Activity Instructions

1. Activity Instructions • Cut up Scenarios • Separate the Exothermic and Endothermic Scenarios (Check with Miss Mahamad) • Glue the Scenarios onto the paper • Identify the System and Surroundings

2. System and Surroundings • Energy can be exchanged between system and surroundings • System: reaction or process being studied • Surroundings: Everything else around the system; environments, world, universe

3. Endothermic vs. Exothermic Rxns • Endothermic = energy is absorbed • Heat Flows from _________ to _________ • Exothermic = energy is released • Heat Flows from _________ to __________ SURR SYSTEM SYSTEM SURR

4. Unit 6 Notes #2:Heat Calculations (Q=mCΔT)

5. Unit 6 Notes #2:Specific Heat Capacity Chem

6. Focus Question: • Now we know that all reactions either release or absorb heat energy from chemical bonds… • How can we as scientists calculate how much energy is being released or absorbed from reactions?

7. How do we measure heat energy? 1. Joules (J) and 2. calories (cal) or kcal = Cal • Interesting Fact: Food is reported in Calories (kcal) because 1 Calorie = 1000 calories

9. Heat Unit Conversions 1 calorie = 4.2 Joule 1kcal = 1000 cal 1 kJ = 1000 joules Often, we will use kilojoules instead of joules • Examples: • Convert 42 Joules = ___calories • Convert 50 calories = ____Joules • Convert 450 kJ = ______Joules • Convert 1000 joules =______ calories =10 cal =210 J

10. Specific Heat Capacity (Cp) • Scientific Definition: The amount of heat required to raise the temperature of 1g of a substance by 1°C. • Easier Definition: A substance’s ability to increase/decrease in temperature: • Large Cp = Hard to heat up/cool down • Small Cp = Easy to heat up / cool down

11. Enthalpy • Enthalpy:heat energy • ∆T = change in temperature/enthalpy Exothermic = -∆T Endothermic = +∆T

12. Which would you rather sit on on a hot day?

14. Explain: You put your hand into the hot oven to remove a batch of cookies and burn your fingers on the pan. The air in the oven is the same temperature as the pan, so why doesn’t it burn your fingers?

16. Explain: • You can bite into the crust of a hot piece of pizza, but as soon as you bite into the cheese, you burn your tongue.

18. Check for Understanding: • Which has the higher specific heat capacity? 1.water 2. metal • Substance1has cp =10, substance 2 has cp =1. Which will heat up more quickly?

19. Specific Heat Capacities of Common Substances Substance Specific Heat Capacity Water (l) liquid 4.18 Aluminum (s) 0.89 Iron (s) 0.45 Mercury (l) 0.14 Carbon (s) 0.73 Silver (s) 0.24 Gold (s) 0.13 Which substance will heat up the fastest? Slowest?

20. Calculating the Heat of Reaction Change In Temp Q = m× Cp × ∆T Heat mass Joules (J) Specific Heat Celsius (°C) Grams (g) grams/Joules deg C (J/g°C)

21. Calculating the Heat of Reaction Change In Temp Q = m× Cp × ∆T Heat mass Specific Heat

22. Q = m× Cp × ∆T • How much energy is required to raise the temperature of 10g of water from 10oC to 15oC? (Cp water = 4.18 J/goC) Practice 1 = 209 J

23. Q = m× Cp × ∆T • How much energy is required to raise the temperature of 10g of water from 10oC to 15oC? (Cp water = 4.18 J/goC) Practice 1 = 209 J

24. Q = m× Cp × ∆T Practice 2 How much total heat is required to raise the temperature of 50g of iron from 10oC to 80 oC? (Cp iron = 0.45 J/goC) 1575 J

25. Q = m× Cp × ∆T Practice 2 How much total heat is required to raise the temperature of 50g of iron from 10oC to 80 oC? (Cp iron = 0.45 J/goC) 1575 J

26. Practice Problem Answers Practice 3 Cp = 0.9 J/goC Practice 4 Cp= = 0.4 J/goC Practice 5 m = 44.44 grams of Iron =44.44 grams of Iron

27. Q = m× Cp × ∆T Practice 3: • A 11 gram sample of pure metal requires 100 J of energy to change its temperature from 20oC to 30oC. Calculate the specific heat and identify the metal. Water (l) liquid 4.18 Water (s) ice 2.03 Water (g) steam 2.0 Aluminum (s) 0.90 Iron (s) 0.45 Mercury (l) 0.14 Carbon (s) 0.73 Silver (s) 0.24 Gold (s) 0.13 Cp = 0.9 J/goC

28. Specific Heat Capacities of Common Substances Substance Specific Heat Capacity Water (l) liquid 4.18 Water (s) ice 2.03 Water (g) steam 2.0 Aluminum (s) 0.90 Iron (s) 0.45 Mercury (l) 0.14 Carbon (s) 0.73 Silver (s) 0.24 Gold (s) 0.13

29. Practice 4 The temperature of a piece of metal with a mass of 100g increases from 25.0oC to 45.0oC when the metal absorbs 800J of heat. What is the specific heat of copper? = 0.4 J/goC

30. Practice 5 A sample of iron requires 500 J to raise its temperature from 25.0 oC to 50.0 oC. What must the mass of the sample be? (Cp for iron = 0.45 J/goC) =44.44 grams of Iron

31. Practice 6 • Suppose 100.0g of H2O releases 4200J of heat. What is the corresponding temperature change (Cp= 4.2 J/goC)? =10oC

32. Practice 7 How many joules of heat energy are required to lower the temperature of 25 g of aluminum by 100oC? (cp = 0.90J/goC) =2250 J

33. Do Now: Convert 6.3 Joules to calories Remember: 1 calorie = 4.2 Joules HW: Calculating Heat Wst Due Wednesday

34. Objective What: Students know energy is released when a material condenses or freezes and is absorbed when a material evaporates or melts. How: Verbally explain how to solve for heat from phase changes with 100%. Why: To be critical thinkers of how heat is involved with phase changes.

35. Practice 6 Suppose 100.0g of H2O releases 4200J of heat. What is the corresponding temperature change (Cp= 4.2 J/goC)? Practice 7 How many joules of heat energy are required to lower the temperature of 25 g of aluminum by 100oC? (cp = 0.90J/goC)