Principles of Technology/Physics in Context (PT/PIC)

1
Principles of Technology/Physics in Context
(PT/PIC)

• Chapter 8
• Internal Energy and
• Properties of Matter 1
• Text p. 150 -154

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Key Objectives

• At the conclusion of this chapter youll be able
  to
• Define the following terms internal energy and
  temperature, absolute zero, heat energy and
  specific heat, and pressure.
• State the fixed points on the Celsius temperature
  scale and the fixed point on the Kelvin
  temperature scale.

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Key Objectives

• At the conclusion of this chapter youll be able
  to
• Relate Kelvin and Celsius temperatures, and solve
  problems involving this relationship.
• State the equation that relates heat energy,
  specific heat, and temperature changes, and solve
  problems using this equation.
• Define the term thermal equilibrium, and solve
  thermal equilibrium problems.

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8.1 INTERNAL ENERGY AND WORK

• Suppose we use a force to move an object along a
  horizontal table at constant speed.
• We know from Chapter 7 that we have done work on
  the object, but what has this work accomplished?
• The kinetic energy has not changed because the
  speed has been kept constant.

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8.1 INTERNAL ENERGY AND WORK

• The gravitational potential energy has not
  changed because the table is horizontal.
• Our work has been used to overcome the friction
  between the object and the table, and, therefore
  we say that the internal energy of the
  object-table system has been increased by the
  work we have done.

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Assessment Question 1

• Suppose we use a force to move an object along a
  horizontal table at constant speed.
• All of the following are true EXCEPT
• No work is done on the object.
• The kinetic energy has not changed because the
  speed has been kept constant.
• The gravitational potential energy has not
  changed because the table is horizontal.
• The internal energy of the object-table system
  has been increased by the work we have done.

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8.1 INTERNAL ENERGY AND WORK

• Roughly speaking, the internal energy (Q) of a
  system is the total kinetic and potential
  energies of the atoms and molecules that make up
  the system.
• A change in the internal energy of an object is
  usually accompanied by a change in its
  temperature.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• Temperature is a measure of the hotness of an
  object with respect to some predefined standard.
• It is a scalar quantity.
• We will see in Section 8.4 that the temperature
  of an object is related to the average kinetic
  energy of its molecules.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• When mercury is placed in a thin tube (whose
  diameter is uniform), the length of the mercury
  column increases with rising temperature.
• Such a device is called a thermometer

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Assessment Question 2

• All of the following are true EXCEPT
• A change in the internal energy of an object is
  usually accompanied by a change in its
  temperature.
• Temperature is a measure of the hotness of an
  object with respect to some predefined standard.
• Temperature is a vector quantity measured by a
  device called a temperaguage.
• The temperature of an object is related to the
  average kinetic energy of its molecules.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• To measure temperature, we need a property that
  changes regularly with changes in temperature.
• One such property is the volume of a liquid such
  as mercury.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• We also need to establish a scale of measurement.
  
• To do this, one or more fixed reference points
  must be set.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• On the Celsius scale, the freezing and boiling
  points of water (at 1 atmosphere of pressure) are
  assigned the respective temperatures of 0C and
  100C.
• (The Fahrenheit scale sets these points at 32F
  and 2 12F.)

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• Another scale is the Kelvin or absolute
  temperature scale.
• Its single reference point is the temperature at
  which water exists simultaneously as a gas,
  liquid, and solid.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• The temperature assigned to this triple point
  is 273.16 K.
• On the Kelvin scale, zero kelvin (0 K) is the
  lowest temperature possible and is known as
  absolute zero.
• At this temperature, molecular motion is at a
  minimum.

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Assessment Question 3

• All of the following are true EXCEPT
• On the Celsius scale, the freezing and boiling
  points of water (at 1 atmosphere of pressure) are
  assigned the respective temperatures of 0C and
  100C.
• On the Fahrenheit scale the freezing and boiling
  points of water (at 1 atmosphere of pressure) are
  assigned the respective temperatures of 32F and
  212F.
• The quadruple point is the temperature at which
  water exists simultaneously as a gas, liquid,
  solid and plasma the temperature is -250 K.
• On the Kelvin scale, zero kelvin (0 K) is the
  lowest temperature possible and is known as
  absolute zero. At this temperature, molecular
  motion is at a minimum.

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• The relationship between the Celsius (TC) and
  Kelvin (TK) temperature scales is as follows

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• PROBLEM
• (a) Convert 37C (TC) to the Kelvin scale (TK)

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• PROBLEM
• (a) Convert 37C to the Kelvin scale.
• SOLUTION
• (a) TKTC 273 37C 273 310K

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Assessment Question 4

• Convert 97C (TC) to the Kelvin scale (TK)
• TKTC 273 97C 273
• -75 K
• 176 K
• 260 K
• 370 K

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• PROBLEM
• (b) Convert 0 K (TK) to the Celsius scale (TC) .

