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Calorimetry

Calorimetry . AP Chemistry. Calorimetry. Calorimetry is the measurement of heat flow. It allows us to calculate the amount of energy required to heat up a substance or to make a substance change states. Calorimetry Terms.

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Calorimetry

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  1. Calorimetry AP Chemistry

  2. Calorimetry • Calorimetry is the measurement of heat flow. • It allows us to calculate the amount of energy required to heat up a substance or to make a substance change states.

  3. Calorimetry Terms • Molar Heat of Fusion— The heat absorbed by one mole of a substance when changing from a solid to a liquid. • For water, it = 6.0 kiloJoules/mole • Heat of solidification is opposite of heat of fusion (heat is released).

  4. Molar Heat of Vaporization— The heat absorbed by one mole of a substance when changing from a liquid to a gas. • For water, it = 40.7 kiloJoules/mole. • Heat of condensation is the opposite of heat of vaporization (heat is released) Heat Required For a Phase Change – Or Latent Heat Process Heat Absorbed or Released = q q = (moles) x (Molar Heat Fusion/Vaporization)

  5. Calculating Heat Required To Change State • Example #1: How much heat is needed to melt 56.0 grams of ice into liquid (the molar heat of fusion for ice is 6.0 kJ/mol)? • 56.0 g 1 mole H2O 6.0 kJ = 18.0 g 1 mole • = 18.7 kJ will be absorbed • q = + 18.7kJ

  6. Example #2 • How much heat energy will be released when 200grams steam condenses back to a liquid water? • Molar Heat condensation = 40.7kJ/mol q = (moles) x (Molar Heat condensation) 200gram 1 mole 40.7 kJ 18gram 1 mole = 452 kJ released So, q = - 452kJ

  7. Heating a Substance with No Phase Change • Also known as a sensible heat process. • Depends on Specific Heat Capacity • Specific Heat Capacity--The amount of energy required to raise one gram of a substance one degree Celcius. • Every substance and every phase of that substance has its own unique specific heat capacity. Water’s Specific Heat (as a liquid) = 4.184 Joules/gram oC or unit cans be J/gK

  8. Specific heat (J/goC) = heat capacity (J/oC) • You will see both terms used. • You can also use Molar Heat Capacity • This is the heat energy required to raise 1 mole of a substance 1oC or 1K. • Molar heat capacity can be determined from the specific heat by multiplying it by the molar mass of the substance.

  9. Energy to Change Temperature q = (mass) ( C ) ( T ) Change in Temperature Tfinal – Tinitial In OCelcius or kelvin Heat Measured in Joules Specific Heat Capacity Mass In grams

  10. Exothermic vs. Endothermic • Exothermic Change – Heat Energy is released to the surroundings. q and ∆H are (-) • Molecules slow down, extra energy is transferred to surrounding. • Cooling phase changes are exothermic • Endothermic Change – Heat energy is absorbed by the system. q and ∆H are (+) • -Molecules move faster as they absorb energy • -Phase changes that require energy (or heat) are endothermic

  11. Changes In State Solid Freezing Sublimation Deposition Melting Vaporization Liquid Gas Condensation

  12. Example #3 • How much energy is needed to heat 80grams of water from 10oC to 55oC at constant pressure? q = m C T = m C (Tfinal – Tinitial ) q = (80grams) ( 4.184 J/goC) (55oC – 10oC) q = + 15062 joules divide by 1000 to get kilojoules 15062 J 1 kJ = 1000J 15.06 kJ absorbed

  13. This problem requires two steps. Since water is solid ice at 0oC, we need to melt the ice and then heat it up to 50oC. • Example #4 -How much energy is needed to change 150grams of ice from 0oC to 50oC? Step 1 – Calculate heat required to melt 150grams ice 150g 1 mole 6.0 kJ = 50 kJ 18grams 1 mole Step 2 - Calculate heat required to heat liquid water from 0oC to 50oC q = mC T = (150g)(4.184 J/goC)(50oC) = 31380 J  convert to kJ = 31.38kJ *Add both heat values together for your final answer 50 kJ + 31.38kJ = 81.38 kJ heat absorbed.

  14. Multiple Step Calorimetry Problems • Add each individual energies together for total Use q = Moles x Molar Heat vap/fus e Gas Heats Vaporization d c Liquid Heats b Solid Heats melting a Use q = mC T qtotal = a + b + c + d + e

  15. Calorimetry • Experimental way of measuring heat generation/consumption by essentially catching all the heat energy in a water bath or water bath + metal apparatus. • Coffee Cup Calorimetry • Styrofoam cup insulates the contents so any heat generated or consumed in the water can be measured by the temperature change • q = -(Hrxn) = mCH2O∆T • Reactions must take place in water, then you measure the change in temperature.

  16. Bomb-Calorimeter • Used to measure heats of reaction. • Usually do combustion reactions in it. • Metal innards absorb heat. You have to keep track of heat absorbed by metal and water. • q = mCH2O∆T + Cp ∆T • qwater + qapparatus • Cp is the heat capacity of the apparatus – determined experimentally.

  17. Conservation of Energy Under conditions of constant pressure Heat Lost = Heat gained qp-lost = qp-gained For a Reaction heat gained by water = heat released by reaction.

  18. Example- When 100mls of 0.1M HCl is mixed with 100mls of 0.1M NaOH in a coffee cup calorimeter, the temperature increases from 25oC to 29.8oC. Assume that the coffee cup is a perfect insulator, pressure is constant and the density of the solution is 1g/ml and specific heat capacity for the mixture is that of water, 4.18 J/gK. • What is the enthalpy change for the reaction? ∆H = q = mC∆T

  19. Since density is 1g/mL , all volumes are gram values. • Total volume must be used = 200mLs = 200grams. • ∆T = Tfinal – Tinitial = 29.8oC – 25oC = 4.8oC • q = mC ∆T = (200g)(4.18J/gK)(4.8K) • q = 4013 J = 4.013 kJ • ∆H = - 4.013 kilojoules • Since temperature of water increased, it is exothermic, which means a - ∆H

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