1. Chapter 2 Energy and Matter

2. Energy Energy makes objects move makes things stop
is needed to “do work”

3. Work Work is done when you climb you lift a bag of groceries
you ride a bicycle
you breathe
your heart pumps blood
water goes over a dam

4. Potential Energy Potential energy is energy
stored for use at a later time.
Examples:
water behind a dam
a compressed spring
chemical bonds in gasoline, coal, or food

5. Kinetic Energy Kinetic energy is the energy of matter in motion.
Examples:
swimming
water flowing over a dam
working out
burning gasoline

6. Chapter 2 Energy and Matter
2.2 Temperature

7. Temperature Temperature
is a measure of how hot or cold an object is compared to another object
indicates that heat flows from the object with a higher temperature to the object with a lower temperature
is measured using a thermometer

8. Temperature Scales Temperature scales
are Fahrenheit, Celsius, and Kelvin
have reference points for the boiling and freezing points of water

9. Units for Measuring Energy or Heat
Heat is measured in joules or calories.
4.184 joules (J) = 1 calorie (cal) (exact)
1 kJ = 1000 J
1 kilocalorie (kcal) = 1000 calories (cal)

10. Fahrenheit Formula
On the Fahrenheit scale, there are 180 °F between the freezing and boiling points, and on the Celsius scale, there are 100 °C.
180 °F = 9 °F = °F
100 °C 5 °C °C
In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0 °C to 32 °F.
TF = 9 (TC) ° 5
or TF = 1.8(TC) °

11. Celsius Formula TC is obtained by rearranging the equation for TF.
TF = (TC) °
Subtract 32 from both sides.
TF – 32 ° = 1.8(TC) + (32 ° – 32 °)
TF – 32 ° = 1.8(TC)
Divide by TF – 32 ° = TC
TF – 32 ° = TC
1.8

12. Solving a Temperature Problem
A person with hypothermia has a
body temperature of 34.8 °C. What
Is that temperature in °F?
TF = 1.8(TC) °
TF = (1.8)(34.8 °C) °
exact tenth’s exact
= 62.6 ° °
= 94.6 °F
tenth’s

13. Kelvin Temperature Scale
The Kelvin temperature
is obtained by adding 273 to the Celsius temperature
TK = TC
In the Kelvin temperature scale:
There are 100 units between the freezing and boiling points of water.
100 K = 100 °C or 1 K = 1 °C
0 K (absolute zero) is the lowest possible temperature.
0 K = –273 °C

14. Chapter 2 Energy and Matter
2.3 Specific Heat

15. Specific Heat 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

16. Example of Calculating Specific Heat
What is the specific heat of a metal if 24.8 g absorbs
65.7 cal of energy and the temperature rises from
20.2 C to 24.5 C?
STEP 1 Given: 24.8 g, 65.7 cal,
ΔT = 20.2 C to 24.5 C
Need: SHmetal = cal/g C
STEP 2 Plan: ΔT = 24.5 C – 20.2 C = 4.3 C
SH = Heat(cal)
g C

17. Example of Calculating Specific Heat (continued)
STEP 3 Set up to calculate SH:
65.7 cal = 0.62 cal/g C (24.8 g)(4.3 C)

18. Heat Equation Rearranging the specific heat expression gives the
Heat = g x  T x cal (or J) = cal (or 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 (cal/g °C) or (J/g °C)

19. Guide to Calculations Using Specific Heat

20. Sample Calculation for Heat
A hot-water bottle contains 750 g of water at 65 °C. If the
water cools to body temperature (37 °C), how many
calories of heat could be transferred to sore muscles?
STEP 1 Given: 750 g of water cools from 65 °C to 37 °C
SHwater = 1.00 cal/g °C
Need: calories of heat transferred
STEP 2 Calculate the temperature change  T:
65 °C – 37 °C = 28 °C

21. Sample Calculation for Heat
STEP 3 Write the heat equation:
Heat (cal) = mass(g) x  T x SH
STEP 4 Substitute given values and solve for heat:
750 g x 28 °C x cal g °C
= cal

22. Learning Check How many calories are obtained from a pat of butter
if it provides 150 J of energy when metabolized?
1) 36 cal
2) 150 cal
3) 630 cal

23. Solution STEP 1 Given: 150 J Need: calories STEP 2 Plan: J cal
STEP 3 Write equality and conversion factors:
1 cal = J
1 cal and J
4.184 J cal
STEP 4 Set up problem:
150 J x 1 cal = 36 cal (1)
4.184 J

24. Learning Check How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0 °C to 77.0 °C?
1) kJ
2) kJ
3) 105 kJ

25. Solution
2) kJ
STEP 1 Given: 325 g of water warms from 15.0 °C to 77.0 °C
SHwater = J/g °C
1 kJ = J
Need: kilojoules of heat needed
STEP 2 Calculate the temperature change  T:
77.0 °C – 15.0 °C = 62.0 °C
STEP 3 Write the heat equation:
Heat (joules) = mass (g) x  T x SH

26. Solution (continued)
STEP 4 Substitute given values and solve for heat: g x °C x J x kJ
g °C J
= 84.3 kJ (2)

27. Chapter 2 Energy and Matter
2.4 Energy and Nutrition

28. Calorimeters A calorimeter is used to measure heat transfer
contains a reaction chamber and thermometer in water
indicates the heat lost by a sample
indicates the heat gained by water

