Chapter 1
Temperature and Heat
1.1 Temperature is a measure of hotness and can be related to the kinetic energy
of molecules of a substance. A number of physical phenomena can be used for
measuring the temperature of an object. An instrument used for measuring temperature
is called a thermometer and is constructed by using one of the following principles:

the change of length, such as length of a mercury column,

the change of volume, such as volume of a fixed mass of gas at constant
pressure,

the change of pressure, such as pressure of a fixed mass of gas at constant
volume,

the change in electric resistance, as in a thermistor,

the flow of electricity due to Seebeck effect, as in a thermocouple,

the radiation, as in radiation pyrometers.
All thermometers require a scale. This scale should be defined by easily repeatable
circumstances or fundamental properties. For instance the Centigrade scale has been
defined from the melting (0 oC) and boiling ( 100 oC) points of pure water at
atmospheric pressure. For temperature the following units can be used: C, K, F. where
symbol C is for Centigrade (or Celsius), K for Kelvin and F for Fahrenheit.
The Celsius, Fahrenheit, and Kelvin temperature scales
It is easy to derive an equation relating the Celsius degree to Kelvin degree, as follows;
T = (TC + 273.15) K.
[1.1a]
Equation 1.1 shows that the Celsius temperature TC is shifted from the absolute Kelvin
temperature T by 273.15 °. Because the size of a degree is the same on two scales,
temperature difference of 5°C is equal to a temperature difference 5K.The two scales
differ only in the choice of the zero point. Thus, the ice point (273.15 K) corresponds to
0°C, and the steam point (373.15 K) is equivalent 100°C.
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The most common temperature scale in everyday use in the United States is Fahrenheit
scale. This scale sets the temperature of the ice point at 32°F and temperature of the
steam point at 212°F. The relationship between the Celsius and Fahrenheit temperature
scales is
TF = (9/5) TC +32 oF
[1.2b]
Equation 1-2 can be used to find a relationship between changes in temperature
on the Celsius and Fahrenheit scales.
∆ TC =∆T=(5/9) ∆TF
[1.2c]
Where as the change in temperature on the Celsius scale equals the change on Kelvin
scale. That is; ∆T = ∆ TC .
Figure 1.1 Comparison between temperature scales
EXAMPLE 1.1 On a day when the temperature reaches 50°F, what is the
temperature in degrees Celsius and in Kelvin?
Solution Substituting TF = 50°F into Equation 1.2, we get
Tc = (5/9)(TF - 32) =(5/9) (50 - 32) = 10°C
From Equation 1.1, we find that
T=Tc+ 273.15 = 283.15 K
EXAMPLE 1.2
A pan of water is heated from 25°C to 80°e. What is the change in its
temperature on the Kelvin scale and on the Fahrenheit scale?
2
Solution from Equation 1.1, we see that the change in temperature on the Celsius
scale equals the change on the Kelvin scale. Therefore,
∆T = ∆Tc = 80 – 25 =55 oC = 55 K
From Equation 1.3, we find
∆TF= (9/5) ∆Tc = (9/5)(80 – 25) = 99°F
EXAMPLE 1.3
The temperature gradient between the skin and the air is regulated by cutaneous (skin)
blood flow. If the cutaneous blood vessels are constricted, the skin temperature and the
temperature of the environment will be about the same. When the vessels are dilated,
more blood is brought to the surface. Suppose that during dilation the skin warms from
72.0oF to 84.0oF. (a) Convert these temperatures to Celsius and find the difference. (b)
Convert the temperatures to Kelvin, again finding the difference.
Solution
(a) Convert the temperatures from Fahrenheit to Celsius and find the difference.
Convert the lower temperature.
Convert the upper temperature
Find the difference of the two temperatures
(b) Convert the temperatures from Fahrenheit to Kelvin and find their difference.
Convert the temperatures, using the previous answers
T1=22+273.15=295.15 K
T1=29+273.15=303.15 K
T=303.15-295.15=7 K
Remark: The change in temperature in Kelvin and Celsius is the same, as it should be.
Exercise 1.1
Core body temperature can rise from 98.6F to 107oF during extreme exercise, such as a
marathon run. Such elevated temperatures can also be caused by viral or bacterial
infections or tumors and are dangerous if sustained.
