Download Presentation
## CH104 Chapter 2: Energy & Matter

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**CH104 Chapter 2: Energy & Matter**Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves**Chapter 2 Energy and Matter**2.2Temperature**Units of Measurement**meter (m) 1 m = 1.09 yd liter (L) 1 L = 1.06 qt gram (g) 1 kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273**Temperature**180o 100o**Learning Check**A. What is the temperature of freezing water? 1) 0 °F 2) 0 °C 3) 0 K B. What is the temperature of boiling water? 1) 100 °F 2) 32 °F 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 100 2) 180 3) 273**Solution**A. What is the temperature of freezing water? 1) 0 °F 2) 0 °C 3) 0 K B. What is the temperature of boiling water? 1) 100 °F 2) 32 °F 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 100 2) 180 3) 273**Trivial Pursuit Question**• At what temperature does oF = oC? - 40 oF = - 40 oC - 40 o**Temperature conversion**• Common scales used • Fahrenheit,CelsiusandKelvin. oF = 1.8 oC + 32 oC = (oF - 32) 1.8 K = oC + 273 SI unit**Solving a Temperature Problem**A person with hypothermia has a body temperature of 34.8 °C. What Is that temperature in °F? F = 1.8(C) + 32 ° F = (1.8)(34.8 °C) + 32 ° exact tenth’s exact = 62.6 ° + 32 ° = 94.6 °F tenth’s**Practice: Temp Conversion**• What is 75.0 º F in ºC? • ºC = (75.0 º F -32 º) = 23.9 ºC 1.8 • What is -12 º F in ºC? • º F = 1.8 (-12) + 32 º F = 10 º C**Learning Check**The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C**Solution**The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C TC = TF – 32 ° 1.8 = (105.8 – 32 °) 1.8 = 73.8 °F = 41.0 °C 1.8 °tenth’s place**Learning Check**A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale? 1) 423 °C 2) 235 °C 3) 221 °C**Solution**A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale? 1) 423 °C 2) 235 °C 3) 221 °C TF – 32 ° = TC 1.8 (455 – 32 °) = 235 °C 1.8one’s place**Learning Check**On a cold winter day, the temperature is –15 °C. What is that temperature in °F? 1) 19 °F 2) 59 °F 3) 5 °F**Solution**On a cold winter day, the temperature is –15 °C. What is that temperature in °F? 1) 19 °F 2) 59 °F 3) 5 °F TF = 1.8TC + 32 ° TF = 1.8(–15 °C) + 32 ° = – 27 + 32 ° = 5 °F one’s place Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign. 1.8 x 15 +/ – = –27**Learning Check**What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K**Solution**What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K TK = TC + 273 = 37 °C + 273 = 310. K one’s place**Chapter 2 Energy and Matter**2.1Energy**Energy**Energy = The capacity to cause change Heat Light Wind**Potential Energy = stored Energy**(Has potential for motion) X Kinetic Energy = Energy in motion (Fulfilling its potential)**Learning Check**Identify each of the following as potential energy or kinetic energy. A. roller blading B. a peanut butter and jelly sandwich C. mowing the lawn D. gasoline in the gas tank**Solution**Identify each of the following as potential energy or kinetic energy. A. roller blading kinetic B. a peanut butter and jelly sandwich potential C. mowing the lawn kinetic D. gasoline in the gas tank potential**Kinetic Energy**KE = 1mv2 2 Which has more E? • Truck moving at 5 mph • Bicycle moving at 5 mph**Units of Measurement**meter (m) 1 m = 1.09 yd liter (L) 1 L = 1.06 qt gram (g) 1 kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273 calorie (cal) 1Kcal = 1000 cal = 1Cal Joule (J) 1 cal = 4.18 J**Learning Check**How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized?**Solution**How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized? Given Need 150 J 1 cal 4.184 J = cal 36 Conversion Factors: 1 cal and 4.184 J 4.184 J 1 cal**Chapter 2 Energy and Matter**2.3Specific Heat**Energy**Units of Energy = calorie cal kilocalorie kcal 1000 cal = 1 kcal Calorie Cal Cal = kcal joule J 4.18 J = 1 cal British Thermal Unit BTU**H2O**Energy • calorie: • E to raise 1 g H20 by 1 oC Specific Heat 1 cal 1g 1oC 1oC**Au**0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Add 1 cal Specific Heat E to raise Temp of 1g substance by 1 oC cal g oC 0oC = start**Au**0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Specific Heat E to raise Temp of 1g substance by 1 oC 30o cal g oC 20o 10o 1o Add 1 cal 0oC = start**Au**0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Specific Heat 30o Low SpHt; Heats quickly 20o 10o 1o High SpHt; Resists change**Sand**0.19 H2O 1.00 Specific Heat Dehydrated person Body temp rises quickly Hydrated person Resists change in body temp Heats quickly Gets hot Resists change Stays cold**Learning Check**A. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day and cool at night. Sand must have a 1) high specific heat 2) low specific heat**Solution**A. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day and cool at night. Sand must have a 1) high specific heat 2) low specific heat**24.8g**Specific Heat Sample Problem: What is the specific heat of a metal if 24.8 g absorbs 65.7 cal of energy and the temp rises from 20.2 C to 24.5 C? 65.7 0.62 cal g oC = cal 24.8g 4.3oC = cal 1g 1oC SpHt = 24.5oC DT = 4.3 Co m = 20.2oC**50g H2O**Specific Heat E = m DT SpHt Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? 87oC DT = 12 Co m = 75oC 1 cal 1g 1oC SpHt =**50g H2O**Specific Heat E = m DT SpHt Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? m DT SpHt 12 Co 1 cal 1g 1oC = cal to heat water 600**750g H2O**Specific Heat E = m DT SpHt Sample Problem: A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? 65oC DT = 28 Co m = 37oC 1 cal 1g 1oC SpHt =**750g H2O**Specific Heat E = m DT SpHt Sample Problem: A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? m DT SpHt 28 Co 1 cal 1g 1oC = cal from cool water 21000**Learning Check**How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C?**325g H2O**Solution E = m DT SpHt How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C? m DT SpHt = KJ 84.3 62 Co 4.184 J 1g 1oC 1 KJ 1000 J 77oC DT = 62 Co 15oC**Chapter 2 Energy and Matter**2.4Energy and Nutrition 1 Cal = 1000 calories 1 Cal = 1 kcal 1 Cal = 4184 J 1 Cal = 4.184 kJ**Energy**Units of Energy = 1 Cal 1000 cal = 1 kcal 1 cal = 4.18 J 1 Cal 4184 J = 4.184 kJ**Calorimeters**A calorimeter • A reaction chamber & thermometer in H2O used to measure heat transfer • indicates the heat lost by a sample • indicates the heat gained by water**Energy from Food**4 kcal g 17 kJ g Carbohydrate 9 kcal g 38 kcal g Fat Protein 4 kcal g 17 kcal g**Energy from Food**Sample Problem: How much energy in kcal (= Cal) would be obtained from 2 Tbl peanut butter containing 6 g carb, 16 g fat, & 7 g protein? 6 g carb 4 kcal 1 g carb = 24 kcal = 196 kcal = 196 Cal 16 g fat 9 kcal 1 g fat = 144 kcal = 200 Cal 7 g protein 4 kcal 1 g protein = 28 kcal