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Heat and Temperature Heat is a form of energy, and is measured in Joules (J). Temperature is different from heat. Temperature is a measure of how hot or cold an object is. It is measured in degrees Celsius (°C). Heat can move from one place to another. The three methods of heat travel are: conduction convection radiation

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barrier HEAT Conduction Experiment ABC

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Pins A, B and C are held in place by vaseline/wax. The barrier stops heat reaching the pins through the air. CONDUCTION occurs in solids when heat passes from particle to particle along the solid.

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Convection Experiment water purple crystal The purple colour allows us to see the heated water moving. HEAT

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The convection currents look like this: hot water rising cold water falling CONVECTION occurs in liquids and gases because their particles can move. Heated air and liquid will rise upwards.

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Day Break Land is warmer than sea. Air heats up and rises over land, cools down and falls over sea. Sunset Sea is warmer than land. Air cools and falls over land, but is warmed over the sea and rises.

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Radiation Experiment A metallic cube known as a Leslie Cube is filled with boiling water.

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All hot objects radiate heat energy. Radiated heat waves are also known as INFRARED RADIATION. It travels at the speed of light (3 x 10 8 ms -1 ). Matt black surfaces radiate more heat than shiny surfaces at the same temperature. Infrared radiation can travel through a vacuum (just as heat from sun is able to travel through space which is a vacuum).

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Nocturnal Lemur Black fur poses no disadvantage as only active at night Lemur Active during the day so has lighter fur. On cold mornings points its belly to the sun.

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Heat Loss

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Reducing Heat Loss Preventative ActionHeat Lost Type cavity wall insulationconduction double glazingconduction carpetsconduction loft insulationconvection draft excludersconvection shiny foil behind radiatorsradiation Heat energy is lost faster when there is a difference in temperature between inside and outside. bigger

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Heat Energy The amount of heat energy required to heat up an object depends on: the mass of the substance ( m ) the material the object is made of the rise in temperature required ( ΔT ) The amount of heat energy is given by:

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x ÷ EHEH c m ΔTΔT QuantityUnit Heat Energy ( E H ) Specific Heat Capacity ( c ) Mass ( m ) Change in Temperature ( ΔT ) Joules ( J ) Joules per kilogram per degrees Celsius ( J kg -1 °C -1 ) Kilograms ( kg ) Degrees Celsius ( °C ) ** Specific Heat Capacities can be found on Data Sheet in exam paper ** Specific Heat Capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C.

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Example 1 Calculate the amount of heat energy required to raise the temperature of 2 kg of water from 20 °C to 100 °C. ** data sheet **

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Example 2 A 3 kg block of copper at 15 °C is heated. Calculate the rise in temperature if 20 kJ of heat energy is transferred to the copper. ** data sheet **

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Example 3 0.5 kg of a liquid is heated from 8 °C to 40 °C by 50 kJ of heat energy. Calculate the specific heat capacity of the liquid.

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Yellow Book Heat Energy – Page 90 Q79 (a) (b) (c) and (d) Q80, Q81, Q82, Q83, Q84, Q85

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Power of a Heater Experiment Joulemeter 12 volt Immersion Heater to 12 V a.c. supply stopwatch

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Zero the joulemeter. Switch on for 20 seconds and record the energy used.

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Specific Heat Capacity Experiment The specific heat capacity of aluminium is measured using the apparatus shown. thermometer aluminium block heater to 12 V a.c. supply stopwatch

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Initial temperature of block °C. The heater is switched on for 5 minutes. Final temperature of block °C. Heat Energy Used

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Specific Heat Capacity The value measured experimentally is higher than 380 J kg -1 °C -1. This is because not all the heat energy is transferred to the aluminium block – some is lost to the surroundings.

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Since E = P x t we can write P x t instead of E. Combining E H = cmΔT and E = Pxt The heat energy equation is: The equation for electrical power is: ** NOT on data sheet **

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Example 1 2 kg of water is heated from 20°C to 60°C by a 2 kW heater. Calculate the time taken.

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Questions Q1.0.5 kg of copper is heated from 15 °C to 50 °C by a 60 W heater. Calculate the time taken. Q2.2 kg of water at 10 °C is heated by a 1.5 kW heater for 3 mins. Calculate the final temperature of the water. Q3.How long will it take a 2.3 kW heater to boil 0.5 kg of water with an initial temperature of 10 °C. 112.6 s ΔT = 32.3 °C so T fin = 42.3 °C 81.8 s

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Measuring & Calculating Time Taken Experiment The time taken for a kettle to boil is measured and calculated. The measured time taken for the kettle to boil is s.

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Variations of E H = cmΔT There are several formulae for energy. These can be used to do heat calculations. ** NOT on data sheet **

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Example 1 A heater operates from the 230 V mains supply and takes a current of 3 A. This heater is used to heat up 2 kg of water at 18 °C. The heater is switched on for 5 minutes. Calculate the rise in temperature of the water and then state the final temperature reached.

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The final temperature of the water is:

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Example 2 A lead bullet of mass 0.05 kg has a speed of 80 ms -1. The bullet hits a wall and stops. Calculate the rise in temperature of the bullet as it does this. (c lead = 130 J kg -1 °C -1 ) All the kinetic energy is transferred into heat energy (E K = E H )

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Question A 2.2 kW kettle contains 1.7 kg of water at 12 °C. The kettle is switched on for 4 minutes. (a)Calculate the maximum temperature reached by the water. (b)In practice, will the maximum temperature reached be bigger or smaller than this calculated value. Explain your answer fully.