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• PROBLEM
• (b) Convert 0 K to the Celsius scale.
• SOLUTION
• (a) TKTC 273
• (a) TC TK - 273 0 K - 273 -273 C

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Assessment Question 5

• Convert 57 K (TK) to the Celsius scale (TC)
• TKTC 273
• TC TK - 273 57 K - 273
• -1550 C
• -216 C
• 220 C
• 330 C

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8.2 TEMPERATURE AND TEMPERATURE SCALES

• Frequently, scientists refer to standard
  temperature, which has a defined value of 273 K
  (the freezing point of water at 1 atmosphere of
  pressure).

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8.3 HEAT AND INTERNAL ENERGY

• An object can change its internal energy by
  absorbing or releasing heat energy.
• If no phase changes (e.g., freezing, boiling) are
  occurring, the absorption or release of heat
  energy is accompanied by a change in temperature.

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8.3 HEAT AND INTERNAL ENERGY

• The amount of heat energy absorbed or released is
  directly related to three factors the substance
  itself, its mass, and the size of the temperature
  change.

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8 HEAT AND INTERNAL ENERGY

• These three quantities can be combined into a
  single equation

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8.3 HEAT AND INTERNAL ENERGY

• where Q stands for the amount of heat energy in
  joules, m is the mass of the substance in
  kilograms, and ?TC is the temperature change of
  the substance in Celsius degrees (C).

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8.3 HEAT AND INTERNAL ENERGY

• The symbol c stands for the specific heat of the
  substance.
• The value of the specific heat identifies the
  substance its units are kilojoules per kilogram
  per Celsius degree (kJ/kg C).

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8.3 HEAT AND INTERNAL ENERGY

• The lower the specific heat of a substance, the
  more readily it changes its temperature in
  response to heat loss or heat gain.
• For example, dry land has a much lower specific
  heat than water.
• For this reason, landlocked areas tend to have
  much larger changes in temperature (warmer
  summers and colder winters) than areas that are
  near large bodies of water.

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Assessment Question 6

• All of the following are true EXCEPT
• An object can change its internal energy by
  absorbing or releasing heat energy. If no phase
  changes (e.g., freezing, boiling) are occurring,
  the absorption or release of heat energy is
  accompanied by a change in temperature.
• The amount of heat energy absorbed or released is
  directly related to three factors the substance
  itself, its mass, and the size of the temperature
  change.
• The lower the specific heat of a substance, the
  more readily it changes its temperature in
  response to heat loss or heat gain.
• Dry land has a much larger specific heat than
  water so landlocked areas tend to have small
  changes in temperature (cool summers and mild
  winters) and areas near large bodies of water
  have large changes in temperature (hot summers
  and cold winters).

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8.3 HEAT AND INTERNAL ENERGY

• A table of specific heats for various substances
  is given below

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8.3 HEAT AND INTERNAL ENERGY

• PROBLEM
• How much heat energy (Q) is needed to raise the
  temperature of 20 kilograms (m) of liquid water
  from 5C to 20C (?T)?

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8.3 HEAT AND INTERNAL ENERGY

• SOLUTIONHow much heat energy (Q) is needed to
  raise the temperature of 20 kilograms (m) of
  liquid water from 5C to 20C (?T)?

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Assessment Question 7

• How much heat energy (Q) is needed to raise the
  temperature of 50 kilograms (m) of liquid water
  from 15C to 40C (?T)?
• Q cm ?T
• Q 4.19 kJ/kgCo)(50 kg) (40C - 15C)
• 50 J
• 250 J
• 5200 J
• 7500 J

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8.3 HEAT AND INTERNAL ENERGY

• The positive sign associated with Q is a result
  of the sign of ?TC and indicates that heat has
  been absorbed.
• If the temperature had decreased, ?TC would have
  been negative, as would Q.
• This would mean that heat had been released.

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8.3 HEAT AND INTERNAL ENERGY

• Suppose two objects with different temperatures
  are brought into contact.
• Experience tells us that heat energy is always
  transferred from the hotter object to the colder
  one until they both reach the same final
  temperature.

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8.3 HEAT AND INTERNAL ENERGY

• At this point, we say that the objects have
  reached thermal equilibrium the final
  temperature is known as the equilibrium
  temperature.
• This will occur, regardless of the nature of the
  objects or their masses.
• (The identity of the objects and their masses
  will determine the value of the final
  temperature, however.)

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8.3 HEAT AND INTERNAL ENERGY

• Problems involving heat-energy transfer utilize
  the relationship Q cm ?TC
• However, this relationship must be used
  twiceonce for the object that releases (loses)
  heat energy, and once for the object that absorbs
  (gains) it.