3
(a) Convert the given temperatures to Celsius and find the difference. (b) Convert the
temperatures to Kelvin, again finding the difference.
Answer (a) 37.0_C, 41.7_C, 4.7_C (b) 310.2 K, 314.9 K, 4.7
1.2 Temperature and the Zeroth Law of Thermodynamics
Scientists have developed a variety of thermometers for making such quantitative
measurements. In order to understand the concept of temperature, it is useful to first
define two often-used phrases, thermal contact and thermal equilibrium. To understand
the meaning of thermal contact, imagine two objects placed in an insulated container so
that they interact with each other but not with the rest of the world. If the objects are at
different temperatures, energy is exchanged between them. The energy exchanged
between objects because of a temperature difference is called heat. For purposes of the
current discussion, we shall assume that two objects are in thermal contact with each
other if heat can be exchanged between them. Thermal equilibrium is a situation in
which two objects in thermal contact with each other cease to have any exchange of
heat. Now consider two objects, A and C, which are not in thermal contact and a third
object, B, which is our thermometer. We wish to determine whether or not A and C are
in thermal equilibrium with each other. The thermometer (object B) is first placed in
thermal contact with A until thermal equilibrium is reached. From that moment on, the
thermometer's reading remains constant, and we record it. The thermometer is then
removed from A and placed in thermal contact with C, and its reading is recorded after
thermal equilibrium is reached. If the two readings are the same, then A
and C are in thermal equilibrium with each other, [Fig 1.2].
[Figure 1.2] The zeroth law of thermodynamics. (a) and (b): If the temperatures of A and B are
found to be the same as measured by object C (a thermometer), no energy will be exchanged
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1.3 Thermal Expansion of Solids and Liquids
Figure 2-1 The change of mercury volume with temperature
Crystal structure of NaCl
At ordinary temperatures, the atoms vibrate about their equilibrium positions with
amplitude of approximately 10-11 and a frequency of 1013 Hz. The average spacing
between the atoms is approximately 10-10 m. As the temperature of the solid increases,
the atoms vibrate with larger amplitudes and the average separation between them
increases. Consequently, the solid expands. If the thermal expansion of an object is
sufficiently small compared with its initial dimensions, then the change in any
dimension is, to a first approximation, dependent on the first power of the temperature
change. This phenomenon, known as thermal expansion. Ordinarily a substance
expands when heated. If an object has an initial length L0 at some temperature and
undergoes a change in temperature T, its linear dimension changes by the amount L,
which is proportional to the object’s initial length and the temperature change:
]1.3a]
The parameter α is called the coefficient of linear expansion.
The change in area of a substance with change in temperature is given by
.
where γ = 2α is the coefficient of area expansion.
5
[1.3b]
Similarly, the change in volume with temperature of most substances is proportional to
the initial volume V0 and the temperature change ∆T;
[1.3c]
where
β=3αis the coefficient of volume expansion.
The expansion and contraction of material due to changes in temperature creates stresses
and strains, sometimes sufficient to cause fracturing.
For isotropic materials: the coefficient of linear expansion, α, is the same in all
directions.
EXAMPLE 1.4 Global Warming and Coastal Flooding
Goal Apply the volume expansion equation together with linear expansion.
Problem (a) Estimate the fractional change in the volume of Earth’s oceans due to an
average temperature change of 1C. (b) Use the fact that the average depth of the ocean
is 4.00 x 103 m to estimate the change in depth.
Strategy In part (a), solve the volume expansion expression, Equation 10.6, for ∆V/V.
For part (b), use linear expansion to estimate the increase in depth. Neglect the
expansion of landmasses, which would reduce the rise in sea level only slightly.
Solution
(a) Find the fractional change in volume.
Divide the volume expansion equation by V0 and
substitute:
(b) Find the approximate increase in depth.
Use the linear expansion equation. Divide the volume expansion coefficient of water by
three to get the equivalent linear expansion coefficient :
Remarks Three-tenths of a meter may not seem significant, but combined with
increased melting of the polar ice caps, some coastal areas could experience flooding.
An increase of several degrees increases the value of ∆L several times and could
significantly reduce the value of waterfront property.
1.4 Thermal Expansion of Gases
The properties of gases are important in a number of thermodynamic processes. Our
weather is a good example of the types of processes that depend on the behavior of gas.
