# Chapter 19

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```Chapter 19
Heat and the First Law of
Thermodynamics
Units of Chapter 19
• Heat as Energy Transfer
• Internal Energy
• Specific Heat
• Calorimetry—Solving Problems
• Latent Heat
• The First Law of Thermodynamics
• The First Law of Thermodynamics Applied;
Calculating the Work
Units of Chapter 19
• Molar Specific Heats for Gases, and the
Equipartition of Energy
• Adiabatic Expansion of a Gas
• Heat Transfer: Conduction, Convection,
19-1 Heat as Energy Transfer
We often speak of heat as though it were a
material that flows from one object to another; it
is not. Rather, it is a form of energy.
Unit of heat: calorie (cal)
1 cal is the amount of heat necessary to raise the
temperature of 1 g of water by 1 Celsius degree.
Don’t be fooled—the calories on our food labels
are really kilocalories (kcal or Calories), the heat
necessary to raise 1 kg of water by 1 Celsius
degree.
19-1 Heat as Energy Transfer
If heat is a form of energy, it ought to be possible
to equate it to other forms. The experiment below
found the mechanical equivalent of heat by using
the falling weight to heat the water:
4.186 J = 1 cal
4.186 kJ = 1 kcal
19-1 Heat as Energy Transfer
Definition of heat:
Heat is energy transferred from one object to
another because of a difference in temperature.
• Remember that the temperature of a gas is
a measure of the kinetic energy of its
molecules.
19-1 Heat as Energy Transfer
Example 19-1: Working off the extra calories.
Suppose you throw caution to the wind and
eat too much ice cream and cake on the order
of 500 Calories. To compensate, you want to
do an equivalent amount of work climbing
stairs or a mountain. How much total height
must you climb?
19-2 Internal Energy
The sum total of all the energy of all the
molecules in a substance is its internal (or
thermal) energy.
Temperature: measures molecules’ average
kinetic energy
Internal energy: total energy of all molecules
Heat: transfer of energy due to difference in
temperature
19-2 Internal Energy
Internal energy of an ideal (atomic) gas:
But since we know the average kinetic energy
in terms of the temperature, we can write:
19-2 Internal Energy
If the gas is molecular
rather than atomic,
rotational and
vibrational kinetic
energy need to be
taken into account as
well.
19-3 Specific Heat
The amount of heat required
to change the temperature
of a material is proportional
to the mass and to the
temperature change:
The specific heat, c, is
characteristic of the
material. Some values are
listed at left.
19-3 Specific Heat
Example 19-2: How heat transferred
depends on specific heat.
(a) How much heat input is needed to
raise the temperature of an empty 20-kg
vat made of iron from 10°C to 90°C?
(b) What if the vat is filled with 20 kg of
water?
19-4 Calorimetry—Solving Problems
Closed system: no mass enters or leaves, but
energy may be exchanged
Open system: mass may transfer as well
Isolated system: closed system in which no
energy in any form is transferred
For an isolated system,
energy out of one part = energy into another part,
or:
heat lost = heat gained.
19-4 Calorimetry—Solving Problems
Example 19-3: The cup cools the tea.
If 200 cm3 of tea at 95°C is poured into a 150-g
glass cup initially at 25°C, what will be the
common final temperature T of the tea and cup
when equilibrium is reached, assuming no heat
flows to the surroundings?
19-4 Calorimetry—Solving Problems
The instrument to the left
is a calorimeter, which
makes quantitative
measurements of heat
exchange. A sample is
heated to a well-measured
high temperature and
plunged into the water,
and the equilibrium
temperature is measured.
This gives the specific
heat of the sample.
19-4 Calorimetry—Solving Problems
Example 19-4: Unknown specific heat
determined by calorimetry.
An engineer wishes to determine the specific
heat of a new metal alloy. A 0.150-kg sample of
the alloy is heated to 540°C. It is then quickly
placed in 0.400 kg of water at 10.0°C, which is
contained in a 0.200-kg aluminum calorimeter
cup. (We do not need to know the mass of the
insulating jacket since we assume the air space
between it and the cup insulates it well, so that
its temperature does not change significantly.)
The final temperature of the system is 30.5°C.
Calculate the specific heat of the alloy.
19-5 Latent Heat
Energy is required for a material to change
phase, even though its temperature is not
changing.
19-5 Latent Heat
Heat of fusion, LF: heat required to change
1.0 kg of material from solid to liquid
Heat of vaporization, LV: heat required to
change 1.0 kg of material from liquid to vapor
19-5 Latent Heat
The total heat required for a phase change
depends on the total mass and the latent heat:
Example 19-5: Will all the ice melt?
A 0.50-kg chunk of ice at -10°C is placed in
3.0 kg of “iced” tea at 20°C. At what
temperature and in what phase will the final
mixture be? The tea can be considered as
water. Ignore any heat flow to the
surroundings, including the container.
19-5 Latent Heat
Problem Solving: Calorimetry
1. Is the system isolated? Are all significant
sources of energy transfer known or
calculable?