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The maximum temperature of the water is: (a)

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(b)In practice, the maximum temperature reached will be less than 74.3 °C. This is because some heat energy is lost to the surroundings.

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Cooling Salol Experiment Liquid salol at 60°C is allowed to cool. The temperature of the liquid salol is measured every minute as it cools. liquid Salol thermometer stopwatch

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Results Time (minutes) Temperature (°C) 1 2 3. 25

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A graph of the temperature against time is plotted for the cooling salol. A B C D E temperature (°C) time 38 change of state During a change of state – NO CHANGE IN TEMPERATURE.

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A B C D E temperature (°C) time 38 StageDescription AB liquid cools down BC CD liquid salol turns to a solid (solidifies) latent (hidden) heat is given out temperature stays the same solid salol cools down to room temp. salol is at room temp. so no further fall in temperature DE

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Change of State The changes of state are: SolidLiquidGas meltingevaporation condensation freezing

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Latent Heat of Fusion (melting) The amount of heat energy required to melt a solid depends on: 1.the mass of the solid 2.what substance the solid is made of. This gives the equation:

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QuantityUnit Heat Energy ( E H ) Mass ( m ) Specific Latent Heat of Fusion ( L f ) Joules ( J ) Kilograms ( kg ) Joules per Kilogram ( J kg -1 ) x ÷ EHEH m LfLf The specific latent heat of fusion for water is 3.34 x 10 5 J kg -1. Specific Latent Heat of Fusion is the amount of energy required to change 1 kg of a substance from solid to liquid

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Example 1 Calculate the energy required to melt 2 kg of ice at 0 °C.

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Example 2 Calculate the time taken by a 500 W heater to completely melt 2 kg of ice at 0 °C.

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Questions 1. Calculate the heat energy required to melt 2.7 kg of ice at 0 °C. 2.A 700 W heater inside a block of ice at 0 °C is switched on for 3 minutes. Calculate the mass of ice which will melt. 3.A 1 kW heater inside a block of ice at 0 °C is switched on. The heater melts 0.65 kg of ice. Calculate the time that the heater was switched on for.

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Measuring Specific Latent Heat of Fusion of Ice Experiment Joulemeter Ice at 0 °C Heater (not switched on) Beaker 2 Control Experiment Water Beaker 1

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The control experiment is not switched on. The control measures the amount of ice that melts due to temperature of the room. The heater connected to the joulemeter is switched on for 5-minutes. The amount of ice melted in beaker 2 is subtracted from the amount melted in beaker 1. Results

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The data booklet value is 3.34 x 10 5 J kg -1. Q.Why is the measured value less than the data book vale? The measured value is less than 3.34 x 10 5 J kg -1 as heat energy is lost to the surroundings.

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Latent Heat of Vaporisation The heat energy needed to boil away a liquid (vaporise) depends on: 1.the mass of liquid 2.what the liquid is. This gives the equation: The specific latent heat of vaporisation of water is 2.26 x 10 6 J kg -1.

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Questions 1.Calculate the energy required to turn 3kg of water at 100 °C into steam at 100 °C. 2.Calculate the time taken by a 2 kW kettle to boil away 0.5 kg of water at 100 °C.

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Measuring Specific Latent Heat of Vaporisation Experiment water 1,850 W heating element The mass of the kettle and water is measured. The kettle is kept on for 5 minutes while the water boils. The new mass of the kettle and boiling water is taken.

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Results

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Applications Involving a Change of State Cool Box frozen pack A frozen pack is placed inside the cool box. It absorbs heat energy from the air and the food inside. The frozen pack melts ( solid to liquid ). The air and food are cooler.

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M compressor Heat The Fridge Heat the motor pumps a liquid through pipe work inside fridge. the liquid absorbs (latent) heat energy, and changes to a gas. air inside fridge becomes cooler. compressor changes gas back to a liquid. gas loses heat energy as travels through pipes at rear.

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Complex Heat Calculations Example 1 (a)Calculate the heat energy required to turn 0.5 kg of ice at 0°C into water at 20°C. c water = 4180 J kg -1 °C -1 L f ice = 3.34 x 10 5 J kg -1 There are 2 parts to this problem: 1.melting ice 2.heating water

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melting the ice (167,000 J) heating the water

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(b)How long would it take a 700 W kettle to do this? x ÷ E P t

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Questions 1.(a)Calculate the heat energy required to turn 1.2 kg of ice at 0°C into water at 18°C. (b)Calculate the time it would take a 600 W heater to do this. 400,800 + 90,288 = 491,088 J 818.5 seconds

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time (s) temperature (°C) 10 40 50250 2.A 0.2 kg mass of a solid substance is heated by a 150 W heater. A graph of temperature against time for the substance as it is heated is shown. (a)What happens between 50 and 250 seconds? (b)What is the melting point (temperature) of the solid? (c)Calculate the specific heat capacity of the solid. (d)Calculate the specific latent heat of fusion of the substance. change of state 40 °C 1250 J kg -1 °C -1 150,000 J kg -1`

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