The equation of state can be very complicated, but is found experimentally to be
relatively simple if the gas is maintained at a low pressure (or a low density). Such a
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low-density gas approximates what is called an ideal gas. Most gases at room
temperature and atmospheric pressure behave approximately as ideal gases. An ideal
gas is a collection of atoms or molecules that move randomly and exert no longrange forces on each other. Each particle of the ideal gas is individually pointlike,
occupying a negligible volume.
To illustrate the thermal expansion in gases, one considers a microscopic model of an
ideal gas to simplify the topic. The model shows that the pressure that a gas exerts on
the walls of its container is a consequence of the collisions of the gas molecules with the
walls.
Avogadro’s number is NA = 6.02 x 1023 particles/mol. A mole of anything, by
definition, consists of an Avogadro’s number of particles. The number is defined so that
one mole of carbon-12 atoms has a mass of exactly 12 g. The mass of one mole of a
pure substance in grams is the same, numerically, as that substance’s atomic (or
molecular) mass.
An ideal gas obeys the equation
[1.4]
where P is the pressure of the gas, V is its volume, n is the number of moles of gas, R is
the universal gas constant
(8.31 J/mol K), and T is the absolute temperature in
kelvins. A real gas at very low pressures behaves approximately as an ideal gas. Solving
problems usually entails comparing two different states of the same system of gas,
dividing the ideal gas equation for the final state by the ideal gas equation for the initial
state, canceling factors that don’t change and solving for the unknown quantity.
EXAMPLE 1.5 An Expanding Gas
Goal Use the ideal gas law to analyze a system of gas.
Problem An ideal gas at 20.0C and a pressure of 1.50 x 105 Pa is in a container having a
volume of 1.00 L. (a) Determine the number of moles of gas in the container. (b) The
gas pushes against a piston, expanding to twice its original volume, while the pressure
falls to atmospheric pressure. Find the final temperature.
Strategy (a) Solve the ideal gas equation of state for the number of moles, n, and
substitute the known quantities. Be sure to convert the temperature from Celsius to
Kelvin! (b) When comparing two states of a gas, it’s often most convenient to divide the
ideal gas equation of the final state by the equation of the initial state. Then quantities
that don’t change can immediately be cancelled, simplifying the algebra.
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Solution
(a) Find the number of moles of gas.
Convert the temperature to kelvins:
Solve the ideal gas law for n and substitute:
(b) Find the temperature after the gas expands to 2.00 L.
Divide the ideal gas law for the final state by the ideal gas law for the initial state:
Cancel the number of moles n and the gas constant R,and solve for Tf
Remark Remember the trick used in part (b), it’s often useful in ideal gas problems.
Notice that it wasn’t necessary to convert units from liters to cubic meters, since the
units were going to cancel anyway.
Exercise 1.2
Suppose the temperature of 4.50 L of ideal gas drops from 375 K to 275 K. (a) If the
volume remains constant and the initial pressure is atmospheric pressure, find the final
pressure. (b) Find the number of moles of gas.
Answer (a) 7.41 x 104 Pa (b) 0.146 mol
1.5 The Kinetic Theory of Gases
Under the conditions of the ideal gas molecules, the equation of pressure, P, of N
molecules contained in a volume V is given by as a function of kinetic energy as follows
[1.5a]
Where
is the average kinetic energy per molecule.
Comparing this equation with the equation of state for an ideal gas in the form of the
equation,
PV = NkBT,
8
we note that the left-hand sides of the two equations are identical. Equating the righthand sides, we obtain
The quantity kB is Boltzmann’s constant (1.38 x10-23 J/K).
This means that the temperature of a gas is a direct measure of the average
molecular kinetic energy of the gas. As the temperature of a gas increases, the
molecules move with higher average kinetic energy.
The internal energy of n moles of a monatomic ideal gas is
[1.5c]
The root-mean-square (rms) speed of the molecules of a gas is
[1.5d]
1.6 Heat and Internal Energy
Internal energy is associated with a system’s microscopic components. Internal energy
includes the kinetic energy of translation, rotation, and vibration of molecules, as well as
potential energy.
Heat is the transfer of energy across the boundary of a system resulting from a
temperature difference between the system and its surroundings. The symbol Q
represents the amount of energy transferred.
