2 Phases of Matter Solid – lattice work, has shape Liquid – fluid, takes shape of containerGas – fluid, needs sealed container(Plasma) – most energetic
3 Temperature Temperature – measure of hotness degrees Celsius or degrees Centigrade or °CFahrenheit scaleKelvin Scale (doesn’t use the symbol)Absolute zeroFreezing of waterBoiling of water-273 °C0 °C100 °C-460 °F32 °F212 °F0 K273 K373 K
4 Temperature and Molecular Motion The higher the temperature, the faster the molecules are moving.The faster the molecules are moving, the more energy they haveFor the same molecule: Gas > Liquid > SolidGas has more energy than a solid
5 Heat Heat is the flow of thermal energy from hot substance to cold substancerelated to the amount of thermal energyboth temperature and mass are importantObjects in contact tend to reach same temperaturethermal equilibrium
6 Heat Measure of energy transferred calorie – cal (little c)Calorie = 1000 calories = kilocalorie (big C)Joule (1 calorie = J)BTU (British Thermal Unit)Transfer of Heat requires temperature differenceHeat moves from high to low temperatureThermal expansionmost materials get bigger as they get hotter
7 Specific Heat Capacity symbol “c”Unit: Energy/mass * temperature differenceJ/kg °C or cal/g °C or BTU/ lb °Fmeasures effect of heat on change in temperaturec of water is 1 cal/g °C this is highc of metals is lower – easier to change temperatureQ = m c T
8 Thermal Equilibrium Problem A 2.0 kg block of aluminum at 60 C is placed in contact with a 3.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks?cal = 900 J/kg Ccsteel = 470 J/kg C
9 Thermal Equilibrium Problem The aluminum block will lose heat to the steel blockThe heat lost by the aluminum equals that gained by the steelThe final temperature will be some place between the initial temperatures of the blocks.
10 Qal + Qsteel = 0 (aluminum loses while steel gains) mal cal Tal + mst cst Tst = 0mal cal (Tf-Ti)al + mst cst (Tf-Ti)st = 02.0(900)(Tf-60) + 3.0(470)(Tf-30) =01,800Tf – 108, ,410Tf – 42,300 = 03,210Tf – 150,300 = 0Tf = 150,300/3,210 = 46.8 C
11 Thermal Equilibrium Problem 2 A 9.0 kg block of aluminum at 60 C is placed in contact with a 1.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks?cal = 900 J/kg Ccsteel = 470 J/kg C
13 Heat Transfer Conduction Convection Radiation Heat transfers along an objectmetals have high conductionConvectionheat transfers when a fluid moveshot air rises and cold air sinkscreates wind and “weather”Radiationany object with thermal energy emits radiationelectromagnetic waveblack absorbs radiant energy better than white
14 Changing Phase Gas Liquid Solid Energy evaporation condensation add heatcondensationremove heatHeat of vaporization540 cal/g for water2257 kJ/kgLiquidmeltingadd heatHeat of fusion80 cal/g for water334 kJ/kgfreezingremove heatSolid
15 Freezing and Melting Freezing Melting temperature remains constant from liquid to solidloses energy (heat) to freeze a liquidMeltingFrom solid to liquidrequires heat to melt solidtemperature remains constant80 cal/g (334 kJ/kg) of heat lost to freeze waterCalled Heat of fusion
16 Evaporation and Condensation Heat required to change liquid to gasboilingCondensationHeat released when gas change to liquidtemperature of solid and liquid are the same540 cal/g (2257 kJ/kg) to evaporated waterCalled Heat of vaporization
18 How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 ° C? How much heat is required to melt 3 g of ice at 0 ° C?How much heat is required to vaporize 4 g of water to steam at100 ° C?How much heat is required to take 2 g of ice from 0 ° C to steam at 100 ° C?
19 How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 °C? Q=mc(Tf-Ti)= 3g x 1 cal/g °C x 16 °C = 48 calHow much heat is required to melt 3 g of ice at 0 °C?3 g x 80 cal/g = 240 calHow much heat is required to vaporize 4 g of water to steam at100 °C?4 g x 540 cal/g = 2160 calHow much heat is required to take 2 g of ice from 0 °C to steam at 100 °C?Ice to water g x 80 cal/g = 160 cal0 °C to 100 °C 2g x 1 cal/g °C x 100 °C = 200 calWater to steam g x 540 cal/g = caltotal cal