1 / 17

Calorimetry

Temperature Change Problems. Unit 7 Honors Chemistry. Phase Change Problems. Calorimetry. How to use math to describe the movement of heat energy. Energy Conversions. Heat is a specific type of energy that can be measured in different ways. The SI unit for heat is Joules

clodia
Download Presentation

Calorimetry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Temperature Change Problems Unit 7 Honors Chemistry Phase Change Problems Calorimetry How to use math to describe the movement of heat energy

  2. Energy Conversions • Heat is a specific type of energy that can be measured in different ways. • The SI unit for heat is Joules • 4.184 Joules = 1 calorie (this will be given) • 1000 calories = 1 kilocalorie • 1000 Joules = 1 kiloJoule

  3. Heat Conversions • How many joules are in 130 calories? • How many calories are in 50 Joules? 4.184 Joules 130 calories = 543.92 Joules (= 540 J (Sig figs!) 1 calorie 50 Joules 1 calorie = 11.95 calories 4.184 Joules

  4. Heat Conversions • How many kilojoules are in 130 Calories? 1 kJ 4.184 Joules 130 Calories 1000J 1 Calorie = 0.54 KiloJoules

  5. Calorimetry • Allows us to calculate the amount of energy required to heat up a substance or to make a substance change states. • Molar Heat of Fusion (Hf)— The heat absorbed by one mole of a substance when changing from a solid to a liquid. • For water, it = 6.0 kiloJoules/mole • or 334 Joules/gram (specific heat of fusion) • Heat of solidification is opposite of heat of fusion (heat is released).

  6. Molar Heat of Vaporization (Hv)— The heat absorbed by one mole of a substance when changing from a liquid to a gas. • For water, it = 40.7 kiloJoules/mole. or 2260 Joules/gram (specific heat of vaporization • Heat of condensation is the opposite of heat of vaporization (heat is released) • Every pure substance will have a unique Molar heat of fusion (Hf) or vaporization (Hv)

  7. Heat Required For a Phase Change • Heat Absorbed or Released = q • For Melting or Freezing use the following: • For Vaporization or Condensation use the following: q = (moles) x Molar Heat Fusion q = (moles) x Molar Heat vaporization

  8. 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 = (moles) x (Hf)

  9. Example #2 • How much heat energy in kJ will be released when 200grams steam condenses back to a liquid water? • Hv = 40.7kJ/mol q = (moles) x (Hv) 200gram 1 mole 40.7 kJ 18gram 1 mole = 452 kJ released Or -452kJ

  10. Heating a Substance with No Phase Change • Specific Heat Capacity--The amount of energy required to raise one gram of a substance one degree Celcius. Water’s Specific Heat (as a liquid) Cp= 4.184 Joules/gram oC *Every pure substance will have its own unique specific heat for every phase!

  11. Heating a Substance with No Phase Change • When you see an increase in the temperature of a sample, the heat is being added to raise the temperature • How much the temperature increases is based upon the heat capacity (Cp) and the massof your sample • The higher the heat capacity number, the longer it takes to heat a substance up and the longer the substance holds on to the heat.

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

  13. Example #3 m How much energy is needed to heat 80 g of water from 10 oC to 55 oC? Tfinal Tinitial q = mCpΔT = m Cp (Tfinal – Tinitial ) = (80g) ( 4.184 J/g C) (55oC – 10oC) q = 15062 joules Is the energy absorbed or released? Absorbed, because temperature inincreasing Final Answer: 15,062 J = 15.06 kJ absorbed/ endothermic

  14. Example #4 m How much energy is needed to cool 150 g of ice from -2 oC to -55 oC? Tfinal Tinitial q = mCpΔT = m Cp (Tfinal – Tinitial ) = (150g) ( 2.06 J/g C) (-55oC – -2oC) q = - 16377 joules Is the energy absorbed or released? Released, because temperature indecreasing Final Answer: - 16377 J = -16.3 kJ released/exothermic

  15. Heat Problem Road Map q = (moles)Hv q = (moles)Hf Gas Heats Vaporization or Condensation Melting or Freezing Liquid Heats q = mCpΔT Solid Heats * Add each individual energies (in kJ) together for total heat energy required for multistep problems (up to 5 steps max!)

  16. 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 #5 -How much energy in kJ 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.

  17. Calorimetry Formula Summary Phase Change • Use Molar Heat constants Melting use q = (moles) x (Hfusion) Vaporize use q = (moles) x (HVaporization) No Phase Change • Use specific heat capacity q = (mass) ( Cp ) ( ΔT )

More Related