gram of the substance (calories/gm). Specific Heat Worksheet Answer Key. The specific latent heat of fusion of ice = 3. • final temperature after ice is melted and knowing the specific heats of aluminum, water, and ice, you will find the latent heat of fusion of ice. 6 kgCorrect answer is option 'C'. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. 0 J of heat when its temperature increases by 120. For the specific heat capacity of water we will use: 4. Calculate the mass of ice that melted at 0oC. •Water vapor content of air may be reported as partial pressure, relative humidity, dew point or frost point, or specific humidity. m s L v + m s c w ΔT = m i L f + m i c w ΔT. Lab 4: Measure the specific latent heat of fusion and the s. Heat given off the steam Q s is equal to the heat taken in by the ice Q i. If you want to find the latent heat of fusion, you take a cup with 0. Prezi’s Big Ideas 2021: Expert advice for the new year; Dec. Fusion is a physical process that is represented as the phase transition of a substance from solid to liquid. Temperature of the each compartment remain homogeneous during whole heat transfer process. Latent Heat of Ice We will first explore the latent heat of ice. If you apply heat to ice, it will melt. The molar latent heat of fusion, which is an alternative name given to the enthalpy of fusion, tells you how much heat is required in order to convert a specific amount of a given substance, either a gram or a mole, from solid at its melting point to liquid Ice is said to have a molar enthalpy of fusion equal to. The specific latent heat of a substance is the amount of heat requires to change the phase of 1 kg of substance at a constant temperature. When 50g of ice has been added and has all melted the temperature of the flask and contents is 40 degree Celsius when a further 80gm of ice has been added and has all melted the temperature of the whole is 10 degree Celsius calculate the specific latent heat of fusion of ice. A determination of the heat of fusion of ice (Lancaster, Pa. the Latent Heat of Fusion and is written as L f. This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the pΔV work). The latent heat of fusion of a substance is. 10 latent heat of fusion (ice to water phase change at ) = 333. The specific latent heat of fusion of water = 334,000J/kg. The amount of latent heat energy required to boil water is 2260 J/g. Specific heat – 1. On freezing, specific heat of foods decreases. The amount of heat required to change the phase of a substance depends on the mass and the type of material of the substance. This means that to convert 1 kg of ice at 0 ºC to 1 kg of water at 0 ºC, 334 kJ of heat must be absorbed by the ice. Specific latent heat is measured in J/kg, if energy is measured in J and mass in kg. 30 kg of ice at 0 °C is added to 1. A 10 g cube of ice at 0 ∘ C is placed outside in direct sunlight on a hot day. (Specific heat capacity of ice = 2. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. vapor --> liquid, latent heat of condensation is released liquid --> ice , latent heat of freezing is released (about 80 cal per gram) ice --> liquid , latent heat of fusion (melting) is added. I managed to obtain the value for steam (2501KJ/kg), but I'm still not successful with ice - there is a multitude of tables and documents that list a variety of materials at 1 bar. An input of 334,000 joules (J) of energy is needed to change 1 kg of ice into 1 kg of water. This is a solid turning into a liquid. How much heat is needed to boil 63 grams of ethanol? Q = (63 g)(838 J/g) = 52. Solution: In this case heat supplied is used into two parts To change the state of ice into the water at 0◦C To increase the temperature of water from 0◦C to 25◦C. This is preferable to the use of a polished calorimeter in an outerjacket owing to the fact that water vapour. In other words, every time the material is heated/cooled, no matter how quickly or by what heating process, the same amount of heat is transferred. 35) How much heat must be added to a 8. In this study, latent heat of fusion of ice in aqueous solutions was measured to understand latent heat of fusion of ice slurries. The temperature stays at 0ºC H 2 O (s) + energy à H 2 O (l) So, in order to calculate the heat of fusion of ice, we need to know how much heat it. Specific Heat Capacity And Specific Latent Heat of Fusion/Vaporization December 26, 2015 November 15, 2010 by Mini Physics Show/Hide Sub-topics (Thermal Physics | A Level). This resources contains the practical for latent heat of fusion of water for the new AQA GCSE. It takes 5. Example 19-6: Determining a latent heat. The specific heats of ice over the temperature range 0° C. : Press of the New era printing company, 1903), by Arthur Whitmore Smith (page images at HathiTrust) Specific heat and heat of fusion of ice, (Washington, Govt. bronze = 435. This Latent Heat of Fusion Lesson Plan is suitable for 10th - 12th Grade. 1 o C just before adding the ice cube. The heat supplied is known as latent heat of fusion (melting) and is stored in the liquid as a kind of potential energy. See Table 2. How much heat is needed to boil 63 grams of ethanol? Q = (63 g)(838 J/g) = 52. When we add ice to an already cold drink, only some of the ice melts, not all of it. Melting Ice on the road by using Salt It is known that the specific latent heat of fusion of salt is higher. Heat given off the steam Q s is equal to the heat taken in by the ice Q i. When a liquid is heated, it starts boiling and is changed into the sc = specific heat capacity of a calorimeter. See full list on en. They can move by vibration, rotation, and translation. 3x10^(5) J/kg calculate the amount of heat energy needed to com - the answers to estudyassistant. 6 kgCorrect answer is option 'C'. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. About this essay More essays like this:. PH – 7 neutral – (less is acid more is base) Mpemba (IM PEM BAH)effect. 0 g of water is cooled from 85. If latent heat of fusion of ice is 80 cals per g at 0°C, calculate molal depression constant for water. The specific latent heat of fusion of ice is sufficiently high, about 336 J g-1. 0005kg Power(W) Temperature. The latent heat of fusion in ice is much higher than its specific heat, which is why it keeps a beverage cool for a longer time. Latent heat is defined for a system with constant temperature. 6 g of water. Fill the insulated cup with exactly 100. It therefore allows us to measure the latent heat of ice by determining the amount of energy gained by ice when it is immersed in a bath of water. Latent heat. This topic will explain the Latent heat of fusion formula with examples. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. Temperature change  Calculate the specific latent heat of ice using the formula: (Initial mass * Latent heat) + (Initial mass * Specific heat capacity * Final temperature) = (Mass of water * Specific heat capacity * Temperature change) (M * L f) + (M * c * T f) = (M w * c * (T. 0 J of heat when its temperature increases by 120. How much heat energy is required to change 0. The use of ice to keep drinks cold in an insulated cooler is a practical example of the use of the latent heat of fusion of water. L f represents the specific latent heat of fusion. How to increase brand awareness through consistency; Dec. 694 W/m-K 6. The formula for Latent heat of fusion: If m kg of the solid changes into the liquid at a constant temperature which is its melting point. 0 1 0 Login to reply the answers Post. The third experiment is about the latent heat of fusion. The latent heat of fusion of water is 144 BTU per pound. l=(m1c(t1-t2))/m. These names describe the direction of energy flow when changing from one phase to the next: from solid to liquid, and liquid to gas. C Latent and Specific Heat: which basically means that the heat absorbed by ice will be the same as the heat released by the steam. 00 C, what is L f? (the specific heat of water is 4186 J/(kg C)). Solution: Concepts: Specific heat, latent heat; Reasoning: The latent heat released as some of the water freezes heats the system. A Lab Experiment to Calculate the Specific Latent Heat of Fusion of Ice PAGES 9. 