1. Chapter 3 Lecture Basic Chemistry Fourth Edition Chapter 3 Matter and Energy 3.5 Specific Heat Learning Goal Calculate specific heat. © 2014 Pearson Education, Inc.

2. Specific heat is different for different substances is the amount of heat, in joules or calories, needed to change 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 © 2014 Pearson Education, Inc. Specific Heat

3. © 2014 Pearson Education, Inc. Calculating Specific Heat To calculate the specific heat (SH) of a substance we measure the heat (q) in joules (J), the mass (m) in grams, and the temperature change, which is written as ΔT. For example, the specific heat of water is

4. © 2014 Pearson Education, Inc. Specific Heat of Some Substances

5. © 2014 Pearson Education, Inc. A large mass of water near a coastal city can absorb or release five times the energy absorbed or released by the same amount of rock near an inland city. Specific Heat of Liquid Water

6. © 2014 Pearson Education, Inc. Guide to Calculating Specific Heat

7. © 2014 Pearson Education, Inc. What is the specific heat of a metal if 24.8 g of the metal absorbs 275 J of energy and the temperature rises from 20.2 ° C to 24.5 ° C? Learning Check

8. © 2014 Pearson Education, Inc. Solution What is the specific heat of a metal if 24.8 g of the metal absorbs 275 J of energy and the temperature rises from 20.2 ° C to 24.5 ° C? Step 1 Given 24.8 g, 275 J, ΔT = 4.3 ° C Need J/g ° C Step 2 Write the relationship for specific heat.

9. © 2014 Pearson Education, Inc. Solution What is the specific heat of a metal if 24.8 g of the metal absorbs 275 J of energy and the temperature rises from 20.2 ° C to 24.5 ° C? Step 3 Substitute the given values into equation.

10. © 2014 Pearson Education, Inc. Heat Equation from Specific Heat Rearranging the specific heat expression gives the heat equation.

11. © 2014 Pearson Education, Inc. Heat Equation from Specific Heat The amount of heat lost or gained by a substance is calculated from the following equation q = m × ΔT × SH where m represents the mass of substance (g) the temperature change is represented by (ΔT) the specific heat, (SH) of the substance is given in units of (J/g ° C)

12. © 2014 Pearson Education, Inc. Guide to Calculations Using Specific Heat

13. © 2014 Pearson Education, Inc. A layer of copper (Cu) on a pan has a mass of 135 g. How much heat (kJ) is needed to raise the temperature of the copper from 26 ° C to 328 ° C? (The specific heat of Cu is 0.385 J/g ° C.) Step 1Given135 g, 26 ° C to 328 ° C, SH = 0.385 J/g ° C Needjoules Sample Problem, Using SH

14. © 2014 Pearson Education, Inc. Sample Problem, Using SH A layer of copper (Cu) on a pan has a mass of 135 g. How much heat (kJ) is needed to raise the temperature of the copper from 26 ° C to 328 ° C? (The specific heat of Cu is 0.385 J/g ° C.) Step 2 Calculate the temperature change. ΔT = 328 ° C – 26 ° C = 302 ° C Step 3 Write the heat equation. q = m × ΔT × SH(Cu)

15. © 2014 Pearson Education, Inc. Sample Problem, Using SH A layer of copper (Cu) on a pan has a mass of 135 g. How much heat (kJ) is needed to raise the temperature of the copper from 26 ° C to 328 ° C? (The specific heat of Cu is 0.385 J/g ° C.) Step 4 Substitute in values and solve.

16. © 2014 Pearson Education, Inc. Learning Check How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 ° C to 77.0 ° C? A. 20.4 kJ B. 77.7 kJ C. 84.3 kJ

17. © 2014 Pearson Education, Inc. Solution How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 ° C to 77.0 ° C? C. 84.3 kJ Step 1Given325 g water SH(water) = 4.184 J/g ° C 15.0 ° C to 77.0 ° C Need kilojoules

18. © 2014 Pearson Education, Inc. Solution How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 ° C to 77.0 ° C? C. 84.3 kJ Step 2 Calculate the temperature change. ΔT = 77.0 ° C – 15.0 ° C = 62.0 ° C Step 3 Write the heat equation. q = m × ΔT × SH(water)

19. © 2014 Pearson Education, Inc. Solution How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 ° C to 77.0 ° C? C. 84.3 kJ Step 4 Substitute in values and solve.

20. © 2014 Pearson Education, Inc. Calculating Heat Loss A 225-g sample of hot tea cools from 74.6 ° C to 22.4 ° C. How much heat, in kilojoules, is lost, assuming that tea has the same specific heat as water? Step 1Given225 g tea (water) SH(water) = 4.184 J/g ° C 74.6 ° C to 22.4 ° C Needkilojoules

21. © 2014 Pearson Education, Inc. Calculating Heat Loss A 225-g sample of hot tea cools from 74.6 ° C to 22.4 ° C. How much heat, in kilojoules, is lost, assuming that tea has the same specific heat as water? Step 2 Calculate the temperature change. ΔT = 74.6 ° C – 22.4 ° C = 52.2 ° C Step 3Write the heat equation. q = m × ΔT × SH(water)

22. © 2014 Pearson Education, Inc. Calculating Heat Loss A 225-g sample of hot tea cools from 74.6 ° C to 22.4 ° C. How much heat, in kilojoules, is lost, assuming that tea has the same specific heat as water? Step 4 Substitute in values and solve.

23. © 2014 Pearson Education, Inc. Learning Check When 8.81 kJ is absorbed by a piece of iron, its temperature rises from 15 ° C to 122 ° C. What is its mass, in grams?

24. © 2014 Pearson Education, Inc. Solution When 8.81 kJ is absorbed by a piece of iron, its temperature rises from 15 ° C to 122 ° C. What is its mass, in grams? Step 1Given8.81 kJ SH(iron) = 0.452 J/g ° C 15 ° C to 122 ° C Needmass of iron

25. © 2014 Pearson Education, Inc. Solution When 8.81 kJ is absorbed by a piece of iron, its temperature rises from 15 ° C to 122 ° C. What is its mass, in grams? Step 2Calculate the temperature change. ΔT = 15 ° C – 122 ° C = 107 ° C Step 3Write the heat equation.

26. © 2014 Pearson Education, Inc. Calculating Heat Loss When 8.81 kJ is absorbed by a piece of iron, its temperature rises from 15 ° C to 122 ° C. What is its mass, in grams? Step 4 Substitute in values and solve.