1. Chapter 10
H E A T

2. Ch 10-1: Temperature and Thermal Equilibrium
Determining an object’s temperature with precision requires
a definition of temperature
established measurements that determine how “hot” or “cold” objects are

3. Heat Energy
Energy must be added to or removed from a substance to change its temperature.
Temperature is PROPORTIONAL to the kinetic energy of atoms.

4. Thermal Energy…
Internal Energy – the energy of a substance due to the random motions of its component particles and equal to the total energy of those particles.

5. Internal Energy
For an ideal gas, the internal energy depends only on the temperature of the gas.
For gases with 2 or more atoms per molecule, as well as for liquids and solids, other properties besides temperature contribute to the internal energy.

6. INTERNAL ENERGY The symbol, U, stands for internal energy
Thus, delta U, stands for the change in internal energy

7. Heating and Cooling
If an object has become hotter, it means that it has gained heat energy.
If an object cools down, it means it has lost energy

8. Thermal Equilibrium
The state in which two bodies in physical contact with each other have identical temperatures

9. Thermal Equilibrium
Ex: When you put a warm coke can into a large beaker of cold water, there is a noticeable temperature difference between the two.
After about 15 minutes, the can of soda will be cooler and the water surrounding it will be slightly warmer. Eventually, both the can and the water will be at the same temperature.
This temperature will not change as long
as conditions remain unchanged in
the beaker. This is Thermal Equilibrium.

10. Thermal Equilibrium is the basis for measuring temperature with thermometers.
This is because the thermometer is at the same temperature, or is in thermal equilibrium with, the object.

11. HEAT ENERGY What is HEAT? Form of energy and measured in JOULES
Particles move about more and take up more room if heated – this is why things expand if heated
It is also why substances change from: solids liquids gases when heated

12. Matter and Temperature
Matter Expands as its temperature increases
This is known as Thermal Expansion

13. Heating and Cooling cont…
Heat energy always moves from: HOT object COOLER object
e.g. Cup of water at 20 °C in a room at 30°C - gains heat energy and heats up – its temperature rises
Cup of water at 20 °C in a room at 10°C loses heat energy and cools down – its temperature will fall.

14. Measuring Temperature
Units depend on scale used:
Fahrenheit, Celsius, and Kelvin (or absolute) scales.
Fahrenheit commonly used in US
Fahrenheit and Celsius temps can be converted…

15. The Fahrenheit temperature scale is an English measurement scale
The Fahrenheit temperature scale is an English measurement scale. FREEZING POINT: Water freezes at 32 oF and BOILING POINT: boils at 2l2 oF. The Fahrenheit degree is smaller than the Celsius degree or Kelvin.

16. On the Celsius temperature scale, (the METRIC SCALE based on the freezing and boiling point of water) water’s freezing point is 0 oC & water’s boiling point is l00 oC

17. The Kelvin (K) scale is based on the boiling and freezing point of water and absolute zero. The Kelvin scale does not use the degree symbol. The units are Kelvins (K), not degrees Kelvin Kelvin unit = 1 oC scale.

18. On this scale the lowest possible temperature is absolute zero, written ”0 K." At absolute zero, the average kinetic energy of particles is zero. Absolute zero on the Kelvin scale is equal to -273 degrees Celsius.

19. Temperature Conversion Formulas: oC. =
Temperature Conversion Formulas: oC = K – 273 K = oC oC = (oF-32) oF = (1.8oC) + 32

20. HOMEWORK: Page 363: #1-3 all THIS WILL BE FOR A GRADE WHEN YOU WALK INTO CLASS TOMORROW!!! Must show all work 

21. QOTD What is the boiling point of water in degrees Fahrenheit?
What is the freezing point of water in degrees Celsius?
What is the value of absolute zero?

22. QOTD What is the boiling point of water in degrees Fahrenheit?
What is the freezing point of water in degrees Celsius?
0 degrees
What is the value of absolute zero?
0 Kelvin

23. 10-1 Review Questions
Which of the following is true for the water molecules inside popcorn kernels during popping?
A: Their Temperature Increases.
B: They are destroyed.
C: Their Kinetic Energy Increases.
D: Their mass changes.

24. HEAT and ENERGY
HEAT: the energy transfer between two objects because of a difference in their temperatures.
Energy transferred as heat ALWAYS transfers from an object of a higher temp to that of a lower temp.

25. Internal Energy
If the can and the water in the glass reach thermal equilibrium, which has greater internal energy?

26. Internal Energy
If the can and the water in the glass reach thermal equilibrium, which has greater internal energy?
A: The glass of water because of its greater mass

27. Thermal energy is unique in that it does NOT exist by itself
Thermal energy is unique in that it does NOT exist by itself. Thermal energy is present ONLY when another form of energy is present.

