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Energy Ability to “do work” or produce a change. Forms: potential energy and kinetic energy Types: mechanical, electrical, nuclear, solar, chemical, etc.

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Presentation on theme: "Energy Ability to “do work” or produce a change. Forms: potential energy and kinetic energy Types: mechanical, electrical, nuclear, solar, chemical, etc."— Presentation transcript:

1 Energy Ability to “do work” or produce a change. Forms: potential energy and kinetic energy Types: mechanical, electrical, nuclear, solar, chemical, etc.

2 Potential Energy -”stored” energy…due to position or composition. Chemical Energy: Energy stored in a substance due to its composition Types of atoms in substance Number and types of bonds Particular way the atoms are arranged

3 Kinetic Energy Energy of motion : Temperature – measures the average KE of the molecules in a sample. ALL matter is in constant motion and therefore has kinetic energy. How do gases, liquids, and solids compare?

4 Law Conservation of Energy In any chemical reaction or physical process, energy can be converted from one form to another, but it cannot be created nor destroyed. Food energy (sugar)  movement H 2 O and CO 2 and light  sugar (C 6 H 12 O 6 ) (C 6 H 12 O 6 )  H 2 O and CO 2 and (energy) Cell Respiration: Photosynthesis: in chloroplasts

5 Heat Energy Heat: “q” or “Q” calorie: amount of energy needed to raise the temperature of 1 gram H 2 O 1 ° C. 1000 cal = 1 kcal Calorie: nutritional or “food” calorie. (1 Cal = 1 kcal = 1000 cal ) Energy flowing from a warmer object to a cooler object. + Q means heat is “added” -Q means heat is “lost”

6 Converting Energy Units Energy Unit Conversions 1 Joule (J) = 0.2390 cal 1 calorie (cal) = 4.184J 1 kilojoule (KJ) = 1000 J 1 Calorie = 1 kcal (nutritional calorie) 1 kcal = 1000 cal

7 Nutritional Calorie A typical breakfast contains 230 nutritional Calories. Convert this to regular calories and Joules. 230 Cal x 1000 cal = 230,000 cal 1Cal 230,000 cal x 4.184 J = 960,000 J 1 cal

8 Specific Heat (C p ) The amount of energy required to raise the temperature of 1 g of ANY substance 1 ºC Measures a substances resistance to temperature change High C p = does NOT heat/cool rapidly Large Q change, small T change Low C p = heats/cools very quickly A little Q changes T a lot!!!

9 Specific Heat (C p ) Higher C p Lower C p Water in the gulf Sand on the beach GrassSidewalk Pizza CrustSauce ClothMetal Buckle

10 Specific Heat (C p ) Formula: Q = m · C p · ΔT Q = heat in Joules M = mass in grams C p = specific heat in J/g · ºC ΔT = change in temp (T final – T initial )

11 Ex: 10.0 g iron changed from 50.4 ºC to 25.0 ºC with a release of 114 J heat. Calculate C p.

12 Given: m = 10.0gWant: C = ? Ti = 50.4 ºC Tf = 25.0 ºC So ΔT = 25 – 50.4 = -25.4 ºC Q = -114 J (negative because it’s lost) Equ: Q = m C ΔT and C = Q m ΔT C = -114J = 0.449 J/g ºC (10.0g)(-25.4 ºC ) (Specific Heat is ALWAYS a (+) number!!)

13 If the temperature of 25g granite, which has a C p of 0.803 J/g · ºC is heated from 25 ºC to 55ºC, how much heat did the granite absorb?

14 If the temperature of 25g granite, which has a C p of 0.803 J/g · ºC is heated from 25 ºC to 55ºC, how much heat did the granite absorb? Given: m = 25gWant: Q = ? C = 0.803J/g ºC (expect + value) Ti = 25 ºC Tf = 55 ºC So ΔT = 55 – 25 = 30. ºC Equ: Q = m · C p ·ΔT Q = (25g)(0.803J/g ºC) (30. ºC) = 602J

15 A piece of metal with a mass of 4.68g absorbs 25 J of heat when its temperature increases by 182 °C. What is the specific heat of the metal?

16 Given: m = 4.68g Want: C = ? Q = 25 J (added) ΔT = 182 °C Equ: Q = m · C p ·ΔT and C = Q m ΔT C = 25 J = 0.029J/g °C (4.68g)(182 °C)


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