336 and the value for aluminium is 2. Nevertheless, the latent heat of sea ice seems negligible to William Connolley and probably also most of his colleagues; they prefer the atmospheric CO2 as the object that everyone should look at. H= 1 × 334,000 = 334,000Joules. In this experiment the heat of fusion of water/ice will be determined by adding ice to water in a calorimeter cup. The SI unit of latent heat is joule. Assume that the stirrer was fully immersed. How many J are needed to melt 30 grams of copper? Q = (30 g)(209 J/g) = 6270 J 5. When we add ice to an already cold drink, only some of the ice melts, not all of it. When all the ice is melted, any further heat transfer causes an increase in temperature of the liquid. It is called specific latent heat of vaporisation when the change of state is between liquid and gas. To measure the specific latent heat of ice by direct heating. Understanding Specific Latent HeatSpecific Latent HeatThe amount of heat required to change the phase of a substance depends on the mass and the type of material of the substance. Ice cubes are placed in warm water and the specific heat of fusion of water is determined from the mix temperature. What are synonyms for heat of fusion?. The specific heat capacity of ice is 2 04 kJ/kg °C · specific latent heat of fusion of ice is 3 (a) How much heat transfer is required to raise the temperature of a 0 750-kg aluminum pot containing 2 50 kg of water from 30 0ºC to the boiling point and then boil away 0 750 kg of water?. of the system?. 25 kgThe ice is at 0º C. Specific heat capacity of aluminium = 910 J/kgK Specific heat capacity of water = 4200 JIkgK Specific latent heat of fusion of ice = 335 000 J/kg Specific latent heat ofevaporation water 2. Dimaunahan Instructor March 22, 2019 SCORE Group Report (40) Analysis (25) Conclusion (25) Presentation (10) Safe Assign Score (30% max) TOTAL. The specific latent heat of fusion of ice determined by the activity is 3. Ice at 0°C -----> Water at 0°C -----> Water at 25°C. (Specific heat of ice = 2090 J/kg °C, Specific heat of water =…. 2 words related to heat of fusion: heat of transformation, latent heat. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. Independent : The Ice Cube used its mass 2. 6 kgCorrect answer is option 'C'. Water 334 0 2260 100. Latent Heat The word ‘latent’ was first used by Joseph Black (1728 – 1799), a French-born Scottish chemist noted for his fundamental work on latent heat and specific heat as well as his discovery of carbon dioxide. Specific latent heat of fusion ice = 336 Jg-1 The amount of heat energy lost by the vessel and water contained in it cooling the water from 50 o C to 5 o C is used in heating ice to melt it and ten rasie its temperature from 0 o C to 5 o C. This article has an unclear citation style. Latent heat of fusion = 3. Calculate the final temperature of water after all the ice has melted. If this occurs in an insulated container then the heat gained by the ice is equal to the heat lost by the container and the water. We got a value of 7, so obviously either the theory is crap or I need a new lab partner. 2 x 103J kg-1oC-1, Latent heat of fusion of ice = 3. At which stages the specific latent heat released? Latent Heat of Fusion Calculate the amount of energy required to melt 35. C Latent and Specific Heat: which basically means that the heat absorbed by ice will be the same as the heat released by the steam. Nevertheless, the latent heat of sea ice seems negligible to William Connolley and probably also most of his colleagues; they prefer the atmospheric CO2 as the object that everyone should look at. What is the Latent Heat of Fusion of a substance? Energy needed to melt or freeze 3. Autores: Se-Yuen Mak , C. But what does all these things mean? Well this is the amount of energy it takes to change or to change a phase of 1 mol of substance so it's going to change a phase so like for example if we're talking about fusing. J would melt 1 kg of ice. Small lumps of ice at 0 degrees Celcius were rapidly dried on filter appaer and dropped into a beaker containing 85g of water at 18. calculate a value for the specific latent heat of fusion for water, stating one assumption made. In this part of the experiment, the heat of fusion of water will be determined. The specific heat capacity of ice is 2 04 kJ/kg °C · specific latent heat of fusion of ice is 3 (a) How much heat transfer is required to raise the temperature of a 0 750-kg aluminum pot containing 2 50 kg of water from 30 0ºC to the boiling point and then boil away 0 750 kg of water?. 3 × 105 J/kgQ. The measurement of the specific latent heat of fusion of ice: two improved methods. The heat that is given off when water freezes keeps the atmospheric temperature higher. How much heat is needed to boil 63 grams of ethanol? Q = (63 g)(838 J/g) = 52. The units for both values are the same. Latent heat is nothing magical but can be very confusing to understand. The heat of fusion for water at 0 °C is approximately 334 joules (79. 3 Bogojawlenskv 143. Latent Heat of Fusion. Draw out a heating curve for mercury and water on the same graph. 1X103 J/kg-K. The experimental value L* F is seen to be too small since the denominator of the first expression for latent heat above contains the entire mass and is therefore too large. Latent Heat and Specific Latent Heat Latent heat is “hidden heat”. The energy released comes from the potential energy stored in the bonds between the particles. Specific Heat is the amount of energy required to raise the 1°C temperature of a body of unit mass. Synonyms for heat of fusion in Free Thesaurus. Q = m L Notice the subscript in L f. 0 g of water at its melting point of 0 °C. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. How much heat in kilojoules is released when 25. (Solved) Explain how specific latent heat of fusion can be determined using mixture method. To measure the specific latent heat of ice by the method of mixtures The first step is to measure the specific heat capacities of copper and water by the methods described earlier in this chapter. 2009's methodology centred around the use of. 00 C and drop 0. Constant : Specific Heat Capacity of ice & water. Using the equation for the heat required for melting, and the value of the latent heat of fusion of water from the previous table, we can solve for part (a). The references used may be made clearer with a different or consistent style of citation and footnoting. The same amount of energy needs to be taken out of the liquid to freeze it. 6 kgCorrect answer is option 'C'. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. There is flexibility in the choice of phase-change temperature range and latent heat of fusion. 02 J/g oC, Specific latent heat of fusion of ice = 334 J/g, Specific latent heat of vaporisation of water = 2260 J/g) 1. Ice has a latent heat of fusion of L = 335,000 J/kg. D: Water boils and absorbs latent heat of vaporization. Latent heat of Fusion : It is the heat required to change 1 kg of a solid substance into liquid state at the melting point of the substance. 48 x 108 g of H. 2 Latent Heat, Specific Heat, and Work Worksheet 12. Specific Heat of Ice Cream at 4°C 2. Water has a known specific heat capacity of 4. What is the Latent Heat of Fusion of a substance? Energy needed to melt or freeze 3. The specific latent heat of fusion of water = 334,000J/kg. Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution. 166 J/g, is called the heat of fusion, it is not called the molar heat of fusion. While the specific heat capacity determines the amount of energy required to increase the temperature of an object by 1°C for every kilogram present through heating, the latent heat of fusion or enthalpy is the energy that is required to change the phase of a matter with no change in. 694 W/m-K 6. Both specific heat and latent heat are properties of a given material. 7915 J/g°C – granite) The specific heat capacity of a compound used in fireworks is 0. A block of ice with a mass of 750 g falls from a plane at 9. Latent heat of fusion. Calculate the specific latent heat of fusion of ice, neglecting any heat loss to the surroundings. The heat of fusion for water at 0 °C is approximately 334 joules (79. Water 334 0 2260 100. When a further 80g of ice is added and all has melted, the temperature of the whole is 10deg. 34 x 105 J kg-1 and. If the power of the heater is P and the time taken to heat the ice = t, then the thermal energy supplied by the heater = thermal energy used to melt ice = Pt. edu The ADS is operated by the Smithsonian Astrophysical Observatory under NASA Cooperative Agreement NNX16AC86A. 100 J of heat energy can melt a large amount of wax, but the same amount of heat can only. If the liquid cools as resolidifies (freezes), the stored potential energy is given back out as heat energy. Extinguishing fire by using boiling water Water that is quickly boiled will become steam which is able to absorb a larger quantity of latent heat from the fire. Solution for Calculate the total heat required to convert 100 g ice at –15°C to steam at 120 °C. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. Fusion and Evaporation Heat of common Materials - Melting points, heat of fusions, boiling points and heat to evaporate common substances - like hydrogen, water, gold and more. The latent heat of melting is the energy absorbed by ice to make water, the latent heat of fusion is the energy released by frozen water turning into ice and the latent heat of vaporization is the thermal energy needed to evaporate water. Place some ice in a beaker of water and keep taking the temperature of the ice-water mixture until it reaches 0 degrees. Specific Heat Capacity Latent Heat (Fusion) Latent Heat (Vaporization) Melting Points Boiling Points Density Air (room conditions) 1012 1. So the formula tells us how much heat energy is needed to go into a material to change it from a solid to a melted liquid. 39 x 105 J kg -1. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. 3 × 105 J kg-1; suggest two reasons why your answer is not this value. 34 x 106 J/kg). 6 kgCorrect answer is option 'C'. 334 kJ is enthalpy of ice or the amount of heat required to melt the ice to. See full list on en. 3 × 105 J kg-1. In most of the experiments in measuring specific latent heat of fusion, I don't remember specific heat capacity of ice is given. This causes ice to float on water. Ice is not very free to move. 0 J of heat when its temperature increases by 120. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. Specific latent heat of fusion of ice = 330 000 Jkg-1 Electrical energy input = 65x5x60 = 19500 J Mass of ice melted = 19500/330 000 = 0. Latent heat of fusion. Check us out at http://physics. 50-kg aluminum calorimeter filled with 1. Heat capacity is defined as the heat required to change the state of a substance. ii) the specific latent heat of fusion of the substance in liquid state. Latent Heat of Fusion Calculator. 20 J g–1 °C–1. Using the digital scale or the triple beam balance determine the mass of the empty. 8X10^5j/kg, it is found that the data from the experiment is relatively accurate. H= 1 × 334,000 = 334,000Joules. 3 × 105 J/kgQ. Included is a video modeling how to carry out the practical, differentiated powerpoint, with keywords, learning objectives, differentiated outcomes, questions on calculating latent heat with model answers for those questions. The same amount of energy needs to be taken out of the liquid to freeze it. Latent heat of Fusion : It is the heat required to change 1 kg of a solid substance into liquid state at the melting point of the substance. 059 kg = 59 g. 0 kg of ice at 0°C to water at 50°C? 2. The purpose of this lab experiment is to measure the specific heat capacity of unknown metal samples and also to determine the latent heat of fusion of water. Specific Latent Heat It takes a certain amount of energy to change the state of 1kg of water from solid to liquid. Specific latent heat (L) of a substance is the amount of thermal energy required for 1kg of a substance to change from one state into another. H= 1 × 334,000 = 334,000Joules. Answer: 1 📌📌📌 question a 20 of of ice at 0c is dropped into water at boiling point, specific heat capacity of water =4200 J/kg•c, sepesific latent heat of fusion (Lf) =3. Find more ways to say latent heat, along with related words, antonyms and example phrases at Thesaurus. A Lab Experiment to Calculate the Specific Latent Heat of Fusion of Ice PAGES 9. Date posted: April 24, 2019. form ice, also at 0 °C. They can move by vibration, rotation, and translation. The specific latent heat of fusion The specific latent heat of fusion of a substance is the quantity of heat required to convert unit mass of the substance from the solid to the liquid state without change in temperature. Suppose you're in a restaurant with a glass of 100. Variables:1. When ice, at zero degrees celsius, is added to warm water, it first melts and then its temperature increases as it takes heat from the water. Suggest reason why the result is different from the accepted value. If latent heat of fusion of ice is 80 cals per g at 0°C, calculate molal depression constant for water. The accepted value for the specific latent heat of fusion of ice is 3. The third experiment is about the latent heat of fusion. How many J are needed to melt 30 grams of copper? Q = (30 g)(209 J/g) = 6270 J 5. 3% of that and apply that to the "heat of melting" of ice. Specific heat capacity, ice is 2. A metal has a latent heat of fusion of 2. Latent heat of. 3 × 105 J kg-1. So – how do you know how much ice has melted? Water ice contracts on melting into liquid. Heat; Fusion; Used for: Latent heat of fusion; Filed under: Fusion, Latent heat of. Latent heat of fusion of ice is the quantity of heat required to convert one gm of ice at 0 0 C into water at 0 0 C. 0 g of water is cooled from 85. What does latent mean? The definition of latent is something present but invisible or hidden. 33 x 10 5 J/Kg. C Latent and Specific Heat: which basically means that the heat absorbed by ice will be the same as the heat released by the steam. Calculate Specific latent heat using heat energy and mass with this online calculator. Calculate the mass of ice that melted at 0oC. Explain why it is important that the ice is crushed so that it melts quickly. See full list on en. (ii) Determine the latent heat of fusion of ice. The heat taken in by the ice Q i will be equal to, Q i = m i L f + m i c w ΔT. 6 kgCorrect answer is option 'C'. Specific Heat. 8X10^5j/kg, it is found that the data from the experiment is relatively accurate. Similarly, while ice melts, it remains at 0 °C (32 °F), and the liquid water that is formed with the latent heat of fusion is also at 0 °C. l=(m1c(t1-t2))/m. Specific Latent Heat Of Fusion of Ice Lice : Specific latent heat of fusion of ice is the quantity of heat energy that is required to change 1kg mass of ice block into liquid at constant temperature Or melting point of the ice block. 2 Aluminium 897 396938. The Latent Heat of Fusion (L f) for water is 3. What is the final temperature, assuming no heat exchange with the surroundings? Take the specific heat capacity of water to be 4200 J kg -1 K -1 and the specific latent heat of fusion of ice to be 3. Specific Latent Heat Of Fusion of Ice L ice = 336 J/g. Repeat the experiment to get 2 sets of results. Calculate the energy needed to melt a 200g ice cream at an initial temperature of -10. Latent heat of fusion = 3. Special heat capacity, you can look at them or the data. If the ice is melting, we. Starting below 0C as heat is added the line rises steeply to 0C where it temporarily levels out. 6 kgCorrect answer is option 'C'. 01 Trial 1 : Specific heat capacity is commonly defined as the amount of heat energy. ” For this paper we will focus on the latent heat of fusion and those high latent heats associated with the phase change between solids to liquids and vice versa. The initial mass of the ice in the container is equal toa)0. C Latent and Specific Heat: which basically means that the heat absorbed by ice will be the same as the heat released by the steam. 3 × 105 J/kgQ. Latent Heat of Fusion and Vaporization Q mc ' T Q f mL f Q v mL v 1. 0 Ammonia has highest ice and steam SH =. I used this formula for latent heat. How much heat is needed to boil 63 grams of ethanol? Q = (63 g)(838 J/g) = 52. Phase diagram of water. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. of vaporization. 1 × 103 J kg–1 K–1 Specific heat capacity of water = 4. How much heat is absorbed by 100 g of ice at –10˚C to become water at 20˚C? 2. The initial mass of the ice in the container is equal toa)0. The problem is the values of latent heat and enthalpy of vaporization wander quite a bit with pressure, and they are quite hard to find for anything else than 1 bar. 1 kJ kg k Density of ice = 920 kg m Markscheme mass = «volume x density» (0. We know: Specific Heat of water = 4186 J/Kg Degree Specific Heat of copper = 387 J/Kg Degree Latent heat of vaporisation of water = 2. Specific heat capacity, water is 4. Nevertheless, the latent heat of sea ice seems negligible to William Connolley and probably also most of his colleagues; they prefer the atmospheric CO2 as the object that everyone should look at. 2016130301 BS ECE-1 Group 3 PHY101L-C2 Engr. 3 x fusion is the 105 J/kg which means that 3. 6 kgCorrect answer is option 'C'. Use Equation 2 and the data in the Data Section to solve for the latent heat of fusion Lf. The heat required to melt (or freeze) a substance is the Latent Heat of Fusion, L F Latent heat of fusion of water/ice = 80 kcal/kg = 333 kJ/kg The heat required to vaporise (condense) a substance is the Latent Heat of Vaporization, L V. The temperature on the y-axis and heat input on the x-axis. 2 kg of water at 20. If you apply heat to ice, it will melt. E h means energy, equals m for mass, times l. Heat can be transferred through convection, radiation, and conduction. Each substance has two specific latent heats:. 