Science league topic 5: Heat and heating curve of substance

1. Science league topic 5: Heat and heating curve of substance Page 477 - 491

2. Energy Exchange • energy is exchanged between the system and surroundings through either heat exchange or work being done Tro, Chemistry: A Molecular Approach

3. How Is Energy Exchanged? • energy is exchanged between the system and surroundings through heat and work • q = heat (thermal) energy • w = work energy • q and w are NOT state functions, their value depends on the process DE = q + w Tro, Chemistry: A Molecular Approach

4. Heat Exchange • heat is the exchange of thermal energy between the system and surroundings • occurs when system and surroundings have a difference in temperature • heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature • thermal equilibrium Tro, Chemistry: A Molecular Approach

5. Quantifying Heat Energy • the heat capacity of an object is proportional to its mass and the specific heat of the material • so we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object Heat = (mass) x (specific heat capacity) x (temp. change) q = (m) x (Cs) x (DT) Tro, Chemistry: A Molecular Approach

6. Specific Heat Capacity • measure of a substance’s intrinsic ability to absorb heat • the specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C • Cs • units are J/(g∙°C) • the molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C • the rather high specific heat of water allows it to absorb a lot of heat energy without large increases in temperature • keeping ocean shore communities and beaches cool in the summer • allows it to be used as an effective coolant to absorb heat Tro, Chemistry: A Molecular Approach

7. Cs m, DT q Example 6.2 – How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from -8.0°C to 37.0°C? • Sort Information Given: Find: T1= -8.0°C, T2= 37.0°C, m=3.10 g q,J q = m ∙ Cs ∙ DT Cs = 0.385 J/g (Table 6.4) • Strategize Concept Plan: Relationships: • Follow the Concept Plan to Solve the problem Solution: • Check Check: the unit and sign are correct

8. Vaporization • molecules in the liquid are constantly in motion • the average kinetic energy is proportional to the temperature • however, some molecules have more kinetic energy than the average if these molecules are at the surface, they may have enough energy to overcome the attractive forces • therefore – the larger the surface area, the faster the rate of evaporation • this will allow them to escape the liquid and become a vapor Tro, Chemistry: A Molecular Approach

9. Condensation • some molecules of the vapor will lose energy through molecular collisions • the result will be that some of the molecules will get captured back into the liquid when they collide with it • also some may stick and gather together to form droplets of liquid • particularly on surrounding surfaces • we call this process condensation Tro, Chemistry: A Molecular Approach

10. Evaporation vs. Condensation • vaporization and condensation are opposite processes • in an open container, the vapor molecules generally spread out faster than they can condense • the net result is that the rate of vaporization is greater than the rate of condensation, and there is a net loss of liquid • however, in a closed container, the vapor is not allowed to spread out indefinitely • the net result in a closed container is that at some time the rates of vaporization and condensation will be equal Tro, Chemistry: A Molecular Approach

11. Energetics of Vaporization • when the high energy molecules are lost from the liquid, it lowers the average kinetic energy • if energy is not drawn back into the liquid, its temperature will decrease – therefore, vaporization is an endothermic process • and condensation is an exothermic process • vaporization requires input of energy to overcome the attractions between molecules Tro, Chemistry: A Molecular Approach

12. Heat of Vaporization • the amount of heat energy required to vaporize one mole of the liquid is called the Molar Heat of Vaporization, DHvap • sometimes called the enthalpy of vaporization • always endothermic, therefore DHvap is + • somewhat temperature dependent • DHcondensation = -DHvaporization Tro, Chemistry: A Molecular Approach

13. kJ mol H2O g H2O Example 11.3 – Calculate the mass of water that can be vaporized with 155 kJ of heat at 100°C Given: Find: 155 kJ g H2O Concept Plan: Relationships: 1 mol H2O = 40.7 kJ, 1 mol = 18.02 g Solution: Check: since the given amount of heat is almost 4x the DHvap, the amount of water makes sense

14. Dynamic Equilibrium • in a closed container, once the rates of vaporization and condensation are equal, the total amount of vapor and liquid will not change • evaporation and condensation are still occurring, but because they are opposite processes, there is no net gain or loss or either vapor or liquid • when two opposite processes reach the same rate so that there is no gain or loss of material, we call it a dynamic equilibrium • this does not mean there are equal amounts of vapor and liquid – it means that they are changing by equal amounts Tro, Chemistry: A Molecular Approach

