1. Unit 12 Study session

2. Practice Specific heat capacity (J/g o C) Initial temperature Final temperature A1540 B1550 C1560 Which metal has the highest specific heat capacity? Why?

3. Practice Specific heat capacity (J/g o C) Initial temperature Final temperature A1540 B1550 C1560 Which metal has the highest specific heat capacity? A, because it is lowest temperature change.

4. Practice Specific heat capacity (J/g o C) Initial temperature Specific heat capacity D200.5 E200.8 F201.2 Which metal has the highest final temperature when you added 100J? Why?

5. Practice Specific heat capacity (J/g o C) Initial temperature Specific heat capacity D200.5 E200.8 F201.2 Which metal has the highest final temperature when you added 100J? D, because it is lowest specific heat capacity.

6. FYI Units for thermochemistry a.Heat/ energy i.Joule ii.Calorie b.Temperature i. O C ii.K c.Specific heat i.J/g o C ii.Cal/g o C

7. 5. When a student dissolved NH 4 NO 3 in water, the solution became colder. What kind of reaction can observe? Endothermic reaction because the solution got cold and system absorbed heat.

8. 5. When a student put hot metal in water, the solution became warmer. What kind of reaction can observe? Exothermic reaction because the solution got warmer and system released heat.

9. Practice system/surrounding Is this an exo or endo? Solution gets hot or cold? Heat will be absorbed by the surroundings

10. 2. Endothermic/Exothermic Endo: absorb heat/∆H=positive/reactant Exo: release heat/ ∆H=negative/product solid liquid gas endo exo

11. FYI Use stoichiometry Consider the equation: 2CH 3 OH (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O (g) + 1500kJ How many grams of CH 3 OH must react to form 15kJ of heat? grammoleheat Energy in an equation  use mole ratio

12. FYI Use stoichiometry Consider the equation: 2CH 3 OH (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O (g) ∆H=-1500kJ How many moles of CH 3 OH must react to form 15kJ of heat? grammoleheat ∆H with kJ/mol  use kJ to 1 mole ratio

13. Consider the equation: 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O (l) ∆H = -1.066x10 4 kJ a.Is this endo or exo? b.exo c.Rewrite the chemical equation including heat. 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O (l) + 1.066x10 4 kJ

14. c. How many joules of heat are produced when 75.0 g of octane (C 8 H 18 ) is burned? 75.0g C 8 H 18 x 1 mol C 8 H 18 x – 1.066x10 4 kJ x 1000J 114.26 g C 8 H 18 2 mol 1kJ = -3.50x10 6 J C 8 H 18 Consider the equation: 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O (l) ∆H = -1.066x10 4 kJ

15. d. How many moles of octane(C 8 H 18 ) must react to form 750 kJ of heat? -750 kJ x 2 mol C 8 H 18 -1.066x10 4 kJ = 0.14 mol C 8 H 18 Consider the equation: 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O (l) ∆H = -1.066x10 4 kJ

16. FYI q= mc∆T 1.q is the amount of heat transferred from an object 2.m is the mass of the object 3.c is the specific heat of the object 4.∆T is the temperature difference of the object (the change in temperature) (J or cal) (gram) (J/g o C) ( o C)

17. How to solve word problems 1.If the question contains either temperature or specific heat, you need to use q=mc∆T 2.If the question contains heat in joule or kilojoule, mole, or mass, you need to use stoichiometry

18. 8. How many joules of heat are needed to raise the temperature of 150 g of water from 25 o C to 45 o C? The specific heat capacity of water is 4.18 J/g o C. Including specific heat  q=mC∆T q=? m=150g c=4.18J/g o C ∆T=45 o C-25 o C=20 o C q=(150g)(4.18J/g o C)(20 o C) =13000J

19. 9.A 20.0g sample of iron was given 1500 calories of heat. What will be the change in temperature? The specific heat capacity of iron is 0.11 cal/g oC. Including specific heat  q=mC∆T q=1500cal m=20.0 c=0.11 cal/g o C ∆T=? o C 1500cal=(20.0g)(0.11cal/g o C)(∆T) ∆T=680 o C

21. 12. How to read the chart Mass of unknown metal 12.50g Mass of water 100.00g Initial temperature of water in calorimeter 15.8 o C Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C Calorimetry lab question needs q=mc∆T Focus on the same substance such as water or metal One is endo, another is exo Follow the questions

22. 12. How to read the chart Focus on the same substance such as water or metal What is ∆T of water? (T f – T i ) 16.5 o C – 15.8 o C = 0.7 o C Mass of unknown metal 12.50g Mass of water 100.00g Initial temperature of water in calorimeter 15.8 o C Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C

23. 12. How to read the chart Calculate the heat absorbed by the water (Use the previous question) Q=mc∆T= (100.00g) (4.184 J/g o C)(0.7 o C) = 293J Mass of unknown metal 12.50g Mass of water 100.00g Initial temperature of water in calorimeter 15.8 o C Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C

24. 12. How to read the chart What is the heat released by the unknown metal? qsurr= - qsys qsurr=293J qsys= - 293J Mass of unknown metal 12.50g Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C

25. 12. How to read the chart What is the ∆T of unknown metal? 16.5 o C – 76.5 o C = -60.0 o C Mass of unknown metal 12.50g Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C

26. 12. How to read the chart Calculate the specific heat of the metal (use previous question) Q=mc ∆T  -293J = (12.50g) x c x (-60.0 o C)  c= 0.39J/g o C Mass of unknown metal 12.50g Final temperature of water + unknown metal in calorimeter 16.5 o C Initial temperature of unknown metal in calorimeter 76.5 o C

27. 12. How to read the chart What is this metal? c= 0.39J/g o C SubstanceSpecific heat capacity (J/g o C) Al (s)0.90 Fe (s)0.45 Cu (s)0.39 Pb (s)0.13

28. How many joules of heat are needed to raise the temperature of 2260 g of water from 28 o C to 48 o C? The specific heat capacity of water is 4.0 J/g o C. How to solve the problem? Including specific heat  q=mC∆T q=? m=200.g c=4.0J/g o C ∆T=48 o C-28 o C=20 o C q=(200g)(4.0J/g o C)(48 o C-28 o C) =16,000J

29. Consider the equation: 2CH 3 OH (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O (g) ∆H = -1500kJ How many kilojoules of heat are produced when 3.20 g of methanol is burned? (molar mass of methanol is 32.05g/mol) How do you solve this problem? Stoich or q=mc∆T? 3.20g CH 3 OH x 1 mol CH 3 OH x – 1500kJ 32.05 g CH 3 OH 1 mol = -150 kJ

30. Practice The specific heat of copper is about 0.40 Joules/gram o C. How much heat in joules is needed to change the temperature of a 10.0 gram sample of copper from 20.0 o C to 60.0 o C? q m c ∆T Q=(10.0g)(0.40J/g o C)(60.0 o C-20.0 o C)