1 / 10

Calorimetry

Calorimetry. Unit 7 Phases of Matter. 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 — The heat absorbed by one mole of a substance when changing from a solid to a liquid .

benny
Télécharger la présentation

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. Calorimetry Unit 7 Phases of Matter

  2. 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— 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).

  3. 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 Heat Absorbed or Released = q q = (moles) x (Molar Heat Fusion/Vaporization)

  4. 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

  5. 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

  6. 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) = 4.184 Joules/gram oC

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

  8. Example #3 • How much energy is needed to heat 80grams of water from 10oC to 55oC? 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

  9. 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.

  10. Multiple Step Calorimetry Problems Use q = Moles x Molar Heat vap/fus Gas Heats Vaporization Liquid Heats Solid Heats melting Use q = mC T * Add each individual energies (in kJ) together for total

More Related