1. Reaction Energy

2. Specific Heat
The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius.
Because different substances have different compositions, each substance has its own specific heat.

3. Exothermic and Endothermic
Exothermic: Heat flows out of the system (to the surroundings). The value of ‘q’ is negative.
Endothermic: Heat flows into the system (from the surroundings). The value of ‘q’ is positive.

4. q = m Cp ΔT Specific Heat q = heat (J) Cp = specific heat (J/(g.°C)
m = mass (g)
ΔT = change in temperature = Tf – Ti (°C)

5. Measuring Heat
Heat changes that occur during chemical and physical processes can be measured accurately and precisely using a calorimeter.
A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

6. A coffee-cup calorimeter made of two Styrofoam cups.

7. Phase Changes Solid Gas Liquid Melting Vaporization Freezing
Condensation

8. Solid Gas Liquid Sublimation Melting Vaporization Freezing
Deposition
Freezing
Condensation

10. Latent Heat q = m Hv q = m Hf Hv = latent heat of vaporization (J/g)
Hf = latent heat of fusion (J/g)

11. Energy and Phase Change
Heat of vaporization (Hv) is the energy required to change one gram of a substance from liquid to gas.
Heat of fusion (Hf) is the energy required to change one gram of a substance from solid to liquid.

12. Specific Heat Problem (25.0 – 50.4) -114 = q = 10.0 m Ciron ∆T
1) The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron?
(25.0 – 50.4)
-114 =
q =
10.0 m
Ciron ∆T
Ciron = J/g°C

13. Specific Heat Problem m = 4.67 g
2) A piece of metal absorbs 256 J of heat when its temperature increases by 182°C. If the specific heat of the metal is J/g°C, determine the mass of the metal.
m = 4.67 g

14. Specific Heat Problem Tf = 58.5 °C
3) If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?
Tf = 58.5 °C

15. Specific Heat Problem q = 70500 J
4) As 335 g of aluminum at 65.5°C gains heat, its final temperature is 300.°C. Determine the energy gained by the aluminum.
q = J

16. Latent Heat Problem q = 12.0 m Hf (334) q = 4010 J
5) How much heat does it take to melt 12.0 g of ice at 0 °C?
q =
12.0 m Hf
(334)
q = 4010 J

17. Latent Heat Problem q = 5.00 m (2260) Hv q = 11300 J
6) How much heat must be removed to condense 5.00 g of steam at 100 °C?
q =
5.00 m
(2260) Hv
q = J

18. Latent Heat Problem Lf = 67 J/g
7) If 335 J of heat are added to melting 5.00 g of gold, what is the latent heat of fusion for gold in J/g?
Lf = 67 J/g

19. Latent Heat Problem m = 6.18 g
8) The latent heat of fusion for platinum is 119 J/g. Platinum absorbs 735 J of heat. What is the mass of platinum?
m = 6.18 g

20. HEATING AND COOLING CURVES

21. The heating curve has 5 distinct regions.

22. The horizontal lines are where phase changes occur.
vaporization
melting

23. During any phase change, temperature is constant.
vaporization
melting

24. Use q = m Cp ΔT for all diagonal lines
Use q = m Cp ΔT for all diagonal lines. Use q = m Hf for melting and q = m Hv for boiling.
vaporization
melting 24

25. Heating Curve for Water
Section D:
Water is
Boiling.
Section A:
Ice is being heated
to the melting
point.
Section C:
Water is
being heated
to the boiling
point.
Section B:
The ice
is melting.
Section E:
Steam is
being heated.

