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Heat in chemical reactions

Heat in chemical reactions. How Do We Measure Heat?. What is Heat?. What will cause more discomfort? bucket of boiling water or drop of boiling water They are at the same temperature but the amount of stored heat energy is different. Temperature. Hotness or coldness

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Heat in chemical reactions

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  1. Heat in chemical reactions How Do We Measure Heat?

  2. What is Heat? What will cause more discomfort? bucket of boiling water or drop of boiling water They are at the same temperature but the amount of stored heat energy is different

  3. Temperature • Hotness or coldness • Proportional to average kinetic energy • Intensive property (does not depend on amount) • Put a thermometer in a hot oven • The fast oven particles bounce off the thermometer • The thermometers particles speed up • The oven particles slow down • Heat is transferred!

  4. Energy • The capacity to do work or produce heat • Energy transfer is measured • Extensive property – depends on amount • units are joules, calories or BTU’s • joules in science (metric unit) • BTU’s for air conditioning, heaters etc • Calories in some science like food • 1 cal = 4.184 j • 1 Calorie = 1 Kcal

  5. Heat transfer (flow) Depends on The capacity of substance to absorb heat • The mass • The change in temperature Why are pots made of metal and handles made of wood or plastic?

  6. Specific heat • amount of energy required to raise the temperature of 1 gram of that substance 1 oC • Specific heat = heat (j or cal) mass (g) x change in temp.(C) • Water has a specific heat of 4.184 J/(g oC) • Copper has a specific heat of 0.387 J/(g oC)

  7. Specific heat problems 1.Calculate the Specific heat of Fe if heat= 145 J mass= 89.3 g Δ temperature= 13.2 oC • Calculate the heat of water if Specific heat= 4.18 J/g C mass= 43.6 g Δ temperature= 8.3 o C

  8. Lab: Specific heat Purpose: To determine the specific heat of Copper and an unknown metal

  9. Measuring heat transfer Calorimeter – well insulated container that minimizes the amount of heat transferred to the surroundings

  10. Their Calorimeter

  11. Our calorimeter

  12. Heat transfer • Use water in calorimeter to absorb or release heat . • We can • measure mass of water in calorimeter • measure the change in temperature of the water in the calorimeter. • know the specific heat of water. • Heat(q)H2O = massH2O∆TH2OSpHeat(S)H2O qsur= -qsystem qH2O = -qreaction

  13. Can’t measure heat directly But can see measure the effects of energy being released or absorbed by the reaction. Water surrounding the rxn chamber will increase or decrease in temperature. Can measure this temperature change.

  14. Thermo Coffee Cup Calorimeter • M = mass of water in cup • SWater =4.184 j/ g C • TAfter – Tbefore = Tchange Tf - Ti = T (for the water) Heat water= mw x Sw x Tw calculate heat collected by calorimeter

  15. Calorimeter Basics Heat absorbed = - Heat released Heat water = heat metal mwater x Swater x Twate = -[mmetal x Smetal x Tmetal] • Final temp. of metal= final temp. of the water

  16. Determine the specific heat of an unknown metal with a mass of 23.8g which is heated to 100.0°C and dropped into a calorimeter containing 50.0g of water at 24.0°C. The final temperature is 32.5°C. H2O: mass = 50.0g T1= 24.0°C, T2= 32.5°C Sp Heat(S)= 4.184.J/g°C Metal: mass= 23.8g T1= 100.0C, T2= 32.5C Sp Heat (S) = ?

  17. Thermo Calorimeter hot metal is added to 100.0 g of 25.6 oC water. The Final temp is 35.9 oC, How much heat is transferred? q = 100.0g x 4.184 J x ( 35.9 – 25.6 C) g C q = 4300 J or 4.30 x 103 J or 4.30 kJ

  18. Specific Heat LabFind the Specific Heat of a metal • Mass metal • Heat metal to 100 °C • Place water in calorimeter • Know mass water • Know Initial temp • Put hot metal in cold water • What happens to temp of water? • What happens to temp of metal? • When does the temp stop changing?

