Chapter 5 Energy and States of Matter. Changes of State. Melting and Freezing. Calculations Using Heat of Fusion

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1 Chapter 5 Energy and States of Matter Changes of State 5.6 Melting and Freezing 5.7 Boiling and Condensation 1 2 Melting and Freezing A substance is melting while it changes from a solid to a liquid. A substance is freezing while it changes from a liquid to a solid. The freezing (melting) point of water is 0 C. Calculations Using Heat of Fusion The heat of fusion is the amount of heat released when 1 gram of liquid freezes at its freezing point. The heat of fusion is the amount of heat needed to melt 1 gram of a solid at its melting point. For water the heat of fusion (at 0 C) is 80. cal 1 g water 3 4 1

2 Calculation Using Heat of Fusion The heat involved in the freezing (or melting) a specific mass of water (or ice) is calculated using the heat of fusion. Heat = g water x 80. cal g water Problem: How much heat in calories is needed to melt 15.0 g of water? Learning Check A. How many calories are needed to melt 5.0 g of ice of 0 C? 1) 80. cal 2) 400 cal 3) 0 cal B. How many calories are released when 25 g of water at 0 C freezes? 1) 80. cal 2) 0 cal 3) 2000 cal 15.0 g water x 80. cal = 1200 cal 1 g water 5 6 Boiling A. How many calories are needed to melt 5.0 g of ice of 0 C? 2) 400 cal 5.0 g x 80. cal 1 g B. How many calories are released when 25 g of water at 0 C freezes? 3) 2000 cal 25 g x 80. cal 1 g Water evaporates when molecules on the surface gain enough energy to form a gas. At boiling, all the water molecules acquire enough energy to form a gas

3 The heat of vaporization Heat of Vaporization Is the amount of heat needed to change 1 g of liquid to gas at the boiling point. Is the amount of heat released when 1 g of a gas changes to liquid at the boiling point. Learning Check How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100 C? 1) 27 kcal 2) 540 kcal 3) kcal Boiling (Condensing) Point of Water = 100 C Heat of Vaporization (water) = 540 cal 1 g water 9 10 Chapter 5 Energy and States of Matter How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100 C? 1) 27 kcal 50.0 g steam x 540 cal x 1 kcal = 27 kcal 1 g steam 1000 cal 5.8 Heating and Cooling Curves

4 Heating Curve A heating curve illustrates the changes of state as a solid is heated. Sloped lines indicate an increase in temperature. Plateaus (flat lines) indicate a change of state. Learning Check A. A flat line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state Cooling Curve A. A flat line on a heating curve represents 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change A cooling curve illustrates the changes of state as a gas is cooled. Sloped lines indicate a decrease in temperature. This cooling curve for water begins at 140 C and ends at -30 C

5 Learning Check Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0 C 2) 50 C 3) 100 C B. At a temperature of 0 C, water 1) freezes 2) melts 3) changes to a gas C. At 40 C, water is a 1) solid 2) liquid 3) gas D. When water freezes, heat is 1) removed 2) added Use the cooling curve for water to answer each. A. Water condenses at a temperature of 3) 100 C B. At a temperature of 0 C, water 1) freezes C. At 40 C, water is a 2) liquid D. When water freezes, heat is 1) removed Combined Heat Calculations Combined Heat Calculations (continued) To reduce a fever, an infant is packed in 250 g of ice. If the ice at 0 C melts and warms to body temperature (37.0 C), how many calories are removed from the body? Step 1: Diagram the changes. 37 C ΔT = 37.0 C 0 C = 37.0 C 0 C S L Step 2: Calculate the heat to melt ice (fusion) 250 g ice x 80. cal = cal 1 g ice Step 3: Calculate the heat to warm the water from 0 C to 37.0 C 250 g x 37.0 C x 1.00 cal = 9250 cal g C Total: Step 2 + Step 3 = cal Final answer = cal

6 Learning Check When a volcano erupts, 150 g of steam at 100 C is released. How many kilocalories are lost when the steam condenses and cools to 15 C? 1) 81 kcal 2) 13 kcal 3) 94 kcal 3) 94 kcal Condense: 150 g x 540 cal x 1 kcal = 81 kcal 1 g 1000 cal Cool: 150 g x 85 C x 1 cal x 1 kcal = 13 kcal g C 1000 cal Total: 81 kcal + 13 kcal = 94 kcal