15. Dynamic Equilibrium Tro, Chemistry: A Molecular Approach

16. Vapor Pressure • the pressure exerted by the vapor when it is in dynamic equilibrium with its liquid is called the vapor pressure • remember using Dalton’s Law of Partial Pressures to account for the pressure of the water vapor when collecting gases by water displacement? • the weaker the attractive forces between the molecules, the more molecules will be in the vapor • therefore, the weaker the attractive forces, the higher the vapor pressure • the higher the vapor pressure, the more volatile the liquid Tro, Chemistry: A Molecular Approach

17. Boiling Point • when the temperature of a liquid reaches a point where its vapor pressure is the same as the external pressure, vapor bubbles can form anywhere in the liquid • not just on the surface • this phenomenon is what is called boiling and the temperature required to have the vapor pressure = external pressure is the boiling point • the normal boiling point is the temperature at which the vapor pressure of the liquid = 1 atm • the lower the external pressure, the lower the boiling point of the liquid Tro, Chemistry: A Molecular Approach

18. Heating Curve of a Liquid • as you heat a liquid, its temperature increases linearly until it reaches the boiling point • q = mass xCsxDT • once the temperature reaches the boiling point, all the added heat goes into boiling the liquid – the temperature stays constant • once all the liquid has been turned into gas, the temperature can again start to rise Tro, Chemistry: A Molecular Approach

19. Melting = Fusion • as a solid is heated, its temperature rises and the molecules vibrate more vigorously • once the temperature reaches the melting point, the molecules have sufficient energy to overcome some of the attractions that hold them in position and the solid melts (or fuses) • the opposite of melting is freezing Tro, Chemistry: A Molecular Approach

20. Energetics of Melting • when the high energy molecules are lost from the solid, it lowers the average kinetic energy • if energy is not drawn back into the solid its temperature will decrease – therefore, melting is an endothermic process • and freezing is an exothermic process • melting requires input of energy to overcome the attractions between molecules Tro, Chemistry: A Molecular Approach

21. Heating Curve of a Solid • as you heat a solid, its temperature increases linearly until it reaches the melting point • q = mass xCsxDT • once the temperature reaches the melting point, all the added heat goes into melting the solid – the temperature stays constant • once all the solid has been turned into liquid, the temperature can again start to rise • ice/water will always have a temperature of 0°C • at 1 atm Tro, Chemistry: A Molecular Approach

22. Heat of Fusion • the amount of heat energy required to melt one mole of the solid is called the MolarHeat of Fusion, DHfus • sometimes called the enthalpy of fusion • always endothermic, therefore DHfus is + • somewhat temperature dependent • DHcrystallization = -DHfusion • generally much less than DHvap • DHsublimation = DHfusion + DHvaporization

23. Heats of Fusion and Vaporization Tro, Chemistry: A Molecular Approach

24. Heating Curve of Water

25. Segment 1 • heating 1.00 mole of ice at -25.0°C up to the melting point, 0.0°C • q = mass xCsxDT • mass of 1.00 mole of ice = 18.0 g • Cs= 2.09 J/mol∙°C Tro, Chemistry: A Molecular Approach

26. Segment 2 • melting 1.00 mole of ice at the melting point, 0.0°C • q = n∙DHfus • n = 1.00 mole of ice • DHfus= 6.02 kJ/mol Tro, Chemistry: A Molecular Approach

27. Segment 3 • heating 1.00 mole of water at 0.0°C up to the boiling point, 100.0°C • q = mass xCsxDT • mass of 1.00 mole of water = 18.0 g • Cs= 2.09 J/mol∙°C Tro, Chemistry: A Molecular Approach

28. Segment 4 • boiling 1.00 mole of water at the boiling point, 100.0°C • q = n∙DHvap • n = 1.00 mole of ice • DHvap= 40.7 kJ/mol Tro, Chemistry: A Molecular Approach

29. Segment 5 • heating 1.00 mole of steam at 100.0°C up to 125.0°C • q = mass xCsxDT • mass of 1.00 mole of water = 18.0 g • Cs= 2.01 J/mol∙°C Tro, Chemistry: A Molecular Approach

30. Homework • Page 513 65, 69, 71, 79, 81 • Cs of water is , Cs of water is 0.449 J/g ∙°C • DHvap of water is 40.7 kJ/mol, DHfusof water is 6.02 kJ/mol