1. Heating & Cooling Curves A STEP BY STEP PRACTICE PROBLEM © Mr. D. Scott; CHS

2. temperature added energy (mass)(molar mass)(ΔH fus ) q = sm  T SOLID MELT (freeze) LIQUID BOIL (condense) GAS Be careful with the units!!! q = smΔT calculations will be in Joules, but ΔH calucaltions will be in kJ….convert before adding them together © Mr. D. Scott; CHS The energy changes for each step are calculated separately and then the separate parts added together. q = sm  T q = sm  (mass)(molar mass)(ΔH vap )

3. 0 o C temperature added energy Start by planning! (how many steps are needed?) 77.0 o C q solid = ? J Melting ΔH = ? kJ © Mr. D. Scott; CHS -14.0 o C q liquid = ? J 0 o C Solve the three steps separately, then add them all together! How much heat is need to change 22.0 grams of water from -14.0 o C to 77.0 o C? (s ice = 2.03 J/g o C, ΔH fus = 6.02 kJ/mol, s water = 4.18 J/g o C) Problem #1

4. How much heat is need to change 22.0 grams of water from -14.0 o C to 77.0 o C? (s ice = 2.03 J/g o C, ΔH fus = 6.02 kJ/mol, s water = 4.18 J/g o C) q solid = ? J © Mr. D. Scott; CHS Problem #1 q liquid = ? J temperature added energy Next, calculate each step. Melting ΔH = ? kJ ΔH = (mass)(molar mass)(ΔH fus ) ΔH = (22.0 g) 1 mol H 2 O 6.02 kJ 18.02 g H 2 O 1 mol H 2 O Melting ΔH = 7.35 kJ q = s ice m ΔT q = (2.03 J/g o C) (22.0 g) (14.0 o C) q solid = 625 J q = s water m ΔT q = (4.18 J/g o C) (22.0 g) (77.0 o C) q liquid = 7,080 J

5. 0 o C temperature added energy Finally, add all the steps together! 77.0 o C q solid = 625 J Melting ΔH = 7.35 kJ © Mr. D. Scott; CHS -14.0 o C q liquid = 7,080 J 0 o C OH NO! They’re not all in the same units! Better convert first How much heat is need to change 22.0 grams of water from -14.0 o C to 77.0 o C? (s ice = 2.03 J/g o C, ΔH fus = 6.02 kJ/mol, s water = 4.18 J/g o C) Problem #1

6. q solid = 625 J 1 kJ. = 0.625 kJ 1000 J q liquid = 7,080 J 1 kJ. = 7.08 kJ 1000 J Melting ΔH = 7.35 kJ Total Energy Change = 15.06 kJ