2. Apply conservation of energy.
3. If no phase changes occur, the heat
transferred will depend on the mass,
specific heat, and temperature change.
(continued)
19-5 Latent Heat
4. If there are, or may be, phase changes,
terms that depend on the mass and the
latent heat may also be present. Determine
or estimate what phase the final system will
be in.
5. Make sure that each term is in the right
place and that all the temperature changes
are positive.
6. There is only one final temperature when
the system reaches equilibrium.
7. Solve.
19-5 Latent Heat
Example 19-6: Determining a latent heat.
The specific heat of liquid mercury is
140 J/kg·°C. When 1.0 kg of solid
mercury at its melting point of -39°C is
placed in a 0.50-kg aluminum
calorimeter filled with 1.2 kg of water at
20.0°C, the mercury melts and the final
temperature of the combination is found
to be 16.5°C. What is the heat of fusion
of mercury in J/kg?
19-5 Latent Heat
The latent heat of vaporization is relevant
for evaporation as well as boiling. The heat
of vaporization of water rises slightly as
the temperature decreases.
On a molecular level, the heat added
during a change of state does not go to
increasing the kinetic energy of individual
molecules, but rather to breaking the close
bonds between them so the next phase
can occur.
19-6 The First Law of Thermodynamics
The change in internal energy of a closed
system will be equal to the energy added to the
system minus the work done by the system on
its surroundings.
This is the law of conservation of energy,
written in a form useful to systems involving
heat transfer.
19-6 The First Law of Thermodynamics
Example 19-7: Using the first law.
2500 J of heat is added to a system, and
1800 J of work is done on the system.
What is the change in internal energy of
the system?
19-6 The First Law of Thermodynamics
The first law can be extended to include
changes in mechanical energy—kinetic energy
and potential energy:
Example 19-8: Kinetic energy transformed to
thermal energy.
A 3.0-g bullet traveling at a speed of 400 m/s
enters a tree and exits the other side with a
speed of 200 m/s. Where did the bullet’s lost
kinetic energy go, and what was the energy
transferred?
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
An isothermal process is one
in which the temperature
does not change.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
In order for an isothermal process to take
place, we assume the system is in contact
with a heat reservoir.
In general, we assume that the system
remains in equilibrium throughout all
processes.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
An adiabatic process is one in which there is
no heat flow into or out of the system.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
An isobaric process (a) occurs at constant
pressure; an isovolumetric one (b) occurs at
constant volume.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
The work done in moving a
piston by an infinitesimal
displacement is:
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
For an isothermal process, P = nRT/V.
Integrating to find the work done in taking the
gas from point A to point B gives:
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
A different path takes the gas first from A to
D in an isovolumetric process; because the
volume does not change, no work is done.
Then the gas goes from D to B at constant
pressure; with constant pressure no
integration is needed, and W = PΔV.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
Conceptual Example 19-9: Work in isothermal
Reproduced here is the PV diagram for a gas
expanding in two ways, isothermally and
adiabatically. The initial volume VA was the same
in each case, and the final volumes were the
same (VB = VC). In which process was more work
done by the gas?
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
Example 19-10: First law in isobaric and isovolumetric processes.
An ideal gas is slowly compressed at a constant pressure of 2.0 atm
from 10.0 L to 2.0 L. (In this process, some heat flows out of the gas
and the temperature drops.) Heat is then added to the gas, holding
the volume constant, and the pressure and temperature are allowed
to rise (line DA) until the temperature reaches its original value (TA =
TB). Calculate (a) the total work done by the gas in the process BDA,
and (b) the total heat flow into the gas.
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
Example 19-11: Work done in an engine.
In an engine, 0.25 mol of an ideal
monatomic gas in the cylinder expands
piston. In the process, the temperature of
the gas drops from 1150 K to 400 K. How
much work does the gas do?
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
The following is a simple summary of the
various thermodynamic processes.
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
For gases, the specific heat depends on the
process—the isothermal specific heat is
different from the isovolumetric one.
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
In this table, we see that the specific heats for
gases with the same number of molecules are
almost the same, and that the difference CP – CV
is almost exactly equal to 2 in all cases.
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
For a gas in a constant-volume process, no
work is done, so QV = ΔEint.
For a gas at constant pressure, QP = ΔEint + PΔV.
Comparing these two processes for a
monatomic gas when the temperature change
is the same gives
which is consistent with the measured values.
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
we expect that
This is also in agreement with
measurements.
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
For a gas consisting of more complex
molecules (diatomic or more), the molar
specific heats increase. This is due to the
extra forms of internal energy that are
possible (rotational, vibrational).
19-8 Molar Specific Heats for Gases, and
the Equipartition of Energy
Each mode of vibration or rotation is called a
degree of freedom. The equipartition theorem
states that the total internal energy is shared
equally among the active degrees of freedom, each
accounting for ½ kT. The actual measurements
show a more complicated situation.
