Leaving Cert Physics - Home



Heat Heat Capacity : Is the amount of heat needed to change the temperature of a body by 1K ( Q = CΔθ )Specific heat capacity : The amount of heat energy( Joules ) needed to raise 1Kg of a substance by 1K unit = J Kg-1 K-1Heat lost Heat gained = Mass (m) x Specific heat capacity (c) x Temperature change (Δθ)Heat supplied Q = m c ΔθNote :If a substance has a high specific heat capacity it will heat up slowly and cool down slowly.Power= Work ( Energy ) Joules Js-1 = Watt Time Second 35407604077335Aim:To find the specific heat capacity of a solidMethod1. Using an electric balance find the mass of the metal block.2. Read the initial temperature on the thermometer.3. Zero the joulemeter.4. Allow the current to flow until the temperature has risen by about 10oC5. Wait a few minutes for the heat to spread throughout the metal block before recording the final temperature. Results Mass of the Aluminium block = 500g ( 0.5 Kg ) Initial Temperature = 120C Final Temperature = 21.5oC ( Temperature change = 9.5oC ) Reading on the Joulemeter = 4,204 J Energy supplied = Heat gained by the Aluminium block Q = m x cAl xΔθ 4204 = 0.5 xcAlx 9.54204 = cAl0.5 x 9.5885 J Kg-1 K-1 = cAl3939540-202565Aim:To find the specific heat capacity of a liquid.Method1. Using an electric balance find the mass of the calorimeter.2. Find the mass of the calorimeter plus the liquid.3.Note the initial temperature of the liquid 4. Zero the joulemeter and the current to flow until the temperature has risen by about 10oC.5. Note the final temperature of the liquid and the final joulemeter readingNote : The temperature of the liquid is the same as the temperature of the calorimeterbecause they are in thermal equilibrium. ( They are in contact with one another)ResultsMass of the copper calorimeter = 50 g ( 0.05 Kg )Mass of the liquid = 80g ( 0.08 Kg ) Mass of the calorimeter + the liquid = 130 g Initial temperature of the liquid = 12oCFinal temperature of the liquid = 21oC Temperature change = 9oCFinal joulemeter reading = 1,985 JSpecific heat capacity of Copper (cCu )=400 J Kg -1 K -1Heat supplied ( written on joulemeter ) = ( Heat gained by the calorimeter ) + ( Heat gained by the liquid ) Q = ( m c Δθ ) Calorimeter + ( m c Δθ ) liquid 1985 = ( 0.05 x 400 x 9 ) + ( 0.08 x c liquid x 9 ) 1985 = 180 + ( 0.72 x c liquid) 1985 - 180 = 0.72 x c liquid1850 = c liquid 0.722,506.9 J Kg -1 K-1 = c liquidNote : 1. The water is heated from 5oC below room temperatureto5oC above room temperature so thattheheat lost to the surroundings is balanced by the heat gained from the surroundings. 2. Use a digital thermometer correct to 2 decimal places to improve accuracy . 3. The heating coil must be covered with water, as the calculations assume that only the water is heated. 4. Stir the liquid to ensure an even distribution of heat.5. A less accurate result would have been obtained using a larger volume of liquid becauseitwould result in a smaller rise in temperature, giving rise to a GREATER % ERRORin the results.Latent Heat : The amount of heat needed to produce a change in state ( solid liquid gas )3914775272415Without a change in temperature.Latent heat of fusion solid liquid Latent heat of vapourisation liquid gas When snow thaws it feel colder because the snow absorbs the latent heat of fusion required from the air.A burn from steam is more painful than those caused by boiling water at the same temperature , because the steam contains the latent heat of vapourisation.Specific Latent Heat : Amount of heat energy(Joules ) needed to produce a change of state in 1 Kg of a substance with NO TEMPERATURE CHANGEsymbol for specific latent heat= L unit = J Kg-1( no temperature change )Latent heat gained (Q) = Mass (m) x Specific latent heat (L)Latent heat lost (Q)Q = m L Heat pump ( Fridge or air conditioning system )The liquid used has (i) a low boiling point (ii) a high specific latent heat of vapourisation.Expansion valvePressure dropsLiquid turns to a vapourDraws the latent heat of vapourisationfrom inside the fridge.Inside the fridge cools 28060651111885Compressor -> pressure increases ->vapour turns to liquid -> latent heat of vapourisation released to the surroundings 4214495-508000Aim :To find the specific latent heat of fusion of ice.Method:1. Using an electric balance record the mass of the calorimeter.2. Record the mass of the calorimeter plus the tepid water.3. Using a thermometer record the temperature of the coldwater in the calorimeter.4. Using a metal tongs transfer the dried ice to the calorimeter.5. When