'The specific latent heat capacity of fusion of ice is 336 J per g'? Solution 15 It means that 1 g of ice at 0 o C absorbs 336 J of heat energy to convert into water at 0 o C. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. 2 Latent Heat, Specific Heat, and Work 1. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. Dimaunahan Instructor March 22, 2019 SCORE Group Report (40) Analysis (25) Conclusion (25) Presentation (10) Safe Assign Score (30% max) TOTAL. In the process of melting. A metal weighing 50. A Lab Experiment to Calculate the Specific Latent Heat of Fusion of Ice PAGES 9. the substance and its specific heat, c, Compute the latent heat of fusion of the ice using Eq. The use of ice to keep drinks cold in an insulated cooler is a practical example of the use of the latent heat of fusion of water. Phase transition has the capacity to store thermal as latent heat of fusion at a constant temperature corresponding to the phase transition temperature of. 2 Latent Heat, Specific Heat, and Work Worksheet 12. 755 kg of water at 30. What is the final temperature, assuming no heat exchange with the surroundings? Take the specific heat capacity of water to be 4200 J kg -1 K -1 and the specific latent heat of fusion of ice to be 3. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. 6 kgCorrect answer is option 'C'. 3 grams of perfectly dry ice would be required for a heat of fusion equal to 80 cal/gm. See Table 2. Thus, the latent heat of fusion encompasses the process of adding heat to melt some solid. com, the world's most trusted free thesaurus. It turns out that the specific heat capacity of water 4186 coul section kilogram degrees Celsius. What is the Latent Heat of Vaporization of a substance? Energy needed to boil or condense 4. temperature. Answer: The latent heat of fusion and vaporization both involve the heat required to change the state of a substance without a change in temperature. 629 J/kg-K 5. This discussion on Ice at 0 degree celcuis is added to 200g of water initially at 70 degree celcuis in a vacuum flask. How many J are needed to melt 30 grams of copper? Q = (30 g)(209 J/g) = 6270 J 5. The most common forms of latent heat are fusion and vaporization. In the case of the latent heat of fusion it is the heat required to change a substance from a solid (ice) to a liquid (water) or vice versa while the latent heat of vaporization from a liquid. Ice can vibrate but ice remains rigid. takes in and its mass. G U D H E I M Lever Brothers Company Today, many problems in oils and fats require a thorough understanding as to the changes taking place during heating and cooling, as well as their specific and latent heats of. Specific heat capacity of. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. example, in this experiment heat added to ice at 0°C causes a change in phase from solid to liquid. Solution for Calculate the total heat required to convert 100 g ice at -15°C to steam at 120 °C. Therefore, the latent heat of fusion of a substance, or enthalpy of fusion, is the change of enthalpy that results from the absorption of energy and that leads the substance in question to pass from solid phase to liquid phase at constant pressure. Water has a very high latent heat for evaporation and fusion, due to the forces between molecules associated with hydrogen bonds (2. 1944-05-01 00:00:00 A. 3x10^(5) J/kg calculate the amount of heat energy needed to com - the answers to estudyassistant. The measurement of the specific latent heat of fusion of ice: two improved methods. 2016130301 BS ECE-1 Group 3 PHY101L-C2 Engr. Latent heat of fusion: If you have a glass of a cool drink, well supplied with ice, you can expect its temperature to drop until it is close to 0 ºC. How long does it takes to freeze 4 kg of water at the temperature 14°C? [Specific latent heat of fusion of ice = 336 J/(g°C), specific heat capacity of water = 4. The specific latent heat of fusion of a substance is defined as the amount of heat required to change 1 kg of the substance from the solid to liquid phase without a change in temperature. Which process, specific latent heat of fusion of ice, or specific latent heat of vaporisation of water requires the least amount of energy to break the intermolecular bonds between any two particles within. The initial mass of the ice in the container is equal toa)0. with mass m soda = 0. At the point when the heat of fusion is referenced to a unit of mass, it is typically called the specific heat of fusion, while the molar heat of fusion alludes to the enthalpy change per measure of substance in moles. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. This heat is called latent heat of fusion. 33 x 10 5 J/Kg. We calculate the mass of ice melted using the the equation: {eq}Q_1 = m_{ice \rightarrow water}L_f {/eq}, where, {eq}L_f = 335000 \ \rm{J/kg} {/eq} is the latent heat of fusion,. Measuring the Latent Heat of Fusion The "calorimeter" is an insulated container which (hopefully) does not allow heat to enter or escape. 'The specific latent heat capacity of fusion of ice is 336 J per g'? Solution 15 It means that 1 g of ice at 0 o C absorbs 336 J of heat energy to convert into water at 0 o C. Check us out at http://physics. Specific heat capacity, ice is 2. Latent heat. (Specific heat of ice = 2090 J/kg °C, Specific heat of water =…. 5 kg of ice at 0°C into water at 25°C? [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/(kg K). Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature. Statement-2: Latent heat refers to change of state without any change in temperature. with mass m soda = 0. The temperature rise in each case is shown in the table. 50-kg aluminum calorimeter filled with 1. How to increase brand awareness through consistency; Dec. 186 J/gCo, while for the aluminum cup and stirrer the specific heat is c Al = 0. Which process, specific latent heat of fusion of ice, or specific latent heat of vaporisation of water requires the least amount of energy to break the intermolecular bonds between any two particles within. Using the Specific Latent Heat Equation. Answer: The latent heat of fusion and vaporization both involve the heat required to change the state of a substance without a change in temperature. Specific Latent Heat¶ The specific latent heat of a substance, is the amount of energy per kilogram required to change the state of said substance. Latent heat of fusion = 3. Ice cubes are placed in warm water and the specific heat of fusion of water is determined from the mix temperature. The initial mass of the ice in the container is equal toa)0. Latent heat of fusion of ice = 335 kJ/kg. temperature. 6 kgCorrect answer is option 'C'. 2 degrees celcius. How much heat does it absorb? 3. 2 x 103 J kg-1 K-1. 2 words related to heat of fusion: heat of transformation, latent heat. Answer: 1 📌📌📌 question a 20 of of ice at 0c is dropped into water at boiling point, specific heat capacity of water =4200 J/kg•c, sepesific latent heat of fusion (Lf) =3. Extinguishing fire by using boiling water Water that is quickly boiled will become steam which is able to absorb a larger quantity of latent heat from the fire. Latent Heat The word ‘latent’ was first used by Joseph Black (1728 – 1799), a French-born Scottish chemist noted for his fundamental work on latent heat and specific heat as well as his discovery of carbon dioxide. The calculation of the specific latent heat of fusion of ice is as follows. change a kilogram of solid ice by one degree Celsius is called the. What is the Latent Heat of Fusion of a substance? Energy needed to melt or freeze 3. Solution for Calculate the total heat required to convert 100 g ice at –15°C to steam at 120 °C. C Latent and Specific Heat: which basically means that the heat absorbed by ice will be the same as the heat released by the steam. The ice will melt, producing ice water (water at 0 oC). Specific latent heat is the latent heat of 1kg of a material, represent by the letter L. 1 kJ kg k Density of ice = 920 kg m Markscheme mass = «volume x density» (0. Given specific heat of ice = 2100 J/kg-K Given specific heat of water = 4200 J/kg-K Latent heat of fusion of ice = 3. In this chemistry lesson, students explain how thermal storage systems work. Heat lost = Heat gained Heat lost by warm water = Heat needed to melt ice + Heat needed to warm the water which was once ice M wCw (T w - T f) = M ice Cw (T f - 0) + M ice Lf Eq. The initial mass of the ice in the container is equal toa)0. Latent heat is the work done in a system in order to hold the atoms or molecules of matter in the same phase.