19-8 Molar Specific Heats for Gases, and
the Equipartition of Energy
For solids at high temperatures, CV is
approximately 3R, corresponding to six degrees
of freedom (three kinetic energy and three
vibrational potential energy) for each atom.
19-9 Adiabatic Expansion of a Gas
For an adiabatic expansion, dEint = -PdV,
since there is no heat transfer.
From the relationship between the change
in internal energy and the molar heat
capacity, dEint = nCVdT.
From the ideal gas law, PdV + VdP = nRdT.
Combining and rearranging gives
(CP/CV)PdV + VdP = 0.
19-9 Adiabatic Expansion of a Gas
Define:
Integration then gives the result:
19-9 Adiabatic Expansion of a Gas
Example 19-12: Compressing an ideal gas.
An ideal monatomic gas is compressed starting at point
A, where PA = 100 kPa, VA = 1.00 m3, and TA = 300 K. The
gas is first compressed adiabatically to state B (PB = 200
kPa). The gas is then further compressed from point B to
point C (VC = 0.50 m3) in an isothermal process. (a)
Determine VB. (b) Calculate the work done on the gas for
the whole process.
19-10 Heat Transfer: Conduction,
Heat conduction can be visualized as
occurring through molecular collisions.
The heat flow per unit time is given by:
19-10 Heat Transfer: Conduction,
The constant k is called the
thermal conductivity.
Materials with large k are
called conductors; those
with small k are called
insulators.
19-10 Heat Transfer: Conduction,
Example 19-13: Heat loss
through windows.
A major source of heat loss
from a house is through the
windows. Calculate the rate of
heat flow through a glass
window 2.0 m x 1.5 m in area
and 3.2 mm thick, if the
temperatures at the inner and
outer surfaces are 15.0°C and
14.0°C, respectively.
19-10 Heat Transfer: Conduction,
Building materials are measured using Rvalues rather than thermal conductivity:
Here, l is the thickness of the material.
19-10 Heat Transfer: Conduction,
Convection occurs when heat flows by the mass
movement of molecules from one place to
another. It may be natural or forced; both these
examples are natural convection.
19-10 Heat Transfer: Conduction,
Radiation is the form of energy transfer we
receive from the Sun; if you stand close to
a fire, most of the heat you feel is radiated
as well.
The energy radiated has been found to be
proportional to the fourth power of the
temperature:
19-10 Heat Transfer: Conduction,
The constant σ is called the Stefan-Boltzmann
constant:
The emissivity ε is a number between 0 and 1
characterizing the surface; black objects
have an emissivity near 1, while shiny ones
have an emissivity near 0. It is the same for
absorption; a good emitter is also a good
absorber.
19-10 Heat Transfer: Conduction,
An athlete is sitting unclothed in a locker
room whose dark walls are at a
temperature of 15°C. Estimate his rate of
heat loss by radiation, assuming a skin
temperature of 34°C and ε = 0.70. Take
the surface area of the body not in
contact with the chair to be 1.5 m2.
19-10 Heat Transfer: Conduction,
If you are in the sunlight, the Sun’s radiation will
warm you. In general, you will not be perfectly
perpendicular to the Sun’s rays, and will absorb
energy at the rate:
19-10 Heat Transfer: Conduction,
This cos θ effect is also
responsible for the
seasons.
19-10 Heat Transfer: Conduction,
Thermography—the detailed measurement of
radiation from the body—can be used in
medical imaging. Warmer areas may be a sign
of tumors or infection; cooler areas on the
skin may be a sign of poor circulation.
19-10 Heat Transfer: Conduction,
The giant star Betelgeuse emits radiant
energy at a rate 104 times greater than our
Sun, whereas its surface temperature is
only half (2900 K) that of our Sun. Estimate
the radius of Betelgeuse, assuming ε = 1 for
both. The Sun’s radius is rS = 7 x 108 m.
Summary of Chapter 19
• Internal energy, Eint, refers to the total energy
of all molecules in an object. For an ideal
monatomic gas,
• Heat is the transfer of energy from one object
to another due to a temperature difference. Heat
can be measured in joules or in calories.
• Specific heat of a substance is the energy
required to change the temperature of a fixed
amount of matter by 1°C.
Summary of Chapter 19
• In an isolated system, heat gained by one part
of the system must be lost by another.
• Calorimetry measures heat exchange
quantitatively.
• Phase changes require energy even though
the temperature does not change.
• Heat of fusion: amount of energy required to
melt 1 kg of material
• Heat of vaporization: amount of energy
required to change 1 kg of material from liquid
to vapor
Summary of Chapter 19
• The first law of thermodynamics:
ΔEint = Q – W.
• Thermodynamic processes: adiabatic (no heat
transfer), isothermal (constant temperature),
isobaric (constant pressure), isovolumetric
(constant volume).
• Work done:
dW = PdV.
• Molar specific heats:
CP – CV = R.
Summary of Chapter 19
• Heat transfer takes place by conduction,