Quantity of heat

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Heat capacity Specific heat capacity Change of states Latent heat of fusion Latent heat of vaporization QUANTITY OF HEAT

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Transcript of Quantity of heat

  • 1. QUANTITY OF HEATHeat capacitySpecific heat capacityChange of statesLatent heat of fusionLatent heat of vaporization

2. HEAT CAPACITY (C) This is the quantity of heat required to raise the temperature of a given mass of a material by one degree Celsius or one Kelvin.It is denoted by Cheat capacity = heat energy absorbed Q/ temperature change C = Q/The SI unit Jk-1 3. Sample Questions1. Calculate the quantity of heat required toraise the temperature of aluminium metalblock with heat capacity of 460 Jk-1 from150 to 450. SolutionQuantity of heat= C = 460 x 30 = 13 800 J. 4. Specific heat capacity (c) When comparing heat capacities of varioussubstances we talk of specific heat capacities .Specific in physics refers to unit quantity of aphysical property. Also called specific heat. Quantity of heat required to raise thetemperature of a unit mass of a substance by onedegree Celsius or Kelvin. It is denoted as c. Si unit is joule per kilogram kelvin. (J/kg K). 5. FORMULAE Specific heat capacity = heat capacity/massc= Q//m =Formulae Q/mQ = mc Materials with high specific heat capacity e.g.water require large amount of heat to changetheir temperatures while those with lessspecific heat capacity requires little heatenergy to change their temperatures e.g..silver 6. Table of specific heat capacitiesJ/kg K.Material Specific HeatMaterial Specific Heat CapacityCapacityAluminium900 Lead 130Brass380 Mercury140Glass(ordinary)400 Methylated spirit 2400Ice670 Sea water 3900Alcohol2100Water 4200Iron 460 Zinc380 7. Sample Question 1How many joules Question given outSample of heat are 1when a piece of iron of mass 50g andspecific heat capacity 460 J/kg K, coolsfrom 80 C t0 20 C? solution Q = mc = 0.05 460(80 20) = 1380 J. 8. Sample question 21. A block of metal of mass 1.5 kg which is suitably Sample is heated from 30 C to 50insulated question 2 C in 8 minutes 20 seconds by an electric heater coil rated 54 watts. Find : a) the quantity of heat supplied by heater. b) The heat capacity of the block.c) Its specific heat capacity. 9. solutiona)Quantity of heat supplied = power x time Solution Q2Q = 54 x 500= 27 000Jb) Heat capacity C = Q/.C= 27000/(50 30)= 1 350J/K.c) specific heat capacity = C/mc = 1 350/1.5 = 900 Jkg-1 K-1 . 10. Sample question 3 What is the final temperature of the mixture if 100g of water at 70 0 Cis added to 200g of cold water at 100 C and well stirred?(neglect heat Sample question 3absorbed by the container) 11. SOLUTIONHeat lost by hot water = Heat received by cold waterLet the final temperature of the mixture = 0 C.change in temperature of hot water = (70 )change in temperature of cold water= ( 10)Thus , using the formula mc , we substitute values in theequation.0.1 x 4200 x (70 ) = 0.2 x 4200 x ( 10) Dividing both sides by 4200 7 0.1 = 0.2 - 2 = 30 0 C. 12. ASSIGNMENT1. The temperature of piece of copper of mass 250gAssignmentis raised t0 100 0 C and is then transferred to awell lagged aluminium can of mass 10.0gcontaining 120g of methylated spirit at 10.00 C.calculate the final steady temperature after thespirit has been well stirred. Neglect the heatcapacity of the stirrer and any loses fromevaporation and use the table of specific heatcapacities for any data required. (32.70 C) 13. Find the final temperature of water if a heatersource rated 42W heats 50g water from 20 0 C infive minutes. ( specific heat capacity of water is4200J/kg/K.The temperature of 500g of a certain metal israised to 100 0 C and is then placed in 200g ofwater at 15 0 C. if the final steady temperaturerises to 21 0 C, calculate the specific heatcapacity of the metal. (128J/kg/K.)Reference physics 5th edition A.F . ABBOTT. Page 198-203. 