Historically, the calorie was the unit used for heat. One calorie is the amount of energy
transfer necessary to raise the temperature of 1 g of water from 14.5oC to 15.5oC. The
“Calorie” used for food is actually 1 kilocalorie
In the US Customary system, the unit is a BTU (British thermal unit).
One BTU is the amount of energy transfer necessary to raise the temperature of 1 lb of
water from 63oF to 64oF. The standard in the text is to use Joules.
The standard in the text is to use Joules. 1 cal = 4.186 J this is known as the mechanical
equivalent of heat.
1.7 Specific heat
The energy required to change the temperature of a substance of mass m by an amount
∆T is
[1.6a]
where c is the specific heat of the substance. In calorimetry problems, the specific heat
of a substance can be determined by placing it in water of known temperature, isolating
the system, and measuring the temperature at equilibrium. The sum of all energy gains
and losses for all the objects in an isolated system is given by
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[1.6b]
where Qk is the energy change in the kth object in the system. This equation can be
solved for the unknown specific heat, or used to determine an equilibrium temperature.
1.8 Latent Heat and Phase Change
The energy required to change the phase of a pure substance of mass m is
[1.7]
where L is the latent heat of the substance. The latent heat of fusion, Lf , describes an
energy transfer during a change from a solid phase to a liquid phase (or vice-versa),
while the latent heat of vaporizaion, Lv, describes an energy transfer during a change
from a liquid phase to a gaseous phase (or vice-versa). Calorimetry problems involving
phase changes are handled with Equation 1.7, with latent heat terms added to the
specific heat terms
1.9 Energy Transfer
Energy can be transferred by several different processes, including work and by
conduction, convection, and radiation.
Conduction can be viewed as an exchange of kinetic energy between colliding
molecules or electrons. The rate at which energy transfers by conduction through a slab
of area A and thickness L is
[1.8a]
where k is the thermal conductivity of the material making up the slab. Some useful
conductivities are given in the following table:
substance
conductivity
air (0 C)
.000057 cal cm / s cm 2 K
H 2 O (20 C) .0014 cal cm / s cm 2 K
Cu
.99 cal cm / s cm 2 K
tissue
18 kcal cm / m 2 hr K
fur/down
.36 kcal cm / m 2 hr K
Convection is the movement of heat by currents in the medium, i.e., the wind. The
convection current in Watts is (empirically)
10
Q / t = 14.5 A Sqrt (v) T .
[1.8b]
when A is measured in square meters and v is the (wind) speed in m / s. Still air actually
has a convection velocity of .23 m / s (called "natural convection") because warm air
rises.
Radiation is the emission of electromagnetic energy (which your body does in the
infrared wavelengths). The radiation current (in Watts) is
Q / t = A (Tb 4 - Ta 4)
[1.8c]
where  is the "emissivity" (a dimensionless radiation "effectiveness", which is .97 for
human skin independent of color, under equilibrium conditions),  is the "StefanBoltzmann" constant (5.67 x 10 - 8 W / m 2 K 4) and the temperature MUST be in K (due
to the fourth power dependence; "b" denotes body, while "a" denotes "ambient", or air,
temperature).
1.10 Work in Thermodynamic Processes
The work done on a gas at a constant pressure is
[1.9]
The work done on the gas is positive if the gas is compressed (∆V is negative) and
negative if the gas expands (∆V is positive). In general, the work done on a gas that
takes it from some initial state to some final state is the negative of the area under the
curve on a PV diagram.
1.11 The First Law of Thermodynamics
According to the first law of thermodynamics, when a system undergoes a change from
one state to another, the change in its internal energy ∆U is
[1.10a]
where Q is the energy transferred into the system by heat and W is the work done on the
system. Q is positive when energy enters the system by heat and negative when the
system loses energy. W is positive when work is done on the system (for example, by
compression) and negative when the system does positive work on its environment.
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The energetic of the body.
Conservation of energy is implicit in all our calculations of energy balance in living
systems. Consider, for example, the energetic for the functioning of an animal. The
body of an animal contains internal thermal energy Et, which is the product of the mass
and specific heat, and chemical energy Ec stored in the tissue of the body. In terms of
energy, the activities of an animal consist of simply eating, working, and rejecting
excess heat by means of various cooling mechanisms (radiation, convection, etc.).
