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Energy

Energy. Also Known As Thermochemistry. Units of Energy and Energy Computations Section 15.1. Joule = SI Calorie = English/US Big C is a food calorie or 1000 kcal Little c is a calorie, the amount of energy required to raise the temperature of one gram of water through one degree Celsius.

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Energy

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  1. Energy Also Known As Thermochemistry

  2. Units of Energy and Energy Computations Section 15.1 • Joule = SI • Calorie = English/US • Big C is a food calorie or 1000 kcal • Little c is a calorie, the amount of energy required to raise the temperature of one gram of water through one degree Celsius.

  3. Calorimetry • Dr. B is watching your calories. • How many calories does it take to raise the temperature of 1.0 gram of water from 20.3 o C to 24.3 o C? • Q = mΔTc where Q is the Quantity of Heat m is the mass of the substance ΔT is the temperature change c is the specific heat capacity of the substance

  4. How many calories does it take to raise the temperature of 1.0 gram of water from 20.3 o C to 24.3 o C? Q = mΔTc Joules • For water, its specific heat (c) is 4.184 J/g. o C • ΔT = 4.0 o C • m = 1.0 g • Q = 1.0g (4.0 o C) 4.184 J/g. o C = • 16.7 J Calories • For water, its specific heat (c) is 1.0 cal/g. o C • ΔT = 4.0 o C • m = 1.0 g • Q = 1.0g (4.0 o C) 1.0 cal/g. o C = • 4.0 cal

  5. Calorimeter Madason burns a chunk of food in a calorimeter that contains 50.0 g of water. Its temperature goes from 23.5 to 36.8 o C . What is Q? in Calories? Joules? Work it in your notes. Solutions follow.

  6. Problem Solution Joules • c = 4.184J/g. o C • ΔT = 36.8-23.5 = 13.3o C • m = 50.0 g • Q = 50.0g (13.3 o C) 4.184 J/g. o C = • 2782 J Calories • c = 1.0 cal/g. o C • ΔT = 36.8-23.5 = 13.3o C • m = 50.0 g • Q = 50.0g (13.3 o C) 1.0 cal/g. o C = • 665 cal

  7. Computing Specific Heat If 5.0 grams of copper cools from 35.0 o C to 22.6 o C and releases 23.6 Joules of heat, what is the specific heat of copper? • 0.076 J/g. o C • 3.8 x 102 J/g. o C • 0.38 J/g. o C • 0.62 J/g. o C • Same formula, different givens and unknowns: Q = mΔTc • Hint: In this problem, solve for c

  8. Solution Q = mΔTc Q = 23.6 Joules (given in problem) m = 5.0 g (given in problem) ΔT = 35.0-22.6 = 12.4o C (given in problem) c = Q/ mΔT c = 23.6 Joules /5.0g. 12.4o C = 0.38 J/g. o C

  9. Do you understand Specific Heat? • The amount of heat required to raise a given amount of substance through a given temperature change. Watch a SHC demonstration on YouTube: • http://www.youtube.com/watch?v=BclB8UaSH4g • Check out this link. 9 minutes.

  10. Test Yourself An equal amount of heat is added to a 10.0 g sample of each of the following metals (shown with its determined c or specific heat capacity). If each metal is initially at 20.0 o C, which metal will reach the highest temperature? • Beryllium 1.82 J/g. o C • Calcium 0.653 J/g. o C • Copper 0.385 J/g. o C • Gold 0.129 J/g. o C

  11. ANSWER, PLEASE? d. GOLD 0.129 J/g. o C

  12. Another Problem One calorie equals 4.184 J. How much energy in joules is supplied by a breakfast bar containing 170 food calories? • 170 J • 711 J • 1.7 x 105 J • 7.11 x 105 J

  13. Answer Please c. 1.7 x 105 J Remember that a food Calorie = 1000 calories or one kilocalorie Copy this next problem in your notebook:

  14. Last Problem* A 5.0 g piece of Aluminum at 100 o C is dropped into a 25.0 g of water in a calorimeter at 20.0 o C. What is the final temperature of the aluminum? *In this section

  15. Think of it as A LAW of Conservation of Energy Problem Q = mΔTc The heat lost by the metal can be computed using the Q formula: You know m = 5.0 g You know c = 0.897 J/g. o C You know ΔT = 100 – X Solve for X Here are the choices: • 16.4 o C • 23.6 o C • 96.4 o C • 103.6 o C

  16. Solution • Eliminate d • Why? You tell me • Problem is Q = mΔTc • You need both Q and ΔT • So you have 2 unknowns, you need 2 equations. • The heat energy gained by the water must equal the heat energy lost by the aluminum. Both will have the same final temperature. • Another way to phrase the Q is, what is the final temperature of the mixture?

  17. The LAW • For Aluminum • Q = mΔTc • m = 5.0 g • c = 0.897 J/g. o C • ΔT = 100 – X • For water • Q = mΔTc • m = 25.0 g • c = 4.184 J/g. o C • ΔT = X – 20 Notice that the temperature of the metal goes down but the water temperature goes up. They both will end at the same temperature so we can say that Q water = Q Aluminum or QAl = Q H2O

  18. And X is the final temperature of both • If QAl = Q H2O • Then mΔTc of the aluminum equals mΔTcof the water so • 5.0g (100-X o C) 0.897 J/g. o C = 25.0g (X-20.0 o C) 4.184 J/g. o C • Use algebra to solve for X Here are the choices again: • 16.4 o C • 23.6 o C • 96.4 o C • 103.6 o C

  19. Go Blue Devils!

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