14. DETERMINATION OF SPECIFIC HEAT CAPACITY HeatDETERMINATIONenergy transfer hence the is a form of OF SPECIFIC HEAT CAPACITYlaw of energy transfer applies. Heat gain equals heat lost. There are various methods of determiningspecific heat capacitya)Mixture method.b)Electrical method.c)Mechanical method. In this coarse we shall look at the first two. 15. Mixture Method1. Solids Experiment 9.2: To determine the specific heat capacity by method of mixture. Apparatus :metal block, beaker ,water, tripod stand, heat source, well lagged calorimeter, stirrer, thermometer and cardboard. procedure : the learner to read and follow the procedure on page 264 secondary klb bk 32. Liquids The learner to follow the procedure on page 265 of secondary physics klb bk 3. 16. Sample Question 11. A lagged copper calorimeter of mass 0.75kg contains 0.9kg of water at 20 0 C. A bolt of mass 0.8kg is transferred from oven at 400 0 C to the calorimeter and a steady temperature of 50 0 C is reached by water after stirring. Calculate the specific heat capacity of the material of the bolt.( specific heat capacity of copper is 400 Jkg-1 K-1 and that of water 4 200Jkg-1 K-1 .2. A block of iron of mass 1.25 kg at 120 0 C was transferred to an aluminium calorimeter of mass 0.3kg containing a liquid of mass 0.6kg at 25 0 C. the block and the calorimeter with its contents eventually reached a common temperature of 50 0 C. given the specific heat capacity of iron is 450 Jkg-1 k-1 ,calculate the specific heat capacity of the liquid. 17. Electrical Method Solids Experiment: to determine the specific heat capacity of material by electrical method. The learner to read the procedure for experiment 9.3 on page 266-268. 18. Sample question 21. A metal cylinder of mass 0.5kg is heated electrically . If the voltmeter reads 15v,the ammeter 3.0v and the temperature of the block rises from 20 0 C to 85 0 C in 10 minutes. Calculate the specific heat capacity of the metal cylinder. heat supplied = heat gainedvIt = mc 15 x 3 x 10 x 60= 0.5 x c x 65c= 831 J/kg/K 19. Sample Question 3In an experiment to determine specific heatcapacity of water , an electrical heater wasused. If the voltmeter reading was 24 V andthat of ammeter reading was 2.0 A. calculatethe specific heat capacity of water if thetemperature of a mass 1.5kg of water in a0.4kg copper calorimeter rose by 6 0 C after13.5 minutes. (specific heat capacity of copperis 400J/kg / K. 20. CHANGE OF STATE Heating leads to a rise in temperature . Sometimes noobservable changes is noted. When the ice is about -10 0 C is heated, heat energy isused in raising its temperature to 0 0 C . Heat energy supplied to the ice at 0 0 C is used tochange ice from solid to liquid. Heat supplied to ice does not change the temperatureof ice but change its state from solid to liquid. 21. LATENT HEAT This is heat involved in change of state of a substance.It can either be heat loss or heat gain. This heat is latent means hidden or concealedbecause it does not show its presence by change intemperature as the extra heat goes into change in state. There two types of latent heat:a) Latent heat of fusionb) Latent heat of vapourization 22. Latent of fusion This is heat required to change the state of a materialfrom solid to liquid or from liquid to solid withoutchange in temperature. As liquid changes to solid latent heat of fusion is givenout and the amount of heat is absorbed when a solidchanges to liquid. The graph below shows temperature vs time(s) 23. Temperature against time.Temp(0 C)(+)Time(s) (-) 24. ExperimentExperiment : To explore the change of state ofnaphthalene using cooling curve.Apparatus: Naphthalene, test tube, water, thermometer the learner to copy fig. 9.8 plus the procedure. Secondary physics klb bk 3 pg 273. 25. The Cooling Curve Of NaphthaleneTemp(0 C)melting pt ----------------------------------- 26. EXPLANATIONOP liquid naphthalene coolingPQ liquid naphthalene changes to solid atconstant temperature( latent heat offusion is given out)this point is also known as meltingpoint.QR solid naphthalene cools to roomtemperature 27. SPECIFIC LATENT HEAT OF FUSION((Lf )This is the quantity of heat required to change the unit mass of a substance from solid to liquid without change in temperature. Lf = Q/mQ = mLf SI unit of specific latent heat is Jkg-1 28. TABLE OF VALUES FOR SPECIFIC HEAT CAPACITIESMATERIAL SPECIFIC LATENT HEAT OF FUSION(X 105 JKg-1 )Copper4.0Aluminium3.9Water (ice) 3.34Wax1.8Naphthalene1.5Solder0.7Lead 0.026mercury 0.013 29. DETERMING SPECIFIC LATENT HEAT OF FUSION OF A MATERIAL