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# Plotting an function defined by an integral [closed]

How can I plot a function defined by an integral.

More specific, I have the following equation:

$$T = \frac{1}{\pi}\int\sqrt{\sin^2\;\arccos\left(-\frac{E}{2}\right)-\sin^2\frac{ϕ}{2}}^{-1}\mathrm{d}ϕ$$

Translating to Mathematica input:

T[E_]:=(1/Pi)*NIntegrate[(Sin[ArcCos[-E]/2]^2 - Sin[ϕ/2]^2)^(-0.5), {ϕ, 0, ArcCos[E]/2}]

Now I need to plot TxE. How do I do it?

-

## closed as off-topic by Pickett, Michael E2, belisarius, RunnyKine, ÖskåAug 25 at 22:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Pickett, Michael E2, belisarius, RunnyKine, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.

You have many trivial problems in your code such as NIntegral instead of NIntegrate, missing square brackets and parenthesis instead of square brackets. Those you should be able to fix yourself if you pay attention to the documentation on those functions. You are on the right site, though. –  Pickett Aug 25 at 18:37
hmm why not? Im trying to plot the function in mathematica –  Geo Aug 25 at 18:37
@Picket, My code is just fine, i just added it here without correct words. I will edit it correctly. –  Geo Aug 25 at 18:38
@Geo If it works for you in Mathematica, better copy and paste it so you don't introduce errors. –  Pickett Aug 25 at 18:39
Using e rather than your E, you can avoid doing any numerical integration by using Integrate to symbolically integrate your expression, and then using Simplify with Assumptions -> e >= -1, to obtain (2 Sqrt[2] EllipticF[ArcCos[e]/4, 2/(1 + e)])/(Sqrt[1 + e] \[Pi]). I used Mathematica 10 to do this. –  Stephen Luttrell Aug 25 at 19:58

Clear[T]

T[e_?NumericQ] := (1/Pi)*NIntegrate[
1/Sqrt[Sin[ArcCos[-e]/2]^2 - Sin[\[Phi]/2]^2],
{\[Phi], 0, ArcCos[e]/2}]

Plot[
{Re[T[e]], Im[T[e]], Abs[T[e]]},
{e, -2, 2},
PlotRange -> All,
WorkingPrecision -> 25,
PlotLegends -> "Expressions",
Frame -> True,
Axes -> False]

EDIT: With the revised formula from your latest edit:

Clear[T]

T[e_?NumericQ] := (1/Pi)*NIntegrate[
Sqrt[Sin[ArcCos[-e/2]]^2 - Sin[\[Phi]/2]^2],
{\[Phi], 0, ArcCos[e]/2}]

Plot[
{Re[T[e]], Im[T[e]], Abs[T[e]]},
{e, -2, 2},
PlotRange -> All,
WorkingPrecision -> 25,
PlotLegends -> "Expressions",
Frame -> True,
Axes -> False]

Perhaps you also meant to change the limits of integration?

Clear[T]

T[e_?NumericQ] := (1/Pi)*NIntegrate[
Sqrt[Sin[ArcCos[-e/2]]^2 - Sin[\[Phi]/2]^2],
{\[Phi], 0, ArcCos[e/2]}]

Plot[
{Re[T[e]], Im[T[e]], Abs[T[e]]},
{e, -2, 2},
PlotRange -> All,
WorkingPrecision -> 25,
PlotLegends -> "Expressions",
Frame -> True,
Axes -> False]

-
uh the first one is the correct. I thought that it should diverge at e=1 but seems it won't. thanks sir. –  Geo Aug 25 at 22:48

format: plain


text:
# Abelian-quotient-pullbackable automorphism-invariant subgroup

BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]
This article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]

## Definition

A subgroup $H$ of an abelian group $G$ is termed an abelian-quotient-pullbackable automorphism-invariant subgroup if, for every abelian-quotient-pullbackable automorphism $\sigma$ of $G$, $\sigma$ restricts to an automorphism of $H$.

## Formalisms

### Function restriction expression

This subgroup property is a function restriction-expressible subgroup property: it can be expressed by means of the function restriction formalism, viz there is a function restriction expression for it.
Find other function restriction-expressible subgroup properties | View the function restriction formalism chart for a graphic placement of this property
Function restriction expression $H$ is a fully invariant subgroup of $G$ if ... This means that full invariance is ... Additional comments
abelian-quotient-pullbackable automorphism $\to$ function every abelian-quotient-pullbackable automorphism of $G$ sends every element of $H$ to within $H$ the invariance property for abelian-quotient-pullbackable automorphisms
abelian-quotient-pullbackable automorphism $\to$ endomorphism every abelian-quotient-pullbackable automorphism of $G$ restricts to an endomorphism of $H$ the endo-invariance property for abelian-quotient-pullbackable automorphisms; i.e., it is the invariance property for abelian-quotient-pullbackable automorphism, which is a property stronger than the property of being an endomorphism
abelian-quotient-pullbackable automorphism $\to$ automorphism every abelian-quotient-pullbackable automorphism of $G$ restricts to an automorphism of $H$ the auto-invariance property for abelian-quotient-pullbackable automorphisms; i.e., it is the invariance property for abelian-quotient-pullbackable automorphism, which is a group-closed property of automorphisms

## Relation with other properties

### Stronger properties

property quick description proof of implication proof of strictness (reverse implication failure) intermediate notions
Characteristic subgroup of abelian group characteristic subgroup of abelian group |FULL LIST, MORE INFO

### Weaker properties

property quick description proof of implication proof of strictness (reverse implication failure) intermediate notions
Subgroup of abelian group subgroup of abelian group (by definition) subgroup of abelian group not implies abelian-quotient-pullbackable automorphism-invariant |FULL LIST, MORE INFO


text:
# [SOLVED]Velocity and acceleration vectors and their magnitudes

#### Dhamnekar Winod

##### Active member

How to answer this question? I am working on this question. Any math help, hint or even correct answer will be accepted.

#### DaalChawal

##### Member
Differentiate them to get velocities and accelerations in x, y, z(i, j, k ) directions. Add them vectorially to get resultant velocity and acceleration.

#### Dhamnekar Winod

##### Active member
$v=\sqrt{37}, a= \sqrt{325}$ at t=0

#### DaalChawal

##### Member
so? what problem are you facing? Once you get velocity and acceleration, calculate the magnitude by $\sqrt[]{a^2 + b^2 +c^2}$ where a vector is given by $a i + b j + c k$

#### Dhamnekar Winod

##### Active member
so? what problem are you facing? Once you get velocity and acceleration, calculate the magnitude by $\sqrt[]{a^2 + b^2 +c^2}$ where a vector is given by $a i + b j + c k$
Answers given are magnitudes of velocity and acceleration vectors at t=0. What are you talking about?