28. Thermal energy is unique in that it does NOT exist by itself
Thermal energy is unique in that it does NOT exist by itself. Thermal energy is present ONLY when another form of energy is present. Ex. Thermal energy like body heat only exist from breaking chemical bonds in food. Ex. Heat from the motion of friction Ex. Electrical items get hot from something resisting the flow of electricity …….Etc.

29. Key Points to remember…
Temperature – measures the average KE of molecules in an object
HEAT – energy transferred from one object to another
Internal Energy – sum of the energies of the molecules.

30. The calorie – the quantity of heat needed to raise the temperature of one gram of water 10C. 1 calorie = 4.18 joules

31. The calorie – The Calorie (also called a kilocalorie) – often measures the energy content in food 1 Calorie = 1000 calories There is a difference!

32. 1 Calorie = 1000 calories There is a difference
1 Calorie = 1000 calories There is a difference! Example item: That means a 190 Calorie bowl of cereal REALLY has 190,000 calories of energy!

33. Heat - units of Energy Heat Energy (heat flow): Q Internal energy: U
calorie (cal) = J
kilocalorie (kcal) = 4186 J
Calorie (dietary Calorie) = 1 kcal = 4186 J
Thus, for every Calorie you consume you actually obtain 4186 J of energy

34. Heat and Work Friction can increase a substances internal energy
For solids, internal energy can be increased by deforming their structure.
EX: Rubber band stretched or a piece of metal bent.

35. Rubber Band Activity…
Purpose: To demonstrate how work increases an object’s internal energy.
Procedure: Hold the rubber band between your thumbs.
Touch the middle section of the rubber band to your lip and note how it feels.
Rapidly stretch the rubber band and keep it stretched.
Touch the middle section of the rubber band to your lip again.
Notice is the rubber band’s temperature has changed.
(you may have to repeat this procedure several times before you can clearly distinguish the temperature difference.)

36. Conservation of Energy
If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property
 PE +  KE +  U = 0

37. Ex: Conservation of Energy
A 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?
ball

38. Ex: Conservation of Energy
A 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?
Givens:
m = 0.10 kg g =  U = ?
h = 10.0 m  PE = ?  KE=?
Formula:  PE +  KE + U = 0 Or
PEi + KEi + Ui = PEf + KEf + Uf

39. Ex: Conservation of Energy
A 0.10 kg ball falls 10 m onto a hard floor and then bounces back up 9.0 m. How much of its mechanical energy is transformed to the internal energy of the ball and the floor?
Givens:
m = 0.10 kg g =  U = ?
h = 10.0 m  PE = ?  KE= ?
Formula: Delta PE + Delta KE + Delta U = 0 or
PEi + KEi + Ui = PEf + KEf + Uf
mg( h) Uf-Ui = 0
mg ( h) = U
0.10 (9.81) (1) = U
0.98 J

40. HOMEWORK
Page 370 Practice: # 1
Page 370 Review: # 1 and 3

42. QOTD
1: A substance’s temperature increases as a direct result of: a. Energy being removed from the particles of the substance b. Kinetic Energy being added to the particles of the substance c: A change in the number of atoms and molecules in a substance

43. QOTD CONT
2: What happens to the internal energy of an ideal gas when it is heated from 0 degrees Celsius to 4 degrees Celsius? a: increases b: decreases c: stays the same d: impossible to determine

44. QOTD cont
Which of the following is proportional to the KE of atoms and molecules?
a: Elastic Energy
b: Thermal Equilibrium
c: Potential Energy
d: Temperature

45. QOTD
Heat Flow occurs between two bodies in thermal contact when they differ in which of the following properties?
A: mass
B: density
C: temperature
D: specific heat

46. QOTD Which of the following is equivalent to 88 degrees Fahrenheit?
A: 49 degrees C
B: 31 degrees C
C: 16 degrees C
D: 58 degrees C

47. 10-3: Changes in temperature and phase
Specific Heat Capacity:
the quantity of energy needed to raise the temperature of 1 kg of substance by 1 degree Celsius at constant pressure.

48. Specific Heat Capacity (c)
Relates mass, temperature change, and energy transferred as heat.
Specific heat capacity equation:
cp = Q/m(T)
Where the subscript p represents heat capacity measured at a constant pressure

49. Specific Heat
Though heat and temperature are not the same thing, there is a correlation between the two, captured in a quantity called specific heat, c.
Specific heat measures how much heat is required to raise the temperature of a certain mass of a given substance.