Without going into detailed calculations, the first law allows us to draw some
conclusions about the energetic of the animal. For example, if the internal temperature
and the weight of the animal are to remain constant (i.e., Ec and Et constant), over a
given period of time the energy intake must be exactly equal to the sum of the work
done and the heat lost by the body. An imbalance between intake and output energy
implies a change in the sum Ec +Et.
1.12 Applications
1.12.a Human metabolism
Metabolism is the set of chemical reactions that
happen in living organisms to maintain life. These
processes allow organisms to grow and reproduce,
maintain their structures, and respond to their
environments.
The chemical energy used by animals is obtained
from the oxidation of food molecules. The glucose
sugar molecule, for example, is oxidized as follows:
C6H12O6 +6O2 →6CO2 +6H2O+energy
For every gram of glucose ingested by the body, 3.81 Cal of energy is released for
metabolic use.
Animals do work and give off energy by heat and this lead us to believe the first
law of thermodynamics can be applied to living organisms to describe them in a general
way.
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The internal energy stored in humans goes into other forms needed for maintaining and
repairing the major body organs and is transferred out of the body by work as a person
walks or lifts a heavy object, and by heat when the body is warmer than its
surroundings. Because the rates of change of internal energy, energy loss by heat, and
energy loss by work vary widely with the intensity and duration of human activity, it’s
best to measure the time rates of change of ∆U, Q, and W. Rewriting the first law, these
time rates of change are related by
[1.12a]
On average, energy Q flows out of the body, and work is done by the body on its
surroundings, so both Q/∆t and W/∆t are negative. This means that ∆U/∆t would be
negative and the internal energy and body temperature would decrease with time if a
human were a closed system with no way of ingesting matter or replenishing internal
energy stores. Because all animals are actually open systems, they acquire internal
energy (chemical potential energy) by eating and breathing, so their internal energy and
temperature are kept constant. Overall, the energy from the oxidation of food ultimately
supplies the work done by the body and energy lost from the body by heat, and this is
the interpretation we give the last equation. That is, ∆U/∆t is the rate at which internal
energy is added to our bodies by food, and this term just balances the rate of energy loss
by heat, Q/∆t, and by work, W/∆t. Finally, if we have a way of measuring ∆U/∆t and
W/∆t for a human, we can calculate Q/∆t and gain useful information on the efficiency
of the body as a machine.
Composition and Energy Content Common of Some Foods
1.12.b Body Temperature Regulation
Since you are a warm-blooded animal, your body attempts to keep its internal
temperature constant. Human life is only compatible with a narrow range of
temperatures:
13
Temperature (C)
Symptoms
28
muscle failure
30
loss of body temp. control
33
loss of consciousness
37
normal
42
central nervous system breakdown
44
death
Humans are constantly generating heat, and so your body must take active steps to lose
that heat. The following table illustrates the power cost of various common activities:
Activity
Energy Cost (Cal/m 2 hr)
sleeping
35
sitting
50
working at a desk
60
standing
85
washing & dressing
100
walking (3 mph)
140
bicycling
250
swimming
350
running
600
Metabolic Rates for Selected Activities
Approximately 80 % of these costs are waste heat. The other side of this coin is cold
weather: your body must then work to stay warm. The mechanisms which either are
used by your body or affect its function are "conduction", "convection", "radiation"
and "evaporation".
Assume that you walk at 2.2 mph on flat ground. At this speed, an average person burns
3.33 kcal / min, 80% of which must be lost in heat. Consider first the conduction of heat
from the center of your body to the skin. Assuming that the average area (inside the
body) through which heat is conducted is 1 square meter and that the average distance
the heat must travel is 10 cm, the temperature difference necessary to maintain normal
body temperature is about 3 oC. Clearly your body cannot rely on conduction for this
service. Now consider the conduction of heat away from the skin. Due to the nature of
the surface of your body, it has a "private climate" about 3 mm deep through which the
temperature changes from skin temperature to the surrounding air temperature. At room
temperature, a person with 2 square meters of body surface area must (when nude) have
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a skin temperature of almost 32 oC when the air is still. This is actually a pretty
reasonable estimate.