There are various methods of determining specific latent heat of fusion of a material a) Mixture Method Apparatus water Ice pieces Calorimeter stirrer 30. PROCEDUREFind the mass of the calorimeter M1 Place water with temperature of about 50 C above the room temperature into the calorimeter. Mass of water + calorimeter M2Record temperature of water in calorimeter 1 Add pieces of ice to the calorimeter Mass of calorimeter and mixture M3Measure the final temp. of the mixture after stirring 2 31. DATA ANALYSISHeat lost by warm water + heat lost by calorimeter = heat gained by ice at 00 C to water at 00 C. + heat gained by water (00 C to final temperature)Let quantity of heat required to melt a unit mass of ice at 00 C to 00 C to water at 00 C be Lf. (m2-m1)Cw (1-2)+ m1Cc (1-2) = (m3- m2) Lf +(m3-m2)Cw (2- 0). 32. ELECTRICAL METHODWith electrical method the quantity of heat is calculated as follows.Heat supplied by the heater = heat gained by the ice.Q = mLf = VItLf = VIt /m 33. SAMPLE QUESTION 1Calculate the quantity of heat required to melt 4 kg of ice and to raise the temperature of the water formed to 50 0 C. take the specific latent heat of ice to be 3.4 x 105 J/kg and the specific heat capacity of water to be 4200 J/kg K. 34. SAMPLE QUESTION 2A beaker contains 200 g of water at 15 0 C. 25 g of ice at 00 C is added to the water which is stirred until the ice is completely melted.a) How much heat is needed to melt all the ice.b) What is the mass of water produced by melting all the ice.c) Calculate the lowest temperature of all mixture, assuming that all the heat to melt the ice is taken ice is taken from the water and no heat enters or leaves the system. ( specific latent heat of fusion of ice 336 000 J / kg. 35. LATENT HEAT OF VAPOURASATIONHeat energy absorbed by a liquid as it changes its state to vapour without change in temperature.OR Heat energy given out by a vapour as it changes its state to liquid without change in temperature. 36. SPECIFIC LATENT HEAT OF VAPORIZATION(Lv) This the heat required to convert unit mass of a liquid, at a boiling point, into vapour without change in temperature. The SI unit Jkg-1Q = m Lv Lv = Q/m 37. Specific Latent Heat of Vaporization Material Specific latent heat of vaporization(x 105 Jkg-1 )Water 22.6Benzene 4.0Petrol8.5Alcohol 8.6Ether 3.5Turpentine2.7Ethanol 8.5 38. SAMPLE QUESTION1. Dry steam is passed into a well-lagged coppercan of mass 250 g containing 400 g of waterand 50 g of ice at 00 C. The mixture is wellstirred and the steam supply cut off when thetemperature of the can and its content reaches20 0 C. Neglect heat losses, find the mass ofsteam condensed. (specific heat capacities:water 4200J/kg K; copper 400 J/kg K ;specific latent heats of fusion of steam 22.6 x105 J/kg.) 39. SOLUTION Using the principle of conservation of energy, we may say heat given out stem = heat received by ice , water and can let the mass of steam condensed = m (g)Heat in joules given out by:Steam condensing to water at 1000 C = m x 226000Condensed steam cooling from 1000 C to 20 0 C 40. SAMPLE QUESTION 2 A jet of dry steam at 1000C is sprayed onto the surface of 100g of dried ice at 00C placed in a plastic container of negligible heat capacity. The temperature of the mixture is 400C when the total mass of the water in the container 120g. Given that the specific heat capacity of water is 4200J/ kg/K and latent heat of fusion of ice is 336KJkg-1; determine the specific latent heat of vaporization of water. 41. Factors affecting melting and boiling points Boiling point There are two factors affecting the boiling point of a liquid.a) Pressureb) ImpuritiesExperiment 9.13: investigates the effects of increased pressure on boiling points. 42. EFFECT OF PRESSURE ON BOILING POINT Increase in pressure increases the boiling point of aliquid. Application of this concept is the pressure cooker. Ithas tight fitting lid which prevents free escape ofsteam thus making the pressure inside to build up.Increase in the boiling point to high temperaturesenables food to cook faster. Decrease in pressure lowers the boiling point of aliquid . 43. Effects of impurities on boiling pointExperiment 9.15: investigates the effects impurities on boiling pointsThe boiling point for of the salt solution is higher than that of the distilled water.The presence of impurities in liquid raises its boiling points. 