#### DaalChawal

##### Member
$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^{-t}$
similarly calculate velocity and acceleration in $y, z$ directions.
Then
You will get velocity as $- e^{-t} i +(-6)sin(3t) j + 6cos(3t) k$
Now put t=0 to get velocity at t=0nand calculate the magnitude.
for acceleration try yourself

#### Dhamnekar Winod

##### Active member
$x= e^{-t}$

$v_x = - e^{-t}$

$a_x = e^t$
similarly calculate velocity and acceleration in $y, z$ directions.
Then
You will get velocity as $- e^{-t} i +(-6)sin(3t) j + 6cos(3t) k$
Now put t=0 to get velocity at t=0nand calculate the magnitude.
for acceleration try yourself
I already computed velocity and acceleration vectors. But i only posted here their magnitudes.

#### DaalChawal

##### Member
Okay then use formula that if a vector is expressed as $x = a i + b j + c k$ then its magnitude is given by $|x| = \sqrt[]{a^2+b^2+c^2}$

#### Country Boy

##### Well-known member
MHB Math Helper
Since this has been sitting a while, yes, the answers given in post #3 are corret.

We have x= e^{-t}, y= 2 cos(3t), and z= 2 sin(3t).
The velocity is given by x'= -e^{-t}, y'= -6 sin(3t), and z'= 6 cos(3t).
The acceleration is given by x''= e^{-t}, y''= -18 cos(3t), and z''= -18 sin(3t).

|v(0)|= sqrt{1+ 0+ 36}= sqrt{37}
|a(0)|= sqrt{1+ 324+ 0}= sqrt{325}


text:
# QM 2011 - XXII International Conference on Ultrarelativistic Nucleus-Nucleus Collisions

May 22 – 28, 2011
Centre Bonlieu
Europe/Zurich timezone

## Open heavy flavor physics in the muon channel with ALICE in pp collisions at 7 TeV and PbPb at 2.76 TeV

May 23, 2011, 5:50 PM
20m
Salon Ravel AB (ImpErial Palace)

### Salon Ravel AB

#### ImpErial Palace

Parallel Heavy flavor and quarkonia production

### Speaker

Xiaoming Zhang (Laboratoire de Physique Corpusculaire, Clermont-Ferrand)

### Description

Heavy-quark production is one of the probes for the investigation of the properties of the high-density medium formed in heavy-ion collisions. This investigation requires also the study of proton-proton collisions. Besides providing the necessary baseline for nucleus-nucleus collisions, proton-proton collisions are of great interest, also in their own right, since they allow to test perturbative QCD at unprecedented low Bjorken-x values. In this contribution, we measure the heavy-flavour production by detecting single muons from semi-leptonic decays. After a description of the ALICE muon spectrometer, we will present the results on the production of single muons from heavy flavour decays at forward rapidity (2.5 < y < 4) in pp collisions at Heavy-quark production is one of the probes for the investigation of the properties of the high-density medium formed in heavy-ion collisions. This investigation requires also the study of proton-proton collisions. Besides providing the necessary baseline for nucleus-nucleus collisions, proton-proton collisions are of great interest, also in their own right, since they allow to test perturbative QCD at unprecedented low Bjorken-x values. In this contribution, we measure the heavy- avour production by detecting single muons from semi-leptonic decays. After a description of the ALICE muon spectrometer, we will present the results on the production of single muons from heavy avour decays at forward rapidity (2:5 < y < 4) in pp collisions at √s = 7 TeV and Pb-Pb collisions at Heavy-quark production is one of the probes for the investigation of the properties of the high-density medium formed in heavy-ion collisions. This investigation requires also the study of proton-proton collisions. Besides providing the necessary baseline for nucleus-nucleus collisions, proton-proton collisions are of great interest, also in their own right, since they allow to test perturbative QCD at unprecedented low Bjorken-x values. In this contribution, we measure the heavy- avour production by detecting single muons from semi-leptonic decays. After a description of the ALICE muon spectrometer, we will present the results on the production of single muons from heavy avour decays at forward rapidity (2:5 < y < 4) in pp collisions at √s = 7 TeV and Pb-Pb collisions at √sNN = 2.76 TeV. In particular, we will show the pt-dierential production cross section in pp collisions and compare it to perturbative QCD predictions, and the nuclear modication factors in Pb-Pb collisions. RAA with respect to a pp reference at √s = 2.76 TeV and RCP in central with respect to peripheral collisions.

### Primary author

Xiaoming Zhang (Laboratoire de Physique Corpusculaire, Clermont-Ferrand)

 Slides


text:
SharedOrbits_and_rotations.ipynbOpen in CoCalc
Author: W. Patrick Hooper
Views : 10

# Orbits of rotations

By default, we will represent points in the circle by numbers in the interval $[0,1)$. The following function converts from these coordinates to the unit circle.

In [4]:
def to_unit_circle(x):
return (cos(2*pi*x), sin(2*pi*x))

Here is an example of the use of the function above:

In [5]:
to_unit_circle(0)

(1, 0)

Note that the output of the last command in a block is automatically printed in Jupyter.

In [7]:
to_unit_circle(1/4)

(0, 1)
In [8]:
to_unit_circle(1)

(1, 0)

The following function returns the rotation of the circle by alpha.

In [9]:
def rotation(alpha):
def T(x):
val = x+alpha
return val - floor(val)
return T

The following constructs the rotation by 2/5:

In [10]:
T = rotation(2/5)

Here we test it:

In [11]:
T(0)

2/5
In [12]:
T(4/5)

1/5

The following is a function that computes the forward orbit of $x$ under $T$ up until time $N$. So, this returns the list of values $[x,T(x), \ldots, T^N(x)].$

In [13]:
def forward_orbit(x, T, N):
orbit = [x] # Start of the orbit.
y = x
for i in range(N):
y = T(y) # Redefine y to be T(y)
orbit.append(y) # Add y at the end of the orbit.
return orbit

Here we see that $0$ has a periodic orbit of least period $5$:

In [14]:
orbit = forward_orbit(0,T,5)
orbit

[0, 2/5, 4/5, 1/5, 3/5, 0]

The following converts orbit into circle coordinates:

In [15]:
circle_orbit = [to_unit_circle(x) for x in orbit]
circle_orbit

[(1, 0), (-1/4*sqrt(5) - 1/4, 1/4*sqrt(-2*sqrt(5) + 10)), (1/4*sqrt(5) - 1/4, -1/4*sqrt(2*sqrt(5) + 10)), (1/4*sqrt(5) - 1/4, 1/4*sqrt(2*sqrt(5) + 10)), (-1/4*sqrt(5) - 1/4, -1/4*sqrt(-2*sqrt(5) + 10)), (1, 0)]

We can plot the circle_orbit:

In [16]:
line2d(circle_orbit, aspect_ratio=1)

Here we plot two orbits:

In [17]:
plt1 = line2d(circle_orbit, aspect_ratio=1)
orbit2 = forward_orbit(1/3, T, 5)
circle_orbit2 = [to_unit_circle(x) for x in orbit2]
plt2 = line2d(circle_orbit2, aspect_ratio=1, color="red")
plt1 + plt2 # Addition combines plots

## An irrational orbit:

Sage can do exact arithmetic in the field AA of all algebraic reals. For more infromation on AA see: http://doc.sagemath.org/html/en/reference/number_fields/sage/rings/qqbar.html.

In [18]:
AA

Algebraic Real Field
In [19]:
alpha = AA(sqrt(2)-1)
alpha

0.4142135623730951?
In [20]:
T = rotation(alpha)

In [21]:
T(0)

0.4142135623730951?
In [22]:
orbit = forward_orbit(0,T,100)

In [23]:
circle_orbit = [to_unit_circle(x) for x in orbit]

In [24]:
line2d(circle_orbit, aspect_ratio=1)

Of course, the full infinitely long forward orbit of any point is dense, as we will prove in class. Also see Devaney's Theorem 3.13.


text:
# Carambis PhotoTrip

## A way to uninstall Carambis PhotoTrip from your PC

You can find on this page detailed information on how to remove Carambis PhotoTrip for Windows. The Windows version was created by MEDIA FOG LTD. More data about MEDIA FOG LTD can be read here. Carambis PhotoTrip is typically installed in the C:\Program Files\Carambis\PhotoTrip folder, but this location can differ a lot depending on the user's option while installing the application. The full command line for removing Carambis PhotoTrip is C:\Program Files\Carambis\PhotoTrip\uninstall.exe. Keep in mind that if you will type this command in Start / Run Note you might get a notification for administrator rights. PhotoTripDeviceServiceInstaller.exe is the Carambis PhotoTrip's main executable file and it occupies about 22.00 KB (22528 bytes) on disk.