The tissue of the body, without blood flowing through it, is a poor conductor. Its thermal
conductivity is comparable to that of cork. (Kc for tissue without blood is 18 Cal /m2-hro
C.) Simple thermal conductivity through tissue is inadequate for elimination of the
excess heat generated by the body. The following calculation illustrates this point.
Assume that the thickness of the tissue between the interior and the exterior of the body
is 3 cm and that the average area through which conduction can occur is 1.5m2. With a
temperature difference T between the inner body and the skin of 2oC, the
heat flow H per hour is,
[1.12b]
In order to increase the conductive heat flow to a moderate level of say 50 Cal/hr, the
temperature difference between the interior body and the skin would have to increase to
about 17 C◦.
For the body in air, convection is in series with private climate conduction. Within the
body, blood convection is used to move the heat from the inside of your body to your
skin. Here the area is the surface area of the capillary bed, which for the average adult
male is about 160 square meters. Using the skin temperature and heat current above, we
see that the blood flow must be around .3 mm / s, which is the correct order of
magnitude. Since the specific heat of blood is larger than that of air, we expect the
thermal current to be larger for blood, and hence the velocity to be smaller than this
estimate.
Q / t = 14.5 A Sqrt (v) T
when A is measured in square meters and v is the (wind) speed in m / s. Still air actually
has a convection velocity of .23 m / s (called "natural convection") because warm air
rises.
Radiation is the emission of electromagnetic energy (which your body does in the
infrared wavelengths). The radiation current (in Watts) is
ΔQ / Δt = ε σ A (Tb 4 - Ta 4)
where ε is the "emissivity" (a dimensionless radiation "effectiveness", which is .97 for
human skin independent of color, under equilibrium conditions), σ is the "StefanBoltzmann" constant (5.67 * 10 - 8 W / m 2 K 4) and the temperature MUST be in K (due
to the fourth power dependence; "b" denotes body, while "a" denotes "ambient", or air,
temperature). In the above scenario, your body's radiation power output is only about
140 W.
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When considering radiation absorbed by the skin from the sun, the emissivity (which is
equal to the absorbency) depends on frequency and therefore on skin color (we know
this is not an equilibrium situation, because many people can get severe sun burns!).
Using data on the reflectance of human skin as a function of wavelength (where
reflectance is 1 - ), we can construct a weighted average emissivity for various skin
colors (weighted by the solar power output as a function of wavelength). Using this
data, Caucasian skin has a weighted average  of .566, while negroid skin has a
weighted average  of .838. Evaporation is of course simply the change of phase of
sweat. The rate of sweat is then related to the thermal current by the latent heat of
vaporization
Q / t = (m / t) L
.
[1.12c]

At body temperature, the latent heat of vaporization of water is 580 cal / g. For short
periods, you can sweat up to four liters per hour; for longer periods (up to 6 hours), 1
liter per hour is common. In addition to sweat, however, your body also loses water
vapor during respiration. The volume of air which you inhale with each breath must be
humidified by your body to saturation in order to be used efficiently. This vapor is then
exhaled, resulting in an evaporative loss which at high altitudes can rival sweat as a
cooling factor. This makes evaporation a major contributor to heat regulation, up to a
point: body functions are severely limited when you have lost 10% of your weight due
to dehydration.
Your body has a number of mechanisms to help it cope with cold weather. Constriction
of surface capillaries is helpful when the ambient temperature is above 19 C (for a nude
person). Shivering raises the average person's metabolic rate about 250 kcal / m2 hr
(relative of course to body surface area). In fact, for any well-insulated animal,
evaporative losses in breathing limits the ability to withstand cold temperatures.
We can summarize the various modes of heat transfer with the following diagram:
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EXAMPLE 1.6 Fighting Fat
PROBLEMS
1-The melting point of gold is l064°C, and the boiling point is 2660°C.
(a) Express these temperatures in Kelvin.
(b) Compute the difference between these temperatures in Celsius degrees and Kelvin.
2- Lake Erie contains roughly 4.00 x1011 m3 of water. (a) How much energy is required
to raise the temperature of that volume of water from 11.0°C to 12.0°C? (b) How many
years would it take to supply this amount of energy by using the 1 000-MW exhaust
energy of an electric power plant?
3- Liquid nitrogen has a boiling point of - 195.81 °C at atmospheric pressure. Express
this temperature in (a) degrees Fahrenheit and (b) Kelvin.
4- How much energy is