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# Lecture 17b- Water Curve Calcs

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The second lecture for chapter 17 (honors and prep chemistry) covers the water curve, specific heat, and the enthalpy of phase changes.

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### Lecture 17b- Water Curve Calcs

1. 1. BELLWORK‐ heat ﬂow in pool When you enter a swimming pool, the water may feel quite cold. ABer a while, though, your body “gets used to it,” and the water no longer feels so cold. Use the concept of heat to explain what is going on.
2. 2. The specific heat capacity (c) of any substance is the amount of heat required to raise the temperature of 1gram of that substance by 1°C. • Every substance has its own specific heat.
3. 3. Water has a high specific heat. c= 4.184 J/(g·°C) Water can absorb or release a lot of heat before changing temperature
4. 4. The heat absorbed during a change in temperature is calculated using the equation Specific Mass of Change Heat heat sample in temp (J) capacity (g) (°C) J/(g·°C) ΔT= Tfinal-Tinitial
5. 5. PracIce How much energy is required to raise 50g of water from 25°C to 40°C?     q = c  m  ΔT     q = unknown     c = 4.18J/(g°C) for water     m = 50g     ΔT = 15°C q = 4.18 x 50 x 15 = 3135 J of energy
6. 6. ∆Hfus = the quanIty of heat absorbed  during melIng    = the quanIty of heat released  during freezing. For water H2O(s)  H2O(l) ∆Hfus = 6.01kJ/mol H2O(l)  H2O(s) ∆Hfus = -6.01kJ/mol
7. 7. How many grams of ice at 0°C will melt if 2.25kJ of heat are added? Known • iniIal and ﬁnal temps are 0 °C •  ΔHfus =  6.01kJ/mol •  ΔH = 2.25kJ 2.25 kJ x 1mol = 0.374 moles would 6.01kJ be melted
8. 8. How much energy is needed to melt 50g of ice? Known The heat of • ΔHfus =  6.01kJ/mol fusion is PER • 50g MOLE so convert grams to moles 50g H2O   x      1mole      2.8 moles x 6.01 kJ =         18g 1mole =16.8 kJ
9. 9. The quanIty of heat absorbed by a evaporaIng liquid is exactly the same as the quanIty of heat released when the vapor condenses = ∆Hvap For water H2O(l)  H2O(g) ∆Hvap = 40.7kJ/mol H2O(g)  H2O(l) ∆Hvap= -40.7kJ/mol
10. 10. Phase change occurs at a speciﬁc temperature (aka the melIng point and boiling point) Temperature is constant during a phase change
11. 11. Temperature is constant during a phase change During a phase change heat is used to overcome intermolecular a`racIons.
12. 12. The ΔHvap is always larger than ΔHfus ∆Hfus  is the energy required to overcome some  intermolecular a`racIons.  ∆Hvap  is the energy required to overcome all intermolecular a`racIons.
13. 13. IN YOUR NOTES-- Include a labeled drawing of the water curve and all of the notes in black and red on the slides to follow.
14. 14. To calculate energy changes (ΔH) for a state change      moles x ΔHfus      OR      moles x ΔHvap
15. 15. To calculate energy required to increase or decrease temperature within a single state q= c m ΔT      Use      the speciﬁc Use the heat of speciﬁc heat Energy steam of liquid water Use the speciﬁc heat of ice
16. 16. To calculate the energy change over more than one secIon of the curve… Calculate the energy for each step and add them together.
17. 17. Calculate the ΔH associated with changing 100g of steam at 125°C to ice at ‐50 °C. 125°C 1.Determine if a phase change occurs within the temperature range. • water will condense at 100°C • water will freeze at 0 °C ‐50°C
18. 18. Calculate the ΔH associated with changing 100g of steam at 125°C to ice at ‐50 °C. 125°C 2. Split problem into steps. 125 °C steam  100 °C steam steam  water 100 °C water  0 °C water ‐50°C water  ice 0 °C ice  ‐50 °C ice
19. 19. 3. Calculate the enthalpy change for each step. 125 °C steam  100 °C steam q = 1.89 J/(g°C) x 100g x ‐25°C = ‐4725J steam  water 100g x 1 mol/18g = 5.55mol x 40.7kJ/mol = ‐226kJ 100 °C water  0 °C water q = 4.18 J/(g°C) x 100g x ‐100°C = ‐41,800J water  ice 5.55mol x 6.01 kJ/mol = ‐33.4kJ 0 °C ice  ‐50 °C ice q = 2.10 J/(g°C) x 100g x ‐50°C = ‐10,500J
20. 20. 4. Add the values for each step to get  the total energy change       ‐4725J = ‐4.725 kJ +      ‐226kJ +    ‐41,800J = ‐41.8 kJ +         ‐33.4kJ +    ‐10,500J = ‐10.5 kJ Whoops! The units aren’t the same!!
21. 21. 4. Add the values for each step to get  the total energy change     ‐4.725 kJ +   ‐226        kJ The total enthalpy +         ‐41.8     kJ change and the +         ‐33.4     kJ changes in each step +         ‐10.5     kJ are all negaIve ΔHtotal = ‐316.4 kJ values because cooling water is an exothermic process
22. 22. EXOTHERMIC STATE CHANGES Heat comes out of substance(system) ∆H is negative Happens in a freezer Freezing(solidification) & condensation ENDOTHERMIC STATE CHANGES Heat goes into substance ∆H is positive Happens on a stove Melting(fusion), vaporization, sublimation
23. 23. The insulated device used to measure heat changes in chemical or physical processes is called a calorimeter. qwater = - qrxn The heat change for the water in the calorimeter is equal to the heat change of the reaction but opposite in sign.
24. 24. CalculaIons for Honors Only