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Unit 2 – Matter and Energy

Unit 2 – Matter and Energy. Mrs. Callender. Lesson Essential Question. How do I calculate energy in a system?. Hydrocarbons and Heat. Most hydrocarbons are used as fuels. Knowing how much energy a fuel provides can tell us if it is useful for a certain application.

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Unit 2 – Matter and Energy

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  1. Unit 2 – Matter and Energy Mrs. Callender

  2. Lesson Essential Question How do I calculate energy in a system?

  3. Hydrocarbons and Heat Most hydrocarbons are used as fuels. Knowing how much energy a fuel provides can tell us if it is useful for a certain application. For example, the amount of energy a food releases when burned, can tell us about it’s caloric content (fats release lots of energy).

  4. Bomb Calorimeter Measures heat released during combustion. Combustion Equation for a peanut: C57H104O6 + 80 O2 ---> 57 CO2 + 52 H2O + energy

  5. Calories and Joules 1 Calorie (food) = 4200 joules 1000 calories = 1 dietary Calorie

  6. Food Energy Problem The fuel value of peanuts is 25 KJ/g. If an average adult needs 2800 kilocalories of energy a day, what mass of peanuts would meet an average adult’s energy needs for the day? Assume all of the fuel value of the peanuts can be converted to useful energy. 1 KJ 1 g 2800 Kcalories 4200 J 1 Calorie X X X X day 1 Calorie 1 Kcalories 1000 J 25 KJ = 470 g of peanuts

  7. Law of Conservation of Energy Reminder: Energy Gained = Energy Lost in a system q gained = q lost

  8. Sample Problem #1 If a piece of aluminum with mass 3.90 g and a temperature of 99.3 oC is dropped into 10.0 cm3 of water at 2.6 oC, what will be the final temperature of the system? (Recall the density of water is 1.00 g/cm3)

  9. System Example #1 LOSS GAINED m = 10.0 g m = 3.90 g Cp = 0.9025 J/goC Cp = 4.184 J/goC Ti = 22.6 oC Ti = 99.3 oC Tf = ? oC Tf = ? oC

  10. qloss = qgained mCpΔT = mCpΔT (3.90 g) (0.9025 J/goC) = (10.0 g) (4.184 J/goC) (Tf - 22.6) (99.3 - Tf) = (41.84 ) (Tf - 22.6) (3.51975) (99.3 - Tf) = 41.84 Tf– 945.584 349.511175 - 3.51975 Tf + 945.584 + 945.584 = 41.84 Tf - 3.51975 Tf 1295.095175 + 3.51975 Tf + 3.51975 Tf = 45.35975 Tf 1295.095175 45.35975 45.35975 = 28.6 oC Tf

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