Presentation on theme: "Phase Changes and their Calculations"— Presentation transcript:
1Phase Changes and their Calculations J. Flint Baumwirt Granada Hills Charter High School
2Symbols used in heat and phase change calculations: - The Greek letter "delta" represents a “change in.”H - Represents enthalpy, which we think of as heat.Q (or q) - Represents heat energy gained or lost.m - Represents mass.T - Represents temperature.Cp - Represents specific heat. NOTE: Cp will differ depending on state (solid, liquid or gas)
3Enthalpy of fusion: Hfus As energy is added to a solid at its melting point, all the energy is used to increase the kinetic energy of the molecules during the phase change.Because of this, the temperature of the melting system remains constant until all of the solid has become liquid. If energy is still being added to the system, the temperature will begin to climb when all of the solid becomes liquid.
4Enthalpy of vaporization: Hvap As energy is added to a liquid at its boiling point, all the energy is used to increase the kinetic energy of the molecules during the phase change.Because of this, the temperature of the boiling system remains constant until all of the liquid has become vapor. If energy is still being added to the system, the temperature will begin to climb when all of the liquid has been vaporized.
5Heat transferThe law of conservation of energy tells us that heat lost by one quantity of matter must be gained by another.As a quantity of matter gains heat energy, its temperature will increase based on its individual specific heat.Specific heat is defined as: the heat required to raise the temperature of one gram of a substance by one Celsius degree.
6OverviewTransitions between solid, liquid, and gaseous phases typically involve large amounts of energy compared to the specific heat. If heat were added at a constant rate to a mass of ice to take it through its phase changes to liquid water and then to steam, the energies required to accomplish the phase changes (called the latent heat of fusion and latent heat of vaporization ) would lead to plateaus in the temperature vs time graph.
7Latent Heat of Fusion and Latent Heat of Vaporization: A Table of Heats of Fusion and Vaporization are provided below for a variety of substances:Notice that the latent heats of vaporization are considerably higher in energy than the latent heats of fusion. This means that the amount of energy required to overcome the attractive forces between the particles of a solid to form the particles of a liquid is small compared to the energy required to pull apart liquid molecules to allow them to form a gas. In other words, it takes MORE energy to turn a liquid to a gas than to melt a solid to form a liquid.SubstanceHfus (kJ/mol)Hvap (kJ/mol)Water5.9840.6Benzene9.9230.7Chloroform12.431.9Diethel Ether6.8626.0Ethanol7.6138.6
8From this information we can now calculate how much energy it takes to melt an ice cube. Notice that the values of the heats of fusion and the heats of vaporization are in kilojoules per mole.Notice also that specific heat is usually given in the unit of Joules/mol
9CalculationsExample: Calculate the energy, in kilojoules necessary to melt 1.00 gram of ice.This is a phase change calculationWhere a solid changes to a liquidThe equation used here is Q = nHfusWhere Q is the energy in Joulesn is the number of moles andHfus is the change in energy for this phase change
10The latent heat of fusion of water = 5.98 kJ/mol. This quantity will be provided for you in the problem or in a table or chartThe latent heat of fusion of water = 5.98 kJ/mol.1.00 gram H2O1 mole5.98 kJ= kJ18.0 grams1 mole1.00 x 5.98 = 0.33218.00
11Calculating Energy Changes: Liquid to Gas Example: Calculate the energy required to heat 25.0 grams of liquid water from 25C to 100C and change it to steam at 100 C.The specific heat of water is J/gC, and the molar heat of vaporization of water is 40.6 kJ/mol.In calculating the energy required to change a substance from one state to another in this problem two different energies must be considered;the energy required to change the substance to a different state ANDthe energy required to raise the temperature to the point of change.Here you must do two problems and then add the energies together. One is a simple energy equation calculation Q=mCpT and the other is like our first example multiplying mass (in the proper units) times the molar heat of vaporization, Q = n Hvap
12Step #1: Q=mCT m = 25.0 grams C=4.184 J/gC T=25C to 100C = 75C = 7.8 x 103 J = 7.8 kJgCStep #2: Vaporization Q = n Hvap25.0 g H2O1 mol H2O40.6 kJ= 57 kJ18.0 gramsStep #3: Add the two energies together (make sure units match): kJ + 57 kJ = 65 kJ
13Molar Heat of Vaporization Sample Heat Calculation Problem:How much heat is needed to convert 250 g of ice at -30 oC to steam at 150 oC? Draw a Phase Diagram showing all possible changes for this problem first.Equations:Specific Heat ValuesMolar Heat of FusionH2O(s) = 2.06 J/gCH2O = 5.98 kJ/molH2O(l) = J/gCMolar Heat of VaporizationH2O(g) = 2.02 J/gCH2O = 40.6 kJ/molQ= mCpT Q=n HfusQ=n Hvap
14vaporizing (boiling)- 30 oC150 oCice waterwater vapor (steam)melting0 oC100 oC43521How much heat is needed to convert 250 g of ice at -30C to steam at 150C?1-Ice temp rising melting 3-water temp rising 4-vaporizing 5-vapor temp risingThis problem requires six individual calculations. There are three in which the phase does not change (yellow), two in which the phase changes (green), and a final calculation to total the energy of the problem. (pink)Specific Heat ValuesMolar Heat of FusionH2O(s) Cp solid = 2.06 J/gCDHfus of H2O = 5.98 kJ/molH2O(l) Cp liquid = J/gCMolar Heat of VaporizationH2O(g) Cp gas = 2.02 J/gCDHvap of H2O = 40.6 kJ/mol
1515.45 kJ + 0.831 kJ + 104.6 kJ + 564 kJ + 25.25 kJ = Total Q = 710 kJ How much heat is needed to convert 250 g of ice at -30C to steam at 150C?This problem requires six individual calculations. There are three in which the phase does not change (yellow), two in which the phase changes (green), and a final calculation to total the energies. (pink)1. Raise the temperature of ice from -30 oC to 0 oC - no phase changeQ = m T Cp solid = (250g) (30 Co) (2.06 J/g . Co) = 15,450 J/1000J kJ-1 =15.45 kJ2. Melt the ice - phase change occurs - temperature does not changeQ = m Hfus = (250g/18.0g mol-1) (5.98 kJ/mol) = kJ3. Raise the temperature of liquid water from 0 oC to 100 oC - no phase changeq = m T Cp liquid= (250g) (100 Co) (4.184 J/g . Co) = 104,600 J/1000J kJ-1 = kJ4. Vaporize the liquid water - phase change occurs - temperature does not changeQ = m Hvap = (250g/18.0g mol-1) (40.6 kJ/mol) = 564 kJ5. Raise the temperature of the water vapor from 100 oC to 150 oC - no phase changeQ = m T Cp gas = (250g) (50 Co) (2.02 J/g . Co) = 25,250 J/1000J kJ-1 = kJ6. Find the total heat needed for the conversion - Add the Q 's from steps 15.45 kJ kJ kJ kJ kJ = Total Q = 710 kJ