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

2. Energy Conversions Heat is a specific type of energy that can be measured in different ways. Heat is a specific type of energy that can be measured in different ways. The SI unit for heat is Joules 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 joules are in 130 calories? How many calories are in 50 Joules? How many calories are in 50 Joules? 130 calories 1 calorie 4.184 Joules 50 Joules 4.184 Joules 1 calorie = 543.92 Joules (= 540 J (Sig figs!) = 11.95 calories

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

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

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

7. Heat Required For a Phase Change Heat Absorbed or Released = q Heat Absorbed or Released = q For Melting or Freezing use the following: For Melting or Freezing use the following: For Vaporization or Condensation 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)? 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 H 2 O 6.0 kJ = 18.0 g 1 mole 56.0 g 1 mole H 2 O 6.0 kJ = 18.0 g 1 mole = 18.7 kJ will be absorbed = 18.7 kJ will be absorbed q = (moles) x (H f )

9. Example #2 How much heat energy in kJ will be released when 200grams steam condenses back to a liquid water? How much heat energy in kJ will be released when 200grams steam condenses back to a liquid water? H v = 40.7kJ/mol H v = 40.7kJ/mol q = (moles) x (H v ) 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. 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) C p = 4.184 Joules/gram o C * 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 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 (C p ) and the mass of your sample How much the temperature increases is based upon the heat capacity (C p ) and the mass of 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. 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) ( C p ) ( T ) Heat Measured in Joules Mass In grams Specific Heat Capacity Change in Temperature T final – T initial In O Celcius

13. Example #3 How much energy is needed to heat 80 g of water from 10 o C to 55 o C? q = mC p ΔT = m C p (T final – T initial ) (55 o C – 10 o C) q = 15062 joules Absorbedincreasing Absorbed, because temperature in increasing m T initial T final = (80g)( 4.184 J/g C) Is the energy absorbed or released? 15,062 J = 15.06 kJ Final Answer: 15,062 J = 15.06 kJ absorbed/ endothermic

14. Example #4 How much energy is needed to cool 150 g of ice from -2 o C to -55 o C? q = mC p ΔT = m C p (T final – T initial ) (-55 o C – -2 o C) q = - 16377 joules Released, because temperature in decreasing m T initial T final = (150g)( 2.06 J/g C) Is the energy absorbed or released? Final Answer: - 16377 J = -16.3 kJ released/exothermic

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

16. Example #5 -How much energy in kJ is needed to change 150grams of ice from 0 o C to 50 o C? Example #5 -How much energy in kJ is needed to change 150grams of ice from 0 o C to 50 o C? This problem requires two steps. Since water is solid ice at 0 o C, we need to melt the ice and then heat it up to 50 o C. Step 1 – Calculate heat required to melt 150grams ice Step 2 - Calculate heat required to heat liquid water from 0 o C to 50 o C q = mC T = (150g)(4.184 J/g o C)(50 o C) = 31380 J  convert to kJ = 31.38kJ 150g 1 mole 6.0 kJ = 50 kJ 18grams 1 mole *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 Use Molar Heat constants Melting use q = (moles) x (H fusion ) Vaporize use q = (moles) x (H Vaporization ) No Phase Change Use specific heat capacity Use specific heat capacity q = (mass) ( C p ) ( ΔT ) q = (mass) ( C p ) ( ΔT )