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

2. is different for different substances
Specific Heat
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.

3. 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. Specific Heat of Some Substances

5. Specific Heat of Liquid Water
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.

6. Guide to Calculating Specific Heat

7. Learning Check
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?

8. 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. 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. Heat Equation from Specific Heat
Rearranging the specific heat expression gives the heat equation.

11. 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. Guide to Calculations Using Specific Heat

13. Sample Problem, Using SH
A layer of copper (Cu) on a pan has a mass of 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 J/g ° C.)
Step 1 Given 135 g, 26 ° C to 328 ° C, SH = J/g ° C
Need joules

14. 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 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. 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 J/g ° C.)
Step 4 Substitute in values and solve.

16. 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 kJ
B kJ
C kJ

17. Solution
How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 ° C to 77.0 ° C?
C kJ
Step 1 Given 325 g water
SH(water) = J/g ° C
15.0 ° C to 77.0 ° C
Need kilojoules

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

19. Solution
How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to °C?
C kJ
Step 4 Substitute in values and solve.

20. 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 1 Given 225 g tea (water)
SH(water) = J/g °C
74.6 °C to 22.4 °C
Need kilojoules

21. 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 = °C – 74.6 °C = °C
Step 3 Write the heat equation.
q = m × ΔT × SH(water)

22. 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. 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. 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 1 Given 8.81 kJ
SH(iron) = J/g ° C
15 ° C to 122 ° C
Need mass of iron

25. 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 2 Calculate the temperature change.
ΔT = 15 ° C – 122 ° C = 107 ° C
Step 3 Write the heat equation.

26. 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.