1. Aim: Students will be able to measure the amount of heat released or absorbed in a chemical reaction. Do Now: What is the difference between potential energy and kinetic energy?

2. ENERGY Kinetic energy – energy of an object in motion. The more an object is moving, the more kinetic energy it has. Potential energy – stored energy. When an object remains still, its potential energy is increasing.

3. Kinetic energy The energy of motion Particles that move faster have greater kinetic energy What are some variables that would increase the kinetic energy of a reaction?

4. Kinetic Energy What makes this pot of still water change to this pot of boiling (full of motion) water?

5. The rise in temperature! When the temperature applied to an object increases, the kinetic energy of the particles in the object also increases. Particles move faster, thus colliding with other particles more. Does this mean particles stop moving when they collide?

6. No! Kinetic energy gets transferred from one particle to another when they collide.

7. What is a good way of measuring the kinetic energy of particles? By using a thermometer. When using a thermometer, you are measuring the average kinetic energy of the water particles.

8. Potential Energy

9. Why is potential energy important in chemistry? Energy stored in chemical bonds is potential energy. Potential energy in chemical bonds are the forces that hold atoms together in compounds.

10. Can you think of any real-world items that have potential energy?!

11. GASOLINE! Energy is stored in this fuel and is released when chemical reactions take place.

12. How do we measure energy? The amount of heat given off or absorbed in a reaction can be calculated using the following equation: q = mC∆T Energy/Heat is always measured in the unit JOULES

13. q = mC∆T q = the amount of heat absorbed or released. m = the mass of the substance C = specific heat capacity of the substance ∆T = the difference between the initial temperature and final temperature.

14. Some practice! How many joules are absorbed when 50.0 grams of water are heated from 30.2°C to 58.6°C?

15. Step 1: Write out the equation you are using. q = mC∆T

16. Step 2: Write out your KNOWN variables: m = 50.0 grams C (specific heat) of water = 4.18 J/g·°C (you can find this value in your reference tables) ∆T = initial temperature – final temperature = 58.6°C - 30.2°C = 28.4 °C

17. Step 3: Fill in the known variables into the equation for measuring the amount of heat absorbed: q = mC∆T q = (50.0 g)(4.18 J/g·°C )(28.4 °C) q = 5936 J So, the amount of energy absorbed is 5936 Joules.

18. Try this on your own When 25.0 grams of water are cooled from 20.0°C to 10.0°C, the number of joules of heat energy released is__________?

19. Remember the steps! q = mC∆T – m = 25.0 g – C = 4.18 J/g·°C – ∆T = 20.0°C - 10.0°C = 10.0°C q = (25.0 g)(4.18 J/g·°C )(10.0°C) q = 1045 Joules of heat were released when 25.0 grams of water were cooled from 20.0°C to 10.0°C

20. Special Situations Recall what we learned about heating curves. At what point does melting occur? At what point does boiling occur? Is the temperature changing at these points in the curves?

21. Special Situations If the temperature is not increasing or decreasing, what can be said about it?

22. Special Situations If the temperature is not increasing or decreasing, what can be said about it? Temperature remains constant!

23. Special Situations During melting and boiling phase changes, temperature is CONSTANT and therefore does not need to be accounted for in the formula for measuring heat. q = mC∆T

24. Remember these KEY FACTS Heat of Fusion = melting Heat of Vaporization = boiling

25. Heat of Fusion Look at the front page of your reference tables! There is a value called Heat of Fusion for water. Heat of Fusion = H f = 334 J/g

26. Heat of Fusion There is a value called Heat of Fusion for water (334 J). You simply substitute this value for the “C” variable in the equation q = mC (Note: The “∆T” part of the equation has been removed because temperature remains constant during melting and boiling and need not be accounted for)

27. Heat of Fusion Example: When 20.0 g of a substance are completely melted at its melting point, 3444 J are absorbed. What is the heat of fusion of this substance?

28. Heat of Fusion Write your equation: q = mC Replace known variables with values given: q = m x H f 3444 J = (20.0 g) (H f ) Divide both sides by 20.0 g in order to find H f

29. Heat of Fusion 3444 J = (20.0 g) (H f ) 20.0 g 20.0 g 172.2 J/g = H f

30. Heat of Vaporization Once again, look at your reference tables. On the first page where you found the value for the Heat of Fusion, you will also find a value for Heat of Vaporization of water. Heat of vaporization = H v = 2260 J/g *Remember – during boiling phase, temperature remains constant and is once again eliminated from the equation.

31. Heat of Vaporization Example: The heat of vaporization of a liquid is 1344 J/g. What is the minimum number of joules needed to change 40.0 g of the liquid to vapor at the boiling point?

32. How to solve: Write equation: q = m x H v

33. How to solve: Replace variables with known values: q = m x H v q = (40.0 g) (1344 J/g) q = 53,800 J

34. Homework In Regents Review Workbook: Page 66 – Answer questions 38-46 ***Remember Midterm Review is everyday from 2:30-4:00 pm