29. Chapter 2 Energy and Matter
2.5
Classification of Matter

30. Matter Matter is the material that makes up a substance
makes up the things we see everyday, such as water, wood, cooking pans, clothes, and shoes

31. Pure Substances A pure substance is classified as
matter with a specific composition
an element when composed of one type of atom
a compound when composed of two or more elements combined in a definite ratio

32. Mixtures A mixture is matter that consists of
two or more substances that are physically mixed, not chemically combined
two or more substances in different proportions
substances that can be separated by physical methods

33. Homogeneous Mixtures In a homogeneous mixture,
the composition is uniform throughout
the different parts of the mixture are not visible

34. Heterogeneous Mixtures
In a heterogeneous mixture,
the composition is not uniform; it varies from one part of the mixture to another
the different parts of the mixture are visible

35. Chapter 2 Energy and Matter
2.6 States and Properties of Matter

36. Solids Solids have a definite shape a definite volume
particles that are close together in a fixed arrangement
particles that move very slowly

37. Liquids Liquids have an indefinite shape but a definite volume
the same shape as their container
particles that are close together but mobile
particles that move slowly

38. Gases Gases have an indefinite shape an indefinite volume
the same shape and volume as their container
particles that are far apart
particles that move very fast

39. Physical Properties Physical properties
are observed or measured without changing the identity of a substance
include shape and color
include melting point and boiling point

40. Physical Change In a physical change,
the identity and composition of the substance do not change
the state can change or the material can be torn into smaller pieces

41. Chemical Properties Chemical properties
describe the ability of a substance to change into a new substance
During a chemical change,
reacting substances form new substances with different compositions and properties
a chemical reaction takes place
Iron Fe
Iron (III) oxide
Fe2O3

42. Chapter 2 Energy and Matter
2.7 Changes of State

43. Melting and Freezing A substance
is melting when it changes from a solid to a liquid
is freezing when it changes from a liquid to a solid
such as water has a freezing (melting) point of 0 °C

44. Heat of Fusion The heat of fusion
is the amount of heat released when 1 gram of liquid freezes (at its freezing point)
is the amount of heat needed to melt 1 gram of a solid (at its melting point)
for water (at 0 °C) is
334 J or 80 cal
1 g of water

45. Guide to Calculations Using Heat of Fusion (or Vaporization)

46. Calculations Using Heat of Fusion
How much heat in calories is needed to melt 15.0 g of
ice at 0 °C ?
STEP 1 Given: 15.0 g of water(s)
change of state: melting at 0 °C
STEP 2 Plan: g of water(s) g of water(l)
STEP 3 Write conversion factors:
1 g of water = cal
1 g of water and cal
80 cal g of water

47. Calculations Using Heat of Fusion (continued)
STEP 4 Set up the problem to calculate calories:
15.0 g water x cal = cal
1 g water

48. Sublimation Sublimation occurs when a solid changes directly to a gas
is typical of dry ice, which sublimes at 78 C
takes place in frost-free refrigerators
is used to prepare freeze-dried foods for long-term storage

49. Evaporation and Condensation
Water
evaporates when molecules on the surface gain sufficient energy to form a gas
condenses when gas molecules lose energy and form a liquid

50. Heat of Vaporization The heat of vaporization is the amount of heat
absorbed to vaporize 1 g of a liquid to gas at the boiling point
released when 1 g of a gas condenses to liquid at the boiling point
Boiling Point of Water = 100 °C
Heat of Vaporization (water) = J or 540 cal
1 g of water

51. Learning Check
Classify each of the following as a 1) physical change or 2) chemical change.
A. ____ burning a candle
B. ____ ice melting on the street
C. ____ toasting a marshmallow
D. ____ cutting a pizza
E. ____ polishing a silver bowl

52. Solution
Classify each of the following as a 1) physical change or 2) chemical change.
A burning a candle
B ice melting on the street
C toasting a marshmallow
D cutting a pizza
E polishing a silver bowl

53. Learning Check How many joules are released when 25.0 g of water
at 0 °C freezes?
1) 334 J
2) J
3) J

54. Solution 3) 8350 J STEP 1 Given: 25.0 g of water(l)
change of state: freezing at 0 °C
STEP 2 Plan: g of water g of water
STEP 3 Write conversion factors:
1 g of water = 334 J
1 g of water and __334 J___
334 J 1 g of water

55. Solution (continued) STEP 4 Set up the problem to calculate joules:
25.0 g water x J = J (3)
1 g water

56. Learning Check How many kilocalories (kcal) are released when 50.0 g
of water(g) as steam from a volcano condenses at
100 °C?
1) 27 kcal
2) 540 kcal
3) kcal

57. Solution 1) 27 kcal STEP 1 50.0 g of water(g) = 50.0 g of water(l)
Change of state: water condensing at 100 °C
STEP 2 Plan: g of water(g) g of water(l)
STEP 3 Write conversion factors:
1 g of water = 540 cal
1 g of water and cal
540 cal g of water
1 kcal = cal
1 kcal and cal
1000 cal kcal