50. See page 372 Table 10-4 Specific Heat capacities
**You must know cp for water =
4.186 x 103 J/(kg * C)

51. Calorimetry
An experimental procedure used to measure the energy transferred from one substance to another as heat.

52. The higher the specific heat, the more energy the material contains when it is at the same temperature as a material with a lower specific heat. Ex. 2 kg of Copper ( 385 J/kg*oC) 2 kg of Glass [664 J/kg*oC ] If both are 20 oC, Glass has more energy!

53. Determining Specific Heat Capacity
For simplicity, a subscript w will always stand for “water” in problems involving specific heat capacities.
Energy absorbed by water = energy released by the substance Or
Qw = Qx
Where Q = mc ΔT

54. Specific Heat Capacity
The equation applies to both substances that absorb energy AND those that transfer energy to their surroundings.
When temp increases; change in T and Q are taken to be positive.
When temp decreases; they are negative and energy is transferred from the substance.

55. Specific Heat Specific heat units: J/kg ·ºC or cal/g ·ºC.
Every substance has a different specific heat, but specific heat is a constant for that substance.
For instance, the specific heat of water, cwater , is 4.19 x 103 J/kg · ºC or 1 cal/g · ºC. That means it takes 4.19 x 103 joules of heat to raise one kilogram of water by one degree Celsius.
Substances that are easily heated, like copper, have a low specific heat, while substances that are difficult to heat, like rubber, have a high specific heat.

56. SPECIFIC HEAT
Specific heat allows us to express the relationship between heat and temperature in a mathematical formula:
Q = mc  T
Q is the heat transferred to a material
m is the mass of the material,
c is the specific heat of the material
T is the change in temperature.

57. Calculating Thermal Energy Q = mc ΔT Q = change in thermal energy Joules ( J ) m = mass (kg ) c = specific heat ( J/kgoC ) ΔT = change in temperature oC

58. Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?
given: formula: sub: answer & unit
Q = Q = mc ΔT m=
c =
ΔT =

59. Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?
given: formula: sub: answer & unit
Q = Q Q = mc ΔT
m= 20 kg
c = 450 J/kgoC
ΔT = = 15 oC

60. Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?
given: formula: sub: answer & unit
Q = Q Q = m c ΔT
m= 20 kg
C = 450 J/kgoC
ΔT = = 15 oC
BE SURE TO WRITE THE UNITS IN YOUR NOTES !

61. Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?
given: formula: sub: answer & unit
Q = Q Q = mc ΔT
m= 20 kg Q = 20 x 450 x 15
C = 450 J/kgoC
ΔT = = 15 oC

62. Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of 20 kg of iron from 35oC up to 50oC if the specific heat for iron is 450 J/kgoC ?
given: formula: sub: answer & unit
Q = Q Q = mc ΔT
m= 20 kg Q = 20 x 450 x = ,000 J
C = 450 J/kgoC
ΔT = = 15 oC

63. A 0. 05 kg metal bolt is heated to an unknown initial temperature
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.
Givens: m metal = 0.05 kg cp, m = 899 J/kg *C m water = 0.15 kg cp, w = 4186 J/kg *C T water = 21.0 C T final = C Unknown T metal = ?

64. mmcp, m ΔTm = mw cp, w ΔTw ΔTm = mw cp, w ΔTw mmcp, m
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.
Givens: m metal = 0.05 kg cp, m = 899 J/kg *C
m water = 0.15 kg cp, w = 4186 J/kg *C
T water = 21.0 C T final = C
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m

65. mmcp, m ΔTm = mw cp, w ΔTw ΔTm = mw cp, w ΔTw mmcp, m
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m
Next step…
Plug and Chug!

66. ΔTm = mw cp, w ΔTw mmcp, m ΔTm = (0.15) (4186) (4) = 56 oC (.05) (899)
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m
ΔTm = (0.15) (4186) (4) = 56 oC
(.05) (899)
Not done yet….looking for INITIAL TEMP OF THE METAL!!!

67. Δ Tm = Tf + Δ Tm Δ Tm = 25.0 C + 56.0 C Δ Tm =81.0 C
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a beaker containing 0.15 kg of water with an initial temperature of 21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If the metal has a specific heat capacity of 899 J/kg *C, find the initial temperature of the metal.
ΔTm = (0.15) (4186) (4) = 56 oC
(.05) (899)
Δ Tm = Tf + Δ Tm
Δ Tm = 25.0 C C
Δ Tm =81.0 C

68. EXAMPLE
4190 J of heat are added to 0.5 kg of water with an initial temperature of 12ºC. What is the temperature of the water after it has been heated?
By rearranging the equation above, we can solve for :
T = Q/(mc)
The temperature goes up by 2 Cº, so if the initial temperature was 12ºC, then the final temperature is 14ºC. Note that when we talk about an absolute temperature, we write ºC, but when we talk about a change in temperature, we write Cº.

69. Homework Part 1
Page 374 #1,2