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8.3 HEAT AND INTERNAL ENERGY

• We assume that the total internal energy is
  conserved therefore, the heat energy lost by
  the hotter object is set equal to the heat energy
  gained by the colder object.

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Assessment Question 8

• Suppose two objects with different temperatures
  are brought into contact.
• All of the following are true EXCEPT
• Heat energy is always transferred from the hotter
  object to the colder one until they both reach
  the same final temperature.
• When both objects are at the same temperature the
  objects have reached thermal equilibrium the
  final temperature is known as the equilibrium
  temperature. This will occur, regardless of the
  nature of the objects or their masses.
• The objects will always reach the same final
  temperature regardless of the identity of the
  objects and their masses.
• The total internal energy is conserved
  therefore, the heat energy lost by the hotter
  object is set equal to the heat energy gained
  by the colder object.

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8.3 HEAT AND INTERNAL ENERGY

• The equations look like this

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8.3 HEAT AND INTERNAL ENERGY

• The vertical bars (J ) indicate that we are
  interested only in the (absolute) value of the
  quantity, not whether it is positive or negative.
  
• The quantity k is evaluated by subtracting the
  smaller temperature from the larger temperature.

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8.3 HEAT AND INTERNAL ENERGY

• PROBLEM
• A 10-kilogram (m) block of copper at 60C is
  placed in contact with an identical 10-kilogram
  (m) block of copper at 20C.
• What is the equilibrium temperature of both
  blocks?

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8.3 HEAT AND INTERNAL ENERGY

• SOLUTION
• We could guess at the answer (40C, the midpoint
  temperature between 20C and 60C), and we would
  be correct! But lets see why this is so.

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8.3 HEAT AND INTERNAL ENERGY

• SOLUTION
• Both blocks have the same mass (10 kg), and both
  are composed of the same substance (copper).
• The change in heat energy should therefore affect
  the temperature of each block equally (but in
  opposite directions).
• If both blocks must reach the same final temp the
  midpoint temperature of 40C is the only
  temperature that satisfies these requirements.

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8.3 HEAT AND INTERNAL ENERGY

• SOLUTION
• Now lets prove that this is the case by solving
  the pair of equations given above

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Assessment Question 9

• A 5.0 kilogram (m) block of copper at 90C is
  placed in contact with an identical 5.0 kilogram
  (m) block of copper at 10C.
• What is both blocks equilibrium temperature?
• Qlost Qgained (cm?T)lost
  (cm?T)gained
• (0.39 kJ/kgC)(5 kg)(90C-Tfinal) (0.39
  kJ/kgC)(5 kg)(Tfinal-10C)
• (90C-Tfinal) (Tfinal-10C)
• Tfinal (90C 10C) / 2
• A. 10C C. 80C
• B. 50C D. 100C

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8.3 HEAT AND INTERNAL ENERGY

• PROBLEM
• Calculate the equilibrium temperature (Tfinal)
  when a 5.0-kilogram (mplatinum) block of platinum
  (specific heat 0.13 kJ/kgC) at 20C
  (Tplatinum) is placed in contact with a
  10.-kilogram (msilver) block of silver (specific
  heat 0.24 kJ/kgC) at 40C (Tsilver).

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8.3 HEAT AND INTERNAL ENERGY
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8.3 HEAT AND INTERNAL ENERGY

• We see that the equilibrium temperature is much
  closer to the initial temperature of the silver.
• This is a direct result of the fact that the mass
  and specific heat of the silver are larger than
  the mass and specific heat of the platinum.

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Conclusion

• Temperature is the quantity used to measure the
  hotness of a body. A thermometer and an
  appropriate scale are needed to make this
  measurement.

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Conclusion

• Heat energy is the energy associated with changes
  in internal energy.
• If no phase changes occur, the transfer of heat
  energy is accompanied by a change in temperature.

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Conclusion

• The specific heat of a sub stance indicates how
  much heat energy is needed to cause a given
  change in the substances temperature.
• When two objects are brought into contact, heat
  energy will be exchanged until the temperatures
  of the objects are equal.

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Assessment Question 10

• Calculate the equilibrium temperature (Tfinal)
  when a 15 kg (mplatinum) block of platinum
  (specific heat 0.13 kJ/kgC) at 10C
  (Tplatinum) is placed in contact with a 25 kg
  (msilver) block of silver (specific heat 0.24
  kJ/kgC) at 90C (Tsilver).
• Qlost Qgained (cm?T)silver
  (cm?T)platinum
• (0.24 kJ/kgC)(25 kg)(90C-Tfinal) (0.13
  kJ/kgC)(15 kg)(Tfinal-10C)
• 6 kJ/C (90C-Tfinal) 2 kJ/C (Tfinal-10C)
• Tfinal 3(90C 10C) / 2
• A. -15C B. 75C C. 150C D. 300C