44. Melting PointThere are two factors that affect the melting point of a substance.a) Pressureb) Impurities 45. PRESSURE EXPERIMENT: To investigate the effect ofpressure on melting point. Apparatus : Block of ice, thin copper wire,two heavy weight, wooded support. Procedure: Attach two heavy weigh to the ends of a thincopper wire. Pass the string over a large block of ice, asshown in the figure. 46. OBSERVATION The wire cuts through the ice block, butleaves it as one piece. This process is known as regelation. 47. Explanation (regelation) Weight exerts pressure on the ice beneath; thispressure makes it melts at a temperature lowerthan its melting point. The water formed loses its latent heat of fusion tothe wire which hence solidifies again as it is nolonger under pressure. The latent heat losed by the water is conducted tothe wire which melts the ice below it. This process continues until the wire cutsthrough leaving the block. 48. CONCLUSION Application of pressure on ice lowers the meltingpoint. NB: If the wire with lower thermal conductivity isused, it will cut through slowly. Poor conductors of heat e.g. cotton will not cutthrough the block at all because it does notconduct heat. 49. APPLICATION OF THE EFFECTS OF PRESSUREON MELTING POINT OF ICE. Ice skating Weight of the skater exerts pressure on the icebelow causing melting at a lowertemperatures. The high pressure reduces melting hencemelting them forming a thin film of waterover which skater slides. 50. Joining ice cubes under pressure By pressing ice cubes hard underpressure, the melting points betweenpoints of contact of the ice is lowered;water recondenses and the two cubes arejoined together. 51. EFFECTS OF IMPURITIESApplication of impurities lowers the melting point the melting point of a substance.Salt is spread on roads during winter to prevent freezing of roads. 52. EVAPORATION Evaporation occurs on the surface of theliquid where molecules escape to air. Molecules at the surface have higher kineticenergy than those ones below hence theybreak from their attractive forces of theneighbouring molecules. Evaporation takes place at all temperatures. 53. Effects of EvaporationThe following are some effects of evaporation Methylated spirit feels cold on the back of your hand than water. This is because hands feel cold as the spirit evaporates from the skin. Evaporating methylated spirit extracts latent heat of vaporisation from the skin making it feel cold. 54. A beaker placed on the film of water on a wooden block as shown on the fig. 9.19 pg 286 klb bk 3. The beaker stucks on the to the wooden block afterthe air is blown through the ether using a footpump. This is because blowing increases the rate ofevaporation of ether forming a layer of icebetween beaker and wood. This shows thatevaporation causes cooling.NB: bubbling increases surface area of etherexposed to air. 55. Factors affecting rate of evaporationTemperature Increase in temperature increases kinetic energy of molecules on the surface; these molecules move faster hence many of them escape ; enhancing evaporation. NB: increase in temperature increases the rate of evaporation. 56. 2. Surface Area Increasing surface area of the liquid exposesmore liquid molecules hence faster moleculesescapes to the environment. Large surface area also clears the way formore molecules to enter the space. 57. 3. Drought Passing air over the liquid sweeps awayescaping vapour molecules e.g. clothesdry faster on a windy day, people takehot beverages by blowing over it. 58. 4. Humidity Humidity is the concentration of watervapour in the atmosphere . High humidity lowers the rate at whichmolecules enter the space hence lowers therate of evaporation. This is why clothes take longer time to dry ona humid day than on dry one. 59. Comparison Between Evaporation And BoilingEvaporationBoilingTakes place at allTakes place at a fixedtemperaturestemperatureTakes place on theTakes place throughoutsurface of the liquid the liquidNo bubbles are formed Bubbles of steam areformedDecreasing atmospheric Decreasing atmosphericpressure increases the pressure lowers therate of evaporationboiling point