  19. Specific Heat LabFind the Specific Heat of a metal • Heat into water = -heat out of metal qwater = -qmetal mwater x Swater x Twater = -[mmetal x Smetal x Tmetal]

  20. Temperature • What happens to the particles of water when I put them into the freezer? • The slow moving air particles hit the water • The fast water molecules move slower • The slow air particles move faster • Heat is transferred!

  21. Energy Change • Exothermic – energy (heat) given off in a reaction • Condensing, freezing Endothermic – energy (heat) absorbed in a reaction Vaporizing, melting

  22. Changes of State • How much energy does it take to heat 10.0 g ice from –20.0 to 100.0o C steam • Specific Heat ice 2.07 j/gC • Heat of fusion 334 j/g • Heat of evaporation 2260 j/g • Specific heat water 4.184 j/gC

  23. Changes of State It takes energy to change from one state to another • Heat of fusion • energy to change a solid into a liquid • Is this process endo or exothermic? • Heat of vaporization • energy to change a liquid to a gas • Is this process endo or exothermic?

  24. Cooling Curve (p.481) Gas Heat of vaporization Liquid Heat of fusion Solid

  25. Heat or Energy Problems heat = mass x specific heat x temp change q (∆H)= m x S x  T joules = g x J x C° g C°

  26. Energy from -20 to 0 q = m x S x T = 10.0g x 2.07j x 20oC goC • Energy from solid to liquid q = 10.0g x 6.7 j/g • Energy from 0 – 100 q = m x S x T = 10.0g x 4.18j x 100oC goC • Energy from liquid to steam q = 10.0 x 41.0 j/g

  27. Heat of reaction A calorimeter is filled with 75.0g of water at an initial temperature of 19.5°C. A 0.50mole sample of solid NaOH is added and the temperature increases to 26.7°C. What is the enthalpy change per mole of NaOH for this solution process. NaOH(s) Na+(aq) + OH-(aq)

  28. We can write expression for heat for both the metal and water Heat(q) H2O = massH2O ∆T H2O Sp Heat(Cp) H2OHeat(q )metal = mass metal ∆T metal Sp Heat(Cp )metalHeat( q) H2O =Heat( q)metal massH2O ∆T H2O Sp Heat(Cp) H2O = massmetal ∆Tmetal Sp HeatCpmetal

  29. Make chart with data H2O MetalmT1T2∆TSp Heat(Cp) 50.0g 23.8g 24.0°C 100.0°C 32.5°C 32.5°C 8.5°C 67.5°C 4.184J/g°C x massH2O ∆TH2O Sp Heat (Cp)H2O = massmetal∆Tmetal Sp Heat(Cp)metal (50.0g)(8.5°C)(4.184J/g°C)= (23.8g)(67.5°C)x x= 1.1 J/g°C

  30. Make chart & put data in chart H2O NaOHm 75.0g (0.050 mol)T1 19.8 °CT2 26.7 °C∆T 6.9 °CCp 4.184 J/g °C Heat(q)H2O = mass ∆T Sp Heat(Cp) = (75.0g)(6.9°C)(4.184J/g°C) = 2200J Heat(q)H2O = Heat(q)NaOH therefore Heat(q)NaOH= 2200J2200J = x x= 44000J 0.050 mol 1 mol ∆H= -44000J/mol NaOH

  31. Calculation of enthalpy changes 2H2O2 2H2O + O2Δ H=-190 kJ What is the heat released if 2.3 g H2O2 react?

  32. Thermo Measuring Heat Transfer • “All” heat went into water • Know heat into water • All came from your experiment. • Heat water = -heat experiment • Minus sign important!

  33. Heat Problem • How much does it take to heat 5.0 g of aluminum (Cp =0.879 j/g C) from 22 C to 100. C? • q = m x Cp x T • q = 5.0g x ( 0.879 j/g C) x (100. – 22) • q = 340 or 3.4 x 102 j are required

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