Presentation is loading. Please wait.

Presentation is loading. Please wait.

Phase Changes and their Calculations J. Flint Baumwirt Granada Hills Charter High School.

Similar presentations


Presentation on theme: "Phase Changes and their Calculations J. Flint Baumwirt Granada Hills Charter High School."— Presentation transcript:

1 Phase Changes and their Calculations J. Flint Baumwirt Granada Hills Charter High School

2 Symbols 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)

3 Enthalpy of fusion:  H fus 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.

4 Enthalpy of vaporization:  H vap 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.

5 Heat transfer The 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.

6 Overview Transitions 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.specific heat heat of fusionheat of vaporization

7 Latent 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. Substance  H fus (kJ/mol)  H vap (kJ/mol) Water Benzene Chloroform Diethel Ether Ethanol

8 From 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

9 Calculations Example: Calculate the energy, in kilojoules necessary to melt 1.00 gram of ice. –This is a phase change calculation –Where a solid changes to a liquid –The equation used here is Q = n  H fus –Where Q is the energy in Joules –n is the number of moles and –  H fus is the change in energy for this phase change

10 1.00 gram H 2 O The latent heat of fusion of water = 5.98 kJ/mol. 1 mole18.0 grams = kJ5.98 kJ1 mole 1.00 x 5.98 = This quantity will be provided for you in the problem or in a table or chart

11 Calculating Energy Changes: Liquid to Gas Example: Calculate the energy required to heat 25.0 grams of liquid water from 25  C to 100  C and change it to steam at 100  C. The specific heat of water is J/g  C, 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 AND –the 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=mC p  T and the other is like our first example multiplying mass (in the proper units) times the molar heat of vaporization, Q = n  H vap

12 Step #1: Q=mC  T m = 25.0 grams C=4.184 J/g  C  T=25  C to 100  C = 75  C 25.0 grams4.184 J75  C= 7.8 x 10 3 J = 7.8 kJ gCgC Step #2: Vaporization Q = n  H vap 25.0 g H 2 O1 mol H 2 O40.6 kJ= 57 kJ 18.0 grams1 mol H 2 O Step #3: Add the two energies together (make sure units match): 7.8 kJ + 57 kJ = 65 kJ

13 Sample Heat Calculation Problem: How much heat is needed to convert 250 g of ice at -30 o C to steam at 150 o C? Draw a Phase Diagram showing all possible changes for this problem first. Specific Heat ValuesMolar Heat of Fusion H 2 O (s) = 2.06 J/g  C H 2 O = 5.98 kJ/mol H 2 O (l) = J/g  C Molar Heat of Vaporization H 2 O (g) = 2.02 J/g  C H 2 O = 40.6 kJ/mol Q= mC p  T Q=n  H fus Q=n  H vap Equations:

14 1-Ice temp rising  2-melting  3-water temp rising  4-vaporizing  5-vapor temp rising Specific Heat ValuesMolar Heat of Fusion H 2 O (s) C p solid = 2.06 J/g  C  H fus of H 2 O = 5.98 kJ/mol H 2 O (l) C p liquid = J/g  C Molar Heat of Vaporization H 2 O (g) C p gas = 2.02 J/g  C  H vap of H 2 O = 40.6 kJ/mol 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 energy of the problem. (pink) vaporizing (boiling) - 30 o C 150 o C ice  water water  vapor (steam) melting 0 o C 100 o C How much heat is needed to convert 250 g of ice at - 30  C to steam at 150  C?

15 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 o C to 0 o C - no phase change Q = m  T C p solid = (250g) (30 C o ) (2.06 J/g. C o ) = 15,450 J/1000J kJ -1 =15.45 kJ 2. Melt the ice - phase change occurs - temperature does not change Q = m  H fus = (250g/18.0g mol -1 ) (5.98 kJ/mol) = kJ 3. Raise the temperature of liquid water from 0 o C to 100 o C - no phase change q = m  T C p liquid = (250g) (100 C o ) (4.184 J/g. C o ) = 104,600 J/1000J kJ -1 = kJ 4. Vaporize the liquid water - phase change occurs - temperature does not change Q = m  H vap = (250g/18.0g mol -1 ) (40.6 kJ/mol) = 564 kJ 5. Raise the temperature of the water vapor from 100 o C to 150 o C - no phase change Q = m  T C p gas = (250g) (50 C o ) (2.02 J/g. C o ) = 25,250 J/1000J kJ -1 = kJ 6. Find the total heat needed for the conversion - Add the Q 's from steps kJ kJ kJ kJ kJ = Total Q = 710 kJ How much heat is needed to convert 250 g of ice at -30  C to steam at 150  C?


Download ppt "Phase Changes and their Calculations J. Flint Baumwirt Granada Hills Charter High School."

Similar presentations


Ads by Google