Chapter 2 Energy and Matter 2.7 Changes of State

Melting and Freezing A substance is melting when it changes from a solid to a liquid is freezing when it changes from a liquid to a solid such as water has a freezing (melting) point of 0 °C

Heat of Fusion The heat of fusion is the amount of heat released when 1 gram of liquid freezes (at its freezing point) is the amount of heat needed to melt 1 gram of a solid (at its melting point) for water (at 0 °C) is 334 J or 80 cal 1 g of water

Guide to Calculations Using Heat of Fusion (or Vaporization)

Calculations Using Heat of Fusion How much heat in calories is needed to melt 15.0 g of ice at 0 °C ? STEP 1 Given: 15.0 g of water(s) change of state: melting at 0 °C STEP 2 Plan: g of water(s) g of water(l) STEP 3 Write conversion factors: 1 g of water = 80 cal 1 g of water and 80 cal 80 cal 1 g of water

Calculations Using Heat of Fusion (continued) STEP 4 Set up the problem to calculate calories: 15.0 g water x 80 cal = 1200 cal 1 g water

Learning Check How many joules are released when 25.0 g of water at 0 °C freezes? 1) 334 J 2) 2000 J 3) 8350 J

Solution 3) 8350 J STEP 1 Given: 25.0 g of water(l) change of state: freezing at 0 °C STEP 2 Plan: g of water g of water STEP 3 Write conversion factors: 1 g of water = 334 J 1 g of water and __334 J___ 334 cal 1 g of water

Solution (continued) STEP 4 Set up the problem to calculate joules: 25.0 g water x 334 J = 8350 J (3) 1 g water

Sublimation Sublimation occurs when a solid changes directly to a gas is typical of dry ice, which sublimes at 78 C takes place in frost-free refrigerators is used to prepare freeze-dried foods for long-term storage

Evaporation and Condensation Water evaporates when molecules on the surface gain sufficient energy to form a gas condenses when gas molecules lose energy and form a liquid

Boiling At boiling, water molecules acquire the energy to form a gas bubbles appear throughout the liquid

Heat of Vaporization The heat of vaporization is the amount of heat absorbed to vaporize 1 g of a liquid to gas at the boiling point released when 1 g of a gas condenses to liquid at the boiling point Boiling Point of Water = 100 °C Heat of Vaporization (water) = 2260 J or 540 cal 1 g of water

Comparing Heat of Fusion and Heat of Vaporization

Learning Check How many kilocalories (kcal) are released when 50.0 g of water(g) as steam from a volcano condenses at 100 °C? 1) 27 kcal 2) 540 kcal 3) 2700 kcal

Solution 1) 27 kcal STEP 1 50.0 g of water(g) = 50.0 g of water(l) Change of state: water condensing at 100 °C STEP 2 Plan: g of water(g) g of water(l) STEP 3 Write conversion factors: 1 g of water = 540 cal 1 g of water and 540 cal 540 cal 1 g of water 1 kcal = 1000 cal 1 kcal and 1000 cal 1000 cal 1 kcal

Solution (continued) STEP 4 Set up the problem to calculate kilocalories: 50.0 g water x 540 cal x 1 kcal = 27 kcal 1 g water 1000 cal

Summary of Changes of State

Heating Curve A heating curve illustrates the changes of state as a solid is heated uses sloped lines to show an increase in temperature uses plateaus (horizontal lines) to indicate a change of state

Learning Check A. A plateau (horizontal 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

Solution A. A plateau (horizontal 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

Cooling Curve A cooling curve illustrates the changes of state as a gas is cooled uses sloped lines to indicate a decrease in temperature uses plateaus (horizontal lines) to indicate a change of state

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, liquid 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

Solution 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, liquid water 1) freezes C. At 40 °C, water is a 2) liquid D. When water freezes, heat is 1) removed

Combined Heat Calculations 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? STEP 1 Given: 250 g ice changes to water at 37.0 °C Need: calories to melt at 0 °C and warm to 37.0 °C STEP 2  T = 37.0 °C – 0 °C = 37.0 °C 37.0 °C 0 °C

Combined Heat Calculations (continued) STEP 3 Conversion factors: 1 g of water (0 °C) = 80 cal 1 g of water (0 °C) and 80 cal 80 cal 1 g of water (0 °C) 1 g of water(l) = 1.00 cal/g °C 1 g of water(l) and 1.00 cal/g °C 1.00 cal/g °C 1 g of water(l)

Combined Heat Calculations (continued) STEP 4 Set up problem: (1) Heat to melt ice (fusion) 250 g ice x 80 cal = 20 000 cal 1 g ice (2) Heat the water(l) from 0 °C to 37.0 °C 250 g x 37.0 °C x 1.00 cal = 9250 cal g °C Total: 20 000 cal + 9250 cal = 29 000 cal (rounded)