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# Quantity of heat

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• A unit mass of a material changing from liquid to solid gives out heat energy equal to specific latent heat of fusion.
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• 1. QUANTITY OF HEATHeat capacitySpecific heat capacityChange of statesLatent heat of fusionLatent 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 C heat capacity = heat energy absorbed Q/ temperature change ∆θ C = Q/∆θThe SI unit Jk-1
• 3. Sample Questions1. Calculate the quantity of heat required to raise the temperature of aluminium metal block with heat capacity of 460 Jk-1 from 150 to 450. Solution Quantity of heat = C∆Ѳ = 460 x 30 = 13 800 J.
• 4. Specific heat capacity (c) When comparing heat capacities of various substances we talk of specific heat capacities . Specific in physics refers to unit quantity of a physical property. Also called specific heat. Quantity of heat required to raise the temperature of a unit mass of a substance by one degree 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/mass c= Q/∆θ/m =Formulae Q/m∆θ Q = mc∆θ• Materials with high specific heat capacity e.g. water require large amount of heat to change their temperatures while those with less specific heat capacity requires little heat energy to change their temperatures e.g.. silver
• 6. Table of specific heat capacities J/kg K. Material Specific Heat Material Specific Heat Capacity CapacityAluminium 900 Lead 130Brass 380 Mercury 140Glass(ordinary) 400 Methylated spirit 2400Ice 670 Sea – water 3900Alcohol 2100 Water 4200Iron 460 Zinc 380
• 7. Sample Question 1How many joules Question given out Sample 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∆θ = 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 50 insulated 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 Q2 Q = 54 x 500 = 27 000Jb) Heat capacity C = Q/∆θ. C= 27000/(50 – 30) = 1 350J/K.c) specific heat capacity = C/m c = 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 C is added to 200g of cold water at 100 C and well stirred?(neglect heat Sample question 3 absorbed by the container)
• 11. SOLUTIONHeat lost by hot water = Heat received by cold waterLet the final temperature of the mixture = θ 0 C.change in temperature of hot water = (70 – θ)change in temperature of cold water = (θ – 10)Thus , using the formula mc∆θ , we substitute values in theequation. 0.1 x 4200 x (70 – θ) = 0.2 x 4200 x (θ – 10) Dividing both sides by 4200 7 – 0.1 θ = 0.2 θ - 2 θ = 30 0 C.
• 12. ASSIGNMENT1. The temperature of piece of copper of mass 250g Assignment is raised t0 100 0 C and is then transferred to a well lagged aluminium can of mass 10.0g containing 120g of methylated spirit at 10.00 C. calculate the final steady temperature after the spirit has been well stirred. Neglect the heat capacity of the stirrer and any loses from evaporation and use the table of specific heat capacities 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 CAPACITY law of energy transfer applies. Heat gain equals heat lost. There are various methods of determining specific 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 gained vIt = mc∆θ 15 x 3 x 10 x 60= 0.5 x c x 65 c= 831 J/kg/K
• 19. Sample Question 3In an experiment to determine specific heat capacity of water , an electrical heater was used. If the voltmeter reading was 24 V and that of ammeter reading was 2.0 A. calculate the specific heat capacity of water if the temperature of a mass 1.5kg of water in a 0.4kg copper calorimeter rose by 6 0 C after 13.5 minutes. (specific heat capacity of copper is 400J/kg / K.
• 20. CHANGE OF STATE Heating leads to a rise in temperature . Sometimes no observable changes is noted. When the ice is about -10 0 C is heated, heat energy is used in raising its temperature to 0 0 C . Heat energy supplied to the ice at 0 0 C is used to change ice from solid to liquid. Heat supplied to ice does not change the temperature of 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 concealed because it does not show its presence by change in temperature 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 material from solid to liquid or from liquid to solid without change in temperature. As liquid changes to solid latent heat of fusion is given out and the amount of heat is absorbed when a solid changes 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 of naphthalene 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 at constant temperature( latent heat of fusion is given out) this point is also known as melting point.QR – solid naphthalene cools to room temperature
• 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/m Q = mLf SI unit of specific latent heat is Jkg-1
• 28. TABLE OF VALUES FOR SPECIFIC HEAT CAPACITIES MATERIAL SPECIFIC LATENT HEAT OF FUSION (X 105 JKg-1 )Copper 4.0Aluminium 3.9Water (ice) 3.34Wax 1.8Naphthalene 1.5Solder 0.7Lead 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. PROCEDUREFind 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 – M2Record temperature of water in calorimeter – θ1 Add pieces of ice to the calorimeter Mass of calorimeter and mixture – M3Measure the final temp. of the mixture after stirring – θ2
• 31. DATA ANALYSISHeat 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 (θ1-θ2)+ m1Cc (θ1-θ2) = (m3- m2) Lf +(m3-m2)Cw (θ2- 0).