26. Heating Curve for Water
Section D
q = mHv
Heating Curve for Water
Section E
q = m Cp ΔT(gas)
Section B
q = mHf
Section C
q = m Cp ΔT(liquid)
Section A
q = m Cp ΔT(solid)

27. + + + + Solving Problems For water: Heat Ice Melt Ice Heat Water
Boil Water
Heat Steam + + + +
Below 0 °C At
0 °C
0 °C -
100 °C At
100 °C
Above
100 °C

28. Specific Heat and Latent Heat
9) How much heat does it take to heat 12 g of ice at – 6 °C to 25 °C water? Round to a whole number.
You begin at ice below 0 °C. Note: The final temperature for this process cannot exceed 0 °C.
q = m 12
(2.05)
Cice ∆T
(0 - – 6)
q = 148 J

29. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 12 g of ice at – 6 °C to 25 °C water? Round to a whole number.
Since the temperature needs to rise to 25 °C, you must melt the ice next.
q = 12 m
(334) Hf
q = 4008 J

30. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 12 g of ice at – 6 °C to 25 °C water? Round to a whole number.
You now have water at 0 °C. The final temperature of the water should be 25 °C.
q = 12 m
Cwater
(4.18)
(25 - 0) ∆T
q = 1254 J

31. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 12 g of ice at – 6 °C to 25 °C water? Round to a whole number.
Finally, add all of the q values together.
q =
148 +
4008 +
1254
q = 5410 J

32. Specific Heat and Latent Heat
10) How much heat does it take to heat 35 g of ice at 0 °C to steam at 150 °C? Round to a whole number.
You begin with ice at 0 °C, so you should melt it first.
q = 35 m
(334) Hf
q = J

33. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 35 g of ice at 0 °C to steam at 150 °C? Round to a whole number.
You now have water at 0 °C. The final temperature of the steam is to be 150 °C. You must take the water to 100 °C before you can even convert it to steam.
q = 35 m
(4.18)
Cwater
( ) ∆T
q = J

34. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 35 g of ice at 0 °C to steam at 150 °C? Round to a whole number.
Now you have water at 100 °C. Convert this to steam.
q = 35 m
(2260) Hv
q = J

35. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 35 g of ice at 0 °C to steam at 150 °C? Round to a whole number.
You must now take the steam at °C to 150 °C.
q = 35 m
(2.02)
Cice ∆T
( )
q = 3535 J

36. Specific Heat and Latent Heat, cont.
How much heat does it take to heat 35 g of ice at 0 °C to steam at 150 °C? Round to a whole number.
Finally, add all of the Q values together.
q =
3535
q = J

37. Specific Heat and Latent Heat
11) How much heat does it take to convert 16.0 g of ice to water at 0 °C?
(5340 J)

38. Specific Heat and Latent Heat
12) How much heat does it take to heat 21.0 g of water at 12.0 °C to water at °C?
(5530 J)

39. Specific Heat and Latent Heat
13) How much heat does it take to heat 14.0 g of water at 12.0 °C to steam at °C?
(37400 J)

40. Heat lost equals heat gained.
Calorimetry
For calorimetry problems, use the equation:
– m Cp ΔT = m Cp ΔT
which is based on the law of conservation of energy.
Heat lost equals heat gained.

41. Calorimetry Problem
14) 125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.

42. 125 g of water at 25. 6°C is placed in a foam-cup calorimeter. A 50
125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.
Find the heat gained by the water.
q =
125 m
(4.18)
Cwater
(29.3 – 25.6) ∆T
q = 1930 J

43. q = m Cmetal ∆T Cmetal = 0.450 J/g°C
125 g of water at 25.6°C is placed in a foam-cup calorimeter. A 50.0 g sample of the unknown metal is heated to a temperature of 115.0°C and placed into the water. Both water and metal have attain a final temperature of 29.3°C. Determine the specific heat of the metal.
Since heat lost equals heat gained, determine the specific heat of the metal.
-1930 =
q =
50.0 m
Cmetal
(29.3 – 115.0) ∆T
Cmetal = J/g°C

44. Calorimetry Problem m = 477 g
15) You put 352 g of water into a foam-cup calorimeter and find that its initial temperature is 22.0°C. What mass of 134°C lead can be placed in the water so that the equilibrium temperature is 26.5°C?
m = 477 g

45. Calorimetry Problem m = 58.0 g
16) You put water into a foam-cup calorimeter and find that its initial temperature is 25.0°C. What is the mass of the water if 14.0 grams of 125°C nickel can be placed in the water so that the equilibrium temperature is 27.5°C?
m = 58.0 g