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NOTES. 3.5 Specific Heat 9.5 Energy Changes in Chemical Reactions 10.5 Changes in State. Teacher Copy. How much heat energy is needed to :. Raise the temperature of the ENTIRE mouse by 1 o C. Raise the temperature of the ENTIRE elephant by 1 o C?. Heat Capacity

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  1. NOTES 3.5 Specific Heat 9.5 Energy Changes in Chemical Reactions 10.5 Changes in State

  2. Teacher Copy

  3. How much heat energy is needed to : Raise the temperature of the ENTIRE mouse by 1oC Raise the temperature of the ENTIRE elephant by 1oC? Heat Capacity the amount of heat energy needed to heat up the ENTIRE object by 1oC

  4. How much heat energy is needed to : Raise the temperature of 1 gram of the mouse by 1oC Raise the temperature of 1 gram of the elephant by 1oC? Specific Heat Capacity the amount of heat energy needed to heat up 1 gram of an object by 1oC

  5. Each substance has its own unique specific heat and, each phase of that substance has its own unique specific heat • Example: specific heats of the 3 phases of water: • Ice = 2.092 J/ g-K • Liquid water = 4.184 J/ g-K • Water vapor = 1.841 J/ g-K

  6. The HEAT EQUATION is used to calculate the amount of heat energy needed to change the temperature of any phase of a substance: q = mSHΔT q = heat energy m = mass (grams) SH = specific heat capacity (most books/ colleges use ‘c’ not ‘SH’) ΔT = change in temperature

  7. Always use this form to calculate ΔT ΔT = Tfinal- Tinitial ΔT should be (–) if the reaction or process was exothermic (ENERGY OUT) Example: Water at 45oC was cooled to 33oC 33oC – 45oC = – 12oC ΔT should be(+) if the reaction or process was endothermic (ENERGY IN) Example: Water at 33oC is warmed to 45oC 45oC – 33oC = + 12oC

  8. Sample Problem: Calculate the heat energy needed to warm 25 g of ice at – 5.0oC to ice at 0.0oC. q = ? m = 25 g SHice= 2.092 J/g-K ΔT = 0.0oC – (– 5.0oC) → 0.0oC + 5.0oC = + 5.0oC q = (25g)(2.092 J/g-K)(+ 5.0oC) → 261.5 = 260 J (2 SF)

  9. Sample Problem: Calculate the heat energy needed to warm 25 g of liquid water at 0.0oC to liquid water at 100.oC q = ? m = 25 g SHliquid H20 = 4.184 J/g-K ΔT = 100.oC – 0.0oC → + 100.oC q = (25g)(4.184 J/g-K)(+100.oC) → 10,460 J = 10,000 J (2 SF)

  10. Sample Problem: Calculate the heat energy needed to warm 25 g of steam at 100.0oC to steam at 105.0oC q = ? m = 25 g SHsteam= 1.841 J/g-K ΔT = 105.0oC – 100.0oC → + 5.0oC q = (25g)(1.841 J/g-K)(+5.0oC) → 230.125 J = 230 J (2 SF)

  11. HEY! how come we can cancel out K with oC ????!!?!?!? Example: q = (25g)(1.841 J/g-K)(+5.0oC) Answer: b/c 1 Kelvin “degree” represents the same ΔT as 1 Celsius degree This would NOT work with Celsius and Fahrenheit nor with Kelvin and Fahrenheit!

  12. How does all of this relate to chemical reactions? The energy change that occurs during a chemical reaction is often called the heat of reaction Another word for heat of reaction is enthalpy(H) which is defined as the energy change which occurs at a constant pressure ΔHRXN= the symbol for the change in enthalpy (heat of reaction) that occurs during a chemical reaction

  13. How ΔHRXN determined? Before we answer that we need to define system and surroundings System = the thing being studied, the thing causing the change in energy Surroundings = all the things that interact with the system, but don’t cause the energy change (note: the surroundings may gain or lose energy - but they are not the cause of the energy change!) in a chemical reaction, the reaction and the chemicals involved in the reaction are considered to be the system

  14. How ΔHRXN determined? Before we answer that we need to define endothermic and exothermic reactions Endothermic Reaction = a chemical reaction in which energy leaves the surroundings and enters the system (the chemicals) Exothermic Reaction = a chemical reaction in which energy leaves the system (the chemicals) and enters the surrounds

  15. How ΔHRXN determined? Before we answer that we need to explain making and breaking chemical bonds Breaking Chemical Bonds = energy leaves the surroundings and enters the system exothermic from the point of view of the surroundings endothermic from the point of view of the system Making Chemical Bonds = energy is released from the system and enters the surroundings exothermic from the point of view the system endothermic from the point of view the surroundings Since we always look at it from the point of view of the chemicals (the system) we say that: • Breaking bonds = an endothermic process • Making bonds = an exothermic process

  16. How ΔHRXN determined? Now we can explain it! Endothermic chemical reaction = more energy goes in to break bonds than comes out when new bonds are made, ΔHRXN = (+) Exothermic chemical reaction= less energy goes in to break bonds than comes out when new bonds are made, ΔHRXN = ( – )