Carambis PhotoTrip is composed of the following executables which occupy 1.58 MB (1656320 bytes) on disk:

• CrashSender1402.exe (920.50 KB)
• dcraw.exe (675.00 KB)
• PhotoTripDeviceServiceInstaller.exe (22.00 KB)

This info is about Carambis PhotoTrip version 2.0.1.2315 only. You can find below a few links to other Carambis PhotoTrip versions:

When planning to uninstall Carambis PhotoTrip you should check if the following data is left behind on your PC.

Folders found on disk after you uninstall Carambis PhotoTrip from your computer:
• C:\Program Files\Carambis\PhotoTrip

The files below are left behind on your disk when you remove Carambis PhotoTrip:
• C:\Program Files\Carambis\PhotoTrip\CrashRpt1402.dll
• C:\Program Files\Carambis\PhotoTrip\CrashSender1402.exe
• C:\Program Files\Carambis\PhotoTrip\dbghelp.dll
• C:\Program Files\Carambis\PhotoTrip\dcraw.exe
• C:\Program Files\Carambis\PhotoTrip\imageformats\qgif4.dll
• C:\Program Files\Carambis\PhotoTrip\imageformats\qico4.dll
• C:\Program Files\Carambis\PhotoTrip\imageformats\qjpeg4.dll
• C:\Program Files\Carambis\PhotoTrip\imageformats\qmng4.dll
• C:\Program Files\Carambis\PhotoTrip\imageformats\qsvg4.dll
• C:\Program Files\Carambis\PhotoTrip\imageformats\qtiff4.dll
• C:\Program Files\Carambis\PhotoTrip\libeay32.dll
• C:\Program Files\Carambis\PhotoTrip\libssl32.dll
• C:\Program Files\Carambis\PhotoTrip\PhotoTripDeviceServiceInstaller.exe
• C:\Program Files\Carambis\PhotoTrip\QtCore4.dll
• C:\Program Files\Carambis\PhotoTrip\QtGui4.dll
• C:\Program Files\Carambis\PhotoTrip\QtNetwork4.dll
• C:\Program Files\Carambis\PhotoTrip\QtScript4.dll
• C:\Program Files\Carambis\PhotoTrip\QtWebKit4.dll
• C:\Program Files\Carambis\PhotoTrip\QtXml4.dll
• C:\Program Files\Carambis\PhotoTrip\ssleay32.dll

Registry keys:
• HKEY_CURRENT_USER\Software\Carambis\PhotoTrip
• HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Uninstall\PhotoTrip

## How to uninstall Carambis PhotoTrip from your PC with Advanced Uninstaller PRO

Carambis PhotoTrip is an application marketed by the software company MEDIA FOG LTD. Sometimes, users decide to erase this program. This is difficult because doing this manually takes some advanced knowledge regarding removing Windows applications by hand. The best EASY practice to erase Carambis PhotoTrip is to use Advanced Uninstaller PRO. Take the following steps on how to do this:

1. If you don't have Advanced Uninstaller PRO already installed on your system, add it. This is good because Advanced Uninstaller PRO is a very potent uninstaller and all around utility to clean your PC.

2. Start Advanced Uninstaller PRO. Take some time to get familiar with Advanced Uninstaller PRO's interface and number of functions available. Advanced Uninstaller PRO is a very good program.

3. Click on the General Tools button

4. Press the Uninstall Programs tool

5. All the applications installed on the PC will be shown to you

6. Scroll the list of applications until you find Carambis PhotoTrip or simply click the Search feature and type in "Carambis PhotoTrip". If it is installed on your PC the Carambis PhotoTrip application will be found automatically. Notice that when you click Carambis PhotoTrip in the list of applications, the following information regarding the program is shown to you:

• Safety rating (in the lower left corner). The star rating explains the opinion other people have regarding Carambis PhotoTrip, from "Highly recommended" to "Very dangerous".
• Opinions by other people - Click on the Read reviews button.
• Technical information regarding the app you want to remove, by pressing the Properties button.
7. Click the Uninstall button. A confirmation window will come up. accept the uninstall by pressing Uninstall. Advanced Uninstaller PRO will automatically uninstall Carambis PhotoTrip.

8. After uninstalling Carambis PhotoTrip, Advanced Uninstaller PRO will ask you to run an additional cleanup. Press Next to go ahead with the cleanup. All the items that belong Carambis PhotoTrip which have been left behind will be detected and you will be asked if you want to delete them. By uninstalling Carambis PhotoTrip with Advanced Uninstaller PRO, you can be sure that no Windows registry entries, files or folders are left behind on your disk.

Your Windows PC will remain clean, speedy and ready to serve you properly.


text: JGAA Volumes Volume 25, no. 1, 2021 Drawing Subcubic 1-Planar Graphs with Few Bends, Few Slopes, and Large Angles Vol. 25, no. 1, pp. 1-28, 2021. Regular paper. Complexity of Geometric k-Planarity for Fixed k Marcus Schaefer Vol. 25, no. 1, pp. 29-41, 2021. Regular paper. Two Results on Layered Pathwidth and Linear Layouts Vol. 25, no. 1, pp. 43-57, 2021. Regular paper. Minimal Embedding Dimensions of Rectangle k-Visibility Graphs Vol. 25, no. 1, pp. 59-96, 2021. Regular paper. Equilateral Spherical Drawings of Planar Cayley Graphs Vol. 25, no. 1, pp. 97-119, 2021. Regular paper. Circumference of essentially 4-connected planar triangulations Vol. 25, no. 1, pp. 121-132, 2021. Regular paper. On the Balance of Unrooted Trees Mareike Fischer and Volkmar Liebscher Vol. 25, no. 1, pp. 133-150, 2021. Regular paper. Multilevel Planarity Vol. 25, no. 1, pp. 151-170, 2021. Regular paper. On Area-Universal Quadrangulations Vol. 25, no. 1, pp. 171-193, 2021. Regular paper. On the Complexity of Some Geometric Problems With Fixed Parameters Vol. 25, no. 1, pp. 195-218, 2021. Regular paper. Fast Generation of Unlabelled Free Trees using Weight Sequences Paul Brown and Trevor Fenner Vol. 25, no. 1, pp. 219-240, 2021. Regular paper. Algebraic Algorithms for Betweenness and Percolation Centrality Vol. 25, no. 1, pp. 241-261, 2021. Regular paper. Heading in the right direction? Using head moves to traverse phylogenetic network space Remie Janssen Vol. 25, no. 1, pp. 263-310, 2021. Regular paper. Schematic Representation of Large Biconnected Graphs Giuseppe Di Battista, Fabrizio Frati, Maurizio Patrignani, and Marco Tais Vol. 25, no. 1, pp. 311-352, 2021. Regular paper. Radial Level Planarity with Fixed Embedding Vol. 25, no. 1, pp. 353-366, 2021. Regular paper. An algorithm for embedding Turán graphs into incomplete hypercubes with minimum wirelength A. Arul Shantrinal, Sandi Klavžar, T.M. Rajalaxmi, and R. Sundara Rajan Vol. 25, no. 1, pp. 367-381, 2021. Regular paper. The number of crossings in multigraphs with no empty lens Vol. 25, no. 1, pp. 383-396, 2021. Regular paper. Least resolved trees for two-colored best match graphs Vol. 25, no. 1, pp. 397-416, 2021. Regular paper. Tree Containment With Soft Polytomies Vol. 25, no. 1, pp. 417-436, 2021. Regular paper. Synchronized Traveling Salesman Problem Gyula Pap and József Varnyú Vol. 25, no. 1, pp. 437-459, 2021. Regular paper. Decycling bipartite graphs Vol. 25, no. 1, pp. 461-480, 2021. Regular paper. On Strict (Outer-)Confluent Graphs Vol. 25, no. 1, pp. 481-512, 2021. Regular paper. Revising the Fellows-Kaschube $K_{3,3}$ Search Vol. 25, no. 1, pp. 513-520, 2021. Concise paper. On 2-Clubs in Graph-Based Data Clustering: Theory and Algorithm Engineering Aleksander Figiel, Anne-Sophie Himmel, André Nichterlein, and Rolf Niedermeier Vol. 25, no. 1, pp. 521-547, 2021. Regular paper.


