1. 1 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Chapter 3 Matter and Energy 3.5 Specific Heat

2. 2 Specific heat Is different for different substances. Is the amount of heat that raises the temperature of 1 g of a substance by 1°C. In the SI system has units of J/g  C. In the metric system has units of cal/g  C. Specific Heat

3. 3 Examples of Specific Heats Table 3.7 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

4. 4 Learning Check What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C?

5. 5 Solution What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C? Given: 24.8 g, 275 J, 20.2  C to 24.5  C Need: J/g  C Plan: SH = Heat/g  C ΔT = 24.5  C – 20.2  C = 4.3  C SH Equation: SH = heat (q) (mass)(  T) Set Up: 275 J = 2.6 J/g  C (24.8 g)(4.3  C)

6. 6 Rearranging the specific heat expression gives the heat equation. Heat(q) = g x °C x J = J g°C The amount of heat lost or gained by a substance is calculated from the Mass of substance (g). Temperature change (  T). Specific heat of the substance (J/g°C). Heat Equation

7. 7 A layer of copper on a pan has a mass of 135 g. How much heat in joules will raise the temperature of the copper from 26°C to 328°C if the specific heat of copper is 0.385 J/g°C? The temperature change is 328°C - 26°C = 302°C. heat (cal) = g x  T x SH(Cu) 135 g x 302°C x 0.385 J g°C = 15 700 J or 1.57 x 10 4 J Using Specific Heat

8. 8 How many kilojoules are needed to raise the temperature of 325 g of water from 15.0°C to 77.0°C? 1) 20.4 kJ 2) 77.7 kJ 3) 84.3 kJ Learning Check

9. 9 How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 3) 84.3 kJ 77.0°C – 15.0°C = 62.0°C 325 g x 62.0°C x 4.184 J x 1 kJ g °C 1000 J = 84.3 kJ Solution

10. 10 Calculating Mass Aluminum is used to make kitchen utensils. What is the mass of an aluminum spatula if 3.25 kJ of heat raise its temperature from 20.0°C to 45.0°C. SH Al = 0. 897 J/g°C? Given: 3.25 kJ (3250 J), 20.0°C to 45.0°C ΔT = 25.0°C Plan: Solve heat equation for mass m = heat ΔT x SH Set Up: 3250 J/g°C = 145 g Al 25.0°C x 0.897 J

11. 11 Transferring Heat Energy Heat energy Flows from a warmer object to a colder object. Provides kinetic energy for the colder object. Lost by the warmer object is equal to the heat energy gained by the colder object.

12. 12 Calorimeters and Heat Transfer A calorimeter Is used to measure heat transfer. Can be made with a coffee cup, water, and a thermometer. Indicates the heat lost by a sample and gained by water. Heat lost (-q) = Heat (q) gained Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

13. 13 Measuring Heat Changes A 50.0-g sample of tin is heated to 99.8°C and dropped into 50.0 g water at 15.6°C. If the final temperature is 19.8°C, what is the specific heat of tin? Heat gain (q) by water = 50.0 g x 4.2°C x 4.184 J/g°C = 880 J Heat loss (-q) by tin = -880 J SH tin = -880 J = 0.22 J/g°C (50.0 g)(-80.0°C)

14. 14 Energy and Nutrition On food labels, energy is shown as the nutritional Calorie, written with a capital C. In countries other than the U.S., energy is shown in kilojoules (kJ). 1 Cal =1000 cal 1 Cal = 1 kcal 1 Cal = 4184 J 1 Cal = 4.184 kJ

15. 15 Caloric Food Values The caloric or energy values for 1 g of a food is given in kJ or kcal (Cal) Table 3.8 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

16. 16 Energy Values for Some Foods Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 3.9

17. 17 Energy Requirements The amount of energy needed each day depends on Age Sex Physical activity Table 3.11 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

18. 18 A cup of whole milk contains 12 g carbohydrate, 9.0 g fat, and 9.0 g protein. How many kcal (Cal) does a cup of milk contain? 1) 48 kcal (48 Cal) 2) 81 kcal (81 Cal) 3) 165 kcal (165 Cal) Learning Check

19. 19 3) 165 kcal 12 g carbohydrate x 4 kcal/g = 48 kcal 9.0 g fat x 9 kcal/g=81 kcal 9.0 g protein x 4 kcal/g=36 kcal Total kcal= 165 kcal = 165 Cal Solution