3.03 Phase Changes of Matter Water crystal Heating-Cooling Curve Dr. Fred Omega Garces Chemistry 100 Miramar College

Phases of Matter: Terminology Energy is required for phase changes to occur. Solid-Liquid-Gas Triangle

Intermolecular Forces At the molecular level: Molecules or matter is held together by “glue” called intermolecular forces Energy added (K.E. increases)

Keeping Matter together Intramolecular Forces - Force which keeps molecule together, i.e., bonds. Intermolecular Forces - Attractive force between molecules. Responsible for keeping matter in solid or liquid phase.

Heating Cooling Curve From Ice to Steam and Vice-versa 6.01 kJ mol Stage1 Stage2 Stage3 Stage4 Stage5 Heat Addition What is the energy needed to take 1g H2O at 0°C to 100°C ? 80 +100+ 540 =720cal 1.84 J g ° 0.43 cal g 6.01 kJ mol 540 cal g 4.184 J g ° 1 cal g o 40.7 kJ mol 80 cal g 2.09 J g ° 0.50 cal g

Heating / Cooling Curves From Steam to Ice and Vice-versa 2.09 J g ° 0.50 cal g 1.84 J g ° 0.43 cal g .334 kJ g 80 cal 4.184 J g ° 1 cal g o 2.26 kJ g 540 cal g What is the energy needed to take 1g H2O at 0°C to 100°C ? 540+100+80=720cal

What is the energy needed to take 1g H2O ice at 0°C to steam 100°C? Heating Cooling Curve From Ice to Steam and Vice-versa Stage1 Stage2 Stage3 Stage4 Stage5 80 cal g 60.2 kJ mol 540 cal g 40.1 kJ mol 1 cal g o 4.184 J g ° What is the energy needed to take 1g H2O ice at 0°C to steam 100°C? 80 +100+ 540 =720cal Units Stage1 S.h. ice Stage2 Heat fusion Stage3 specific heat (s.h.) water Stage4 Heat vaporization Stage5 S.h. gas cal /g 0.50 cal / g K 80 cal/g 1.0 540 cal / g J g-1 K-1 2.11 J / g K 334 J/g 4,18 2257 2.08 cal mol-1 9.01 cal / mol K 1441 18.102cal / mol K 9728 J mol-1 K-1 38.1 J / mol K 6.02 kJ / mol K 75.33 40.1 kJ/mol 37.5 J /mol K

SOLUTION STUDY CHECK SAMPLE PROBLEM Heat of Fusion Ice cubes at 0 °C with a mass of 26 g are added to your soft drink. a. How much heat (joules) must be added to melt all the ice at 0 °C? b. What happens to the temperature of your soft drink? Why? calorie STEP 1 Given 26 g of H2O(s) Need joules to melt ice SOLUTION cal STEP 2 calorie STEP 3 Equalities/Conversion Factors 80 cal 80 cal 80 cal STEP 4 Set Up Problem 80 cal 2080 calorie b. The soft drink will be colder because heat from the soft drink is providing the energy to melt the ice. STUDY CHECK In a freezer, 150. g water at 0 °C is placed in an ice cube tray. How much heat, in kilojoules, must be removed to form ice cubes at 0 °C?

SOLUTION STUDY CHECK SAMPLE PROBLEM Heat removed from water in sauna. In a sauna, 150 g of water is converted to steam at 100 °C. How many kilocalories of heat are needed? STEP 1 Given 150 g of H2O(l) to H2O(g) Need kilocalories of heat to change state SOLUTION STEP 2 STEP 3 Equalities/Conversion Factors STEP 4 Set Up Problem STUDY CHECK When steam from a pan of boiling water reaches a cool window, it condenses. How much heat, in kilocalories (kcal), is released when 25 g of steam condenses at 100 °C?

SAMPLE PROBLEM 2.76 Raising temperature of water A pitcher containing 0.75 L (750 g) of water at 4°C is removed from the refrigerator. How many Kcal (and kJ) are needed to warm the water to a room temperature of 22 °C First calculate for 1 g water, then multiply by 750 gram H2O 22°C – 4°C = 18 °Temp change Therefore it will cost 18 cal for 1 gram For 750 grams : 750 g * (18cal/g) = 13500 cal or 13.5 Cal or 58.2 KJ Stage1 Stage2 Stage3 Stage4 Stage5 4.184 J g ° 1 cal g o - 22°C - 4°C Heat Addition

Answer: Calculate all for 1 g then multiply final answer by mass SAMPLE PROBLEM Combining Heat Calculations Calculate in calories and Joule the following: a) Energy required to heat 25.0 g of water from 12.5°C to 25.7°C b) Energy required to heat 38.0 g of water from 1°C to 100°C steam. c) Energy lose when 15.0 g of water cools from 75 °C to 0°C solid ice. d) Final temperature when 600. cal is added to 6.00 g ice at starting at 0°C. Answer: Calculate all for 1 g then multiply final answer by mass 13. 2 cal / g x 25.0 g = 330 cal = 1420 J (99 cal + 540 cal ) x 38.0 g = 24300 cal = 24.3 kcal =105 kJ (75 cal + 80 cal ) x 15.0 g = 2330 cal = 2.33 kcal = 10.0 kJ Cost (80 cal) x 6 g = 480 cal. There fore 600 – 480 = 120 cal 120 cal / 6 g = 20° Final temperature = 20° C Double Check: (80 cal + 20 cal ) x 6.00 g = 600. cal. This checsks

SOLUTION Answer: SAMPLE PROBLEM 2.14 Combining Heat Calculations Using Table 2.13 and the specific heat of ethanol (2.46 J/g °C), calculate the total heat, in joules, needed to convert 15.0 g of ethanol at 25.0 °C to gas at 78.0 °C. STEP 1 Given 15.0 g of ethanol at 25.0 °C Need heat (J) needed to warm the ethanol and change it to gas SOLUTION STEP 2 When several changes occur, draw a diagram of heating and changes of state. Total heat = joules needed to warm ethanol from 25.0 °C to 78.0 °C + joules to change liquid to gas at 78.0 °C Answer: PART 1: Energy to Warm Ethanol 25° to 88° Temperature change = 78°– 25° = 53° for 1 gram ethanol, cost 2.46 J for every ° So 2.46 J/° x 53° = 130.38 J for 1 gram 15.0 g x (130.38 J / g) = 1955.7 J Part 2: Energy to convert ethanol liquid to gas Need heat vaporization ethanol = 841 J for 1 g Therefore for 15.0 g, that is 841 J x 15.0 g = 12615 J Part 1 + part 2 = 1955.7 + 12315 = 14.6 kJ = 3.38 Kcal

Summary The equilibria between the three phase of a substance is a function of the pressure and the temperature as shown in a phase diagram. Equilibria between any two phase are indicated by a line. The line through the melting point usually slopes slightly to the right as pressure increases (solid is more dense than liquid). For water this slope is to the left. Features of the phase diagram is the normal melting point (freezing point) and the normal boiling point. The triple point is the condition in which all three phases of matter exist and the critical point is the condition in which the liquid and the gas phase are indistinguishable.