  17. Energy as a Reactant or Product When ΔHRXN = (+) , then energy is a reactant Example: Energy + N2(g) + O2(g) → 2 NO(g) ΔHRXN = +90.2 kJ (Endothermic) When ΔHRXN = ( – ) , then energy is a product Example: C(s) + O2(g) → CO2(g) + Energy ΔHRXN = − 393.5 kJ (Exothermic)

  18. So far all we have been doing is heating things up….. …….But, what about actually changing the phase of the water? Heat of fusion the energy needed to melt or freeze (fuse)a substance Heat of vaporization the energy needed to vaporize or condense a substance

  19. Each substance has its own unique heat of fusion and heat of vaporization • Example: • Heats of fusion and vaporization for water: • Heat of fusion = 333.41 J/g • Heat of vaporization = 2,257 J/g

  20. To calculate the amount of heat energy needed to change the phase of a substance q = mΔHfus Or q = mΔHVap q = heat energy m = mass (grams) ΔHfus= heat of fusion ΔHvap = heat of vaporization

  21. Notice that ΔT plays no part in heats of fusion and vaporization. This is because, at a phase change the temperature does NOT change!

  22. Sample Problem: Calculate the heat energy needed to melt 25 g of ice at 0.0oC. q = ? m = 25 g ΔHfus= 333.41 J/g q = (25g)(333.41 J/g) → 8,335.25 = 8,300 J (2 SF)

  23. Sample Problem: Calculate the heat energy needed to vaporize 25 g of liquid water at 100.oC. q = ? m = 25 g ΔHvap= 2257 J/g q = (25g)(2257 J/g) → 56,425 = 56,000 J (2 SF)

  24. By summing all the energy changes we can calculate the total amount of energy needed to take ice from – 5oC to steam at 105oC. 1st Step = warm up ice by 5oC to mp = 260 J 2nd Step = melt the ice = 8,300 J 3rd Step = warm up liquid water to bp = 10,000 J 4th Step = vaporize the water = 56,000 J 3rd Step = warm up steam by 5oC = 230 J Total Energy Needed = 74,790 J 75,000 J

  25. This can be shown graphically on a Heating / Cooling Curve (this one is for water) Boiling Point liquid/vapor Gas Only Liquid Only Ice/liquid Melting Point Ice Only Yes, I know the scale is off! 

  26. This can be shown graphically on a Heating / Cooling Curve (this one is for water) liquid/vapor present Heat ice = EK =↑ and T↑ Melt Ice = EKand T unchanged, EP ↑ Heat liquid = EK =↑ and T↑0 Ice/liquid present Vaporize liquid = EKand T unchanged, EP ↑ Heat vapor = EK =↑ and T↑

  27. pressure: The process of measuring these heats is know as calorimetry Calorie Get it???? Calorimeter A device used to measure quantities of heat Calorimeters are filled with water and chemical reactions are carried out inside the calorimeter

  28. Here’s an example of a much less expensive, and so much less accurate, ‘foam cup’ calorimeter Here’s an example of a expensive and highly accurate ‘bomb calorimeter’ Guess which one we will use ! ? !

  29. The principle behind the calorimeter is actually quite simple Heat lost by the reaction (– qrxn= mrxnSHrxnΔTrxn) …….must be equal to…… Heat gained by the water (+qwater= mwaterSHwaterΔTwater)

  30. Sample Problem: 30.0 g of water at Ti of 19.9oC is placed in a coffee cup calorimeter. A 5.37 g sample of solid NH4NO3 is added to the water in the calorimeter. Tf of the resulting solution is 8.1oC, what is the heat of solution for this process? NOTE: You consider the heat exchange process to be ‘over’ when the temperature on the thermometer begins to return to room temperature NOTE: the calorimeter and the water are the surroundings and the ammonium nitrate is the system – we will ignore the small amount of heat transferred to the calorimeter

  31. 1st: Calculate ΔT ΔT = Tfinal – Tinitial 8.1oC – 19.9 oC = – 11.8oC the (–) sign indicates that the surroundings (the water) LOST heat to the system (the solid) 2nd Calculate ΔHsoln q = mSHwaterΔT m = mass water + mass solid = 35.4 g SHwater = 35.4 g (4.184 J/g-K)( – 11.8oC) q = – 1740 J 3rd Change the sign of q Heat lost by the water = heat gained by the solid, so the sign of q must be (+) = +1740 J

  32. Heats of Solution are usually given in kJ/ mol so…. Divide q by the grams of solid 1740 J 5.37 g solid = 324 J/g Multiply the J/g by the molar mass of the solid 324 J/ g x 80.04 g/ mol = 25, 900 J/ mol Divide J/ mol by 1000 to change it to kJ 25,900 J/ mol 1000 J/ kJ = 25.9 kJ/mol

  33. Divide q by the grams of solid 1740 J 5.37 g solid = 324 J/g Multiply the J/g by the molar mass of the solid 324 J/ g x 80.04 g/ mol = 25, 900 J/ mol Divide J/ mol by 1000 to change it to kJ 25,900 J/ mol 1000 J/ kJ = 25.9 kJ/mol

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