text:
[texhax] Including pictures from MS Powerpoint into FoilTeX slides

Patric Glöde Patric.Gloede at t-online.de
Sun Apr 10 12:27:40 CEST 2005

Dear TeX support,

I have some MS PowerPoint slides with diagrams which I want to paste
into my FoilTeX slides. (The PPT slides consist of nothing but the
diagrams. The diagrams, however, are made up out of different lines and
other elements, so that it is not possible to work wit ONE single picture.)
What is the most efficient way to get these diagrams into FoilTeX?
I have tried to convert the PPT  slides into pdf and than use the
\includegraphics command. This theoretically works, but of course the
format of the then pdf slides is not right. They are too high, so that
the picture needs several FoilTeX slides. Just defining  another scale
or height  in the option of \includegraphics did not work --> I did not
manage to position the picture afterwards and in fact, what I need, is
not to resize the picture itself but to cut away the "white parts" of
the pdf slide so that just the original picture appears.
I should add, that I do not dispose of  Adobe Acrobat which  might
enable me to tailor the pdf slides with the pictures. Neither do I
dispose of devices to convert the ppt pictures into eps and tailor the
eps pictures.

I would be very grateful for advices how to tackle this not so trivial task!

Best regards

Patric Gloede


text:
PRESENTED BY
PRESENTED BY

# 4/25/12 - Morning OT

Finals are approaching...bleh.

If there is somebody that is really good with the ArcGIS API for Flex, let me know. I have a project due tomorrow and I'm stuck on a couple of things. If I have to write another damn program I might go postal and mail a couple of letters or something (that's what that means, right?).

Now to the discussion topic of the day: What is your opinion of network marketing? I just signed up about a week ago with a company called Vemma and they make wellness products. The particular product I'm selling is an energy drink, so naturally, my market is high school/college kids. Have you participated in network marketing? If so, were you successful or was it a complete flop for you?

150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150 words, 150!

### The Next FanPosts

Shortcuts to mastering the comment thread. Use wisely.

C - Next Comment


text:
# A residue formula

I have strong evindence to believe that the following identity holds: $$\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}} = d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right),\quad a,d,n+1 \in \mathbb{Z}_{\ge 1}.$$

Have you seen this formula before? Any ideas on how to prove it?

Any suggestions are welcome. I'll describe my partial success in the next section. I'm including the "combinatorics" tag because both of my methods got stuck in a combinatorial statement.

## Verification for a finite sample and proof for small $n$

I verified this for thousands of particular values of $(a,d,n)$ and proved it for all $(a,d)$ with small $n$.

To get a proof for small $n$ one can write out the power series expansion in order to find the coefficient of $(z-1)^{-1}$. However, this gets increasingly complicated for higher $n$ and I don't think a proof for all $n$ will emerge from this line of attack unless you are a brilliant combinatorialist.

In order to check this equality for particular values of $(a,d,n)$, it makes a huge difference in computation time to combine Cauchy's integral formula with Faà di Bruno's formula for expanding higher derivatives of compositions. Despite the advantage of quick evaluation, these formulas seem too unwieldy to get a proof for arbitrary $(a,d,n)$.

• Do you mean $|z-1|=\epsilon$? – Brendan McKay Feb 5 '18 at 12:32
• Is it not just a matter of changing variables $w=z^d$ and using straightforward Cauchy? – thedude Feb 5 '18 at 12:49
• There is the following formula $$\sum_{n=-\infty}^\infty \left({a\atop b+\alpha n}\right)v^{b+\alpha n}=\frac{1}{\alpha}(1+v)^a,\quad |v|=1,~ |\arg v|<\pi,~ 0<\alpha\le 1$$ proved in the article T.J. Osler, Taylor's series generalized for fractional derivatives and applications, SIAM J. Math. Anal. 2, 37-48 (1971) that formally can be written as a fractional residue formula. – user82588 Feb 8 '18 at 17:21

Let us consider the function $w=z^d$ around $z=1$. It is holomorphic and locally invertible with a holomorphic inverse. So we can think of $z$ as $w^{1/d}$ where the latter symbol means a branch of the $d$-th root function taking $1$ at $1$. With this notation, the integral can be written as $$\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{n!}{2\pi i}\oint\frac{1}{d}\frac{w^{(a-d)/d} \mathrm{d}w}{(w-1)^{n+1}}.$$ Here, we understand $w^{(a-d)/d}$ as the $(a-d)$-th power of $z=w^{1/d}$, and the integration is on the closed curve $\{w=(1+\epsilon\cdot e^{i t})^d:\ 0\leq t\leq2\pi\}$ with winding number $1$ around $w=1$. Hence, by Cauchy's theorem, the integral equals the $n$-th derivative of $w^{(a-d)/d}$ at $w=1$: $$\frac{n!}{2\pi i}\oint_{|z-1|=\epsilon} \frac{z^{a-1} \mathrm{d}z}{(z^d-1)^{n+1}}=\frac{1}{d}\prod_{j=1}^n\left(\frac{a-d}{d}+1-j\right)=d^{-n-1}\prod_{j=1}^{n}\left(a-jd\right).$$


text:
# Time dependence in non-dispersive waves

I'm reading the book "A Student's Guide to Waves" by Daniel A. Fleisch and Laura Kinnaman in an attempt to get quickly up to speed in some of the terminology and maths of waves.

At some point it mentions the following equation $$\psi = y = f(x,t)$$ And that the easiest way to think about the shape of the wave is by considering a snapsnot, i.e. $$y=f(x,0)$$ Then it goes on saying "Many waves maintain the same shape over time - the wave moves in the direction of propagation, but all peaks and troughs move in unison, so the shape does not change as the wave moves. For such "non-dispersive" waves, $$f(x,0)$$ can be written as $$f(x)$$, since the shape of the wave doesn't depend on when you take the snapshot."

I don't understand why this would be true, because the phase would still be dependent on when the snapshot was taken? And if it were to be considered true, couldn't you in the same way consider a dispersive wave as a function of only location if you are talking about a snapshot only?

A one-dimensional wave that keeps its form over time has the general form $$y(x, t)= f(x\pm ct)$$ where $$c$$ is the speed of the wave. At $$t=0$$ one has $$y(x, 0)=f(x)$$. For more complicated waves this is not true: the form of the wave will be some function of two variables $$g(x, t)$$. While it is true that you could write $$g(x, 0) = g(x)$$ for some function $$g(x)$$, at later times the function $$g(x, t)$$ does not need to resemble $$g(x)$$. You might like to play around with this graph to get a better idea of what this means.