• 32. ELECTRICAL METHODWith electrical method the quantity of heat is calculated as follows.Heat supplied by the heater = heat gained by the ice.Q = mLf = VItLf = VIt /m
• 33. SAMPLE QUESTION 1Calculate 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 VAPOURASATIONHeat 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-1 Q = m Lv Lv = Q/m
• 37. Specific Latent Heat of Vaporization Material Specific latent heat of vaporization (x 105 Jkg-1 )Water 22.6Benzene 4.0Petrol 8.5Alcohol 8.6Ether 3.5Turpentine 2.7Ethanol 8.5
• 38. SAMPLE QUESTION1. Dry steam is passed into a well-lagged copper can of mass 250 g containing 400 g of water and 50 g of ice at 00 C. The mixture is well stirred and the steam supply cut off when the temperature of the can and its content reaches 20 0 C. Neglect heat losses, find the mass of steam condensed. (specific heat capacities: water 4200J/kg K; copper 400 J/kg K ; specific latent heats of fusion of steam 22.6 x 105 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) ImpuritiesExperiment 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 a liquid. Application of this concept is the pressure cooker. It has tight fitting lid which prevents free escape of steam thus making the pressure inside to build up. Increase in the boiling point to high temperatures enables food to cook faster. Decrease in pressure lowers the boiling point of a liquid .
• 43. Effects of impurities on boiling pointExperiment 9.15: investigates the effects impurities on boiling pointsThe 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 PointThere are two factors that affect the melting point of a substance. a) Pressure b) Impurities
• 45. PRESSURE EXPERIMENT: To investigate the effect of pressure 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 thin copper wire. Pass the string over a large block of ice, as shown in the figure.
• 46. OBSERVATION The wire cuts through the ice block, but leaves it as one piece. This process is known as regelation.
• 47. Explanation (regelation) Weight exerts pressure on the ice beneath; this pressure makes it melts at a temperature lower than its melting point. The water formed loses its latent heat of fusion to the wire which hence solidifies again as it is no longer under pressure. The latent heat losed by the water is conducted to the wire which melts the ice below it. This process continues until the wire cuts through leaving the block.
• 48. CONCLUSION Application of pressure on ice lowers the melting point. NB: If the wire with lower thermal conductivity is used, it will cut through slowly. Poor conductors of heat e.g. cotton will not cut through the block at all because it does not conduct heat.
• 49. APPLICATION OF THE EFFECTS OF PRESSURE ON MELTING POINT OF ICE. Ice skating Weight of the skater exerts pressure on the ice below causing melting at a lower temperatures. The high pressure reduces melting hence melting them forming a thin film of water over which skater slides.
• 50. Joining ice cubes under pressure By pressing ice cubes hard under pressure, the melting points between points of contact of the ice is lowered; water recondenses and the two cubes are joined together.
• 51. EFFECTS OF IMPURITIESApplication 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 the liquid where molecules escape to air. Molecules at the surface have higher kinetic energy than those ones below hence they break from their attractive forces of the neighbouring molecules. Evaporation takes place at all temperatures.
• 53. Effects of EvaporationThe 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 after the air is blown through the ether using a foot pump. This is because blowing increases the rate of evaporation of ether forming a layer of ice between beaker and wood. This shows that evaporation causes cooling. NB: bubbling increases surface area of ether exposed to air.
• 55. Factors affecting rate of evaporation Temperature 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 exposes more liquid molecules hence faster molecules escapes to the environment. Large surface area also clears the way for more molecules to enter the space.
• 57. 3. Drought Passing air over the liquid sweeps away escaping vapour molecules e.g. clothes dry faster on a windy day, people take hot beverages by blowing over it.
• 58. 4. Humidity Humidity is the concentration of water vapour in the atmosphere . High humidity lowers the rate at which molecules enter the space hence lowers the rate of evaporation. This is why clothes take longer time to dry on a humid day than on dry one.
• 59. Comparison Between Evaporation And Boiling Evaporation BoilingTakes place at all Takes place at a fixedtemperatures temperatureTakes place on the Takes place throughoutsurface of the liquid the liquidNo bubbles are formed Bubbles of steam are formedDecreasing atmospheric Decreasing atmosphericpressure increases the pressure lowers therate of evaporation boiling point