all the ice has melted stir the water and record the final temperature of the water in the calorimeter.6. Record the final mass of the calorimeter + water + melted ice .ResultsMass of Copper Calorimeter= 50g ( 0.05Kg )Mass of calorimeter + water = 145g ( 0.145 Kg )Mass of calorimeter + water + melted ice= 170g ( 0.170 Kg ) Initial temperature of water = 25oCFinal temperature of water + melted ice = 5oCGivencCu = 400 J Kg-1 K-1cwater = 4200 J Kg-1 K-1 Heat gained = Heat lostHeat gained by + Heat to raise melted = Heat lost by + Heat lost by the cold water ice in melting ice from 0oC to 5oC calorimeter in the calorimeter ( mL ) ice + ( m cwaterΔθ ) melted ice = ( m cCuΔθ ) copper calorimeter + ( m cwaterΔθ ) coldwater( 0.025L ) + ( 0.025 x 4200 x 5) = ( 0.05 x 400 x 20 ) + ( 0.095 x 4200 x 20 ) 0.025L+ 525 = 400 + 7980 0.025L + 525 = 8,380 0.025L = 8,380 - 525 0.025L = 7,855L = 7,855 0.025L = 314,200 J Kg-1Note : The ice was dried because the calculations assume that only ice at 0oC was added to the water mass of the ice = ( mass of calorimeter + cold water + melted ice ) – (mass of calorimeter + cold water )Aim: To find the specific latent heat of vapourisation of water4253230212090Allow the steam to condense into the cold water in the calorimeter until the temperature has risen by about 20oCResults Mass of the copper calorimeter = 50g ( 0.05 Kg )Mass of the calorimeter + water = 145g ( 0.145 Kg ) Mass of the calorimeter + water + steam ) = 148.5g ( 0.1485 Kg )Initial temperature of water = 22oCFinal temperature of water + condensed steam = 43oCGivencCu = 400 J Kg-1 K-1cwater = 4200 J Kg-1 K-1Heat lost = Heat gained Heat lost by steam + Heat lost by condensed = Heat gained by + Heat gained by in condensing steam from 100oC to 43oC calorimeter cold water ( mL )steam + ( m cwaterΔθ )condensed steam = ( m cCuΔθ ) copper calorimeter + ( m cwaterΔθ ) cold water( 0.0035L )+ ( 0.0035 x 4200 x 57 ) = ( 0.05 x 400 x 21 ) + ( 0.095 x 4200 x 21 )( 0.0035L ) + ( 837.90 ) = ( 420 ) + ( 8,379 ) 0.0035 L + 837.90 = 8,799 0.0035 L = 8,799 - 837.90 0.0035 L = 7961.1 L = 7961.1 0.0035L = 2,274.60J Kg-1Note : The steam trap was used to dry the steam . The calculations assume that only steam at 100oC was added to the water.mass of steam = ( mass of calorimeter + water + condensed steam ) – ( mass of calorimeter + water ) Heat can travel by one of three methods1.Conduction : Is the movement of heat through a solid with no overall movement of the molecules.2. Convection : Is the transfer of heat through a liquid or a gas due to the movement of the molecules3. Radiation : Is the transfer of heat by means of electromagnetic waves ( heat transfer in a vacuum ) Note : A black surface heats quickly and cools quickly. A white ( or shiny ) surface heats slowly and cools slowly U value : The amount of heat conducted per second(energy per second = watt ) through 1 m2 of a substance when a temperature difference of 1 K exists between its ends.Unit = Watt meter -2 Kelvin -1( Wm-2 K -1 )Note : When a building is well insulated it has a very lowUvalueSolar constant ( irradiance ) : The average amount of the suns energy falling per secondperpendicularly on 1m2 of the earth`s atmosphere .Wm-2Heat is a form of energy. ( unit = Joule )Temperature is a measure of how hot or cold an abject is .( S.I. unit = Kelvin )100 oC373.15 K 0oC273.15K-273.15oC0K (absolute zero )Note : A temperature change of x oCcorresponds to a temperature change of x K A thermometric property : Is any property that changes measurably with temperatureThermometer Thermometric propertyMercury ( or alcohol ) in glass Liquids expand when heatedResistance of a metallic conductorResistance increases with temperatureResistance in a semiconductorResistance decreases with temperature Volume of gas at constant pressure Gases expand when heated Disagreement between thermometers Generally thermometers based on different thermometric properties only agree at 0oC and 100oCTwo different types of thermometer will give slightly different readings at the same temperature Becausedifferent thermometric propertiesreact differently to the same temperature changeThis is why we need a standard thermometer . The mercury in glass is used as a standard .36601408225790Thermoelectric effect ( A thermocouple )When the two junctions are at different temperatures.anemf ( voltage ) is generated and a current flows.The bigger the temperature difference between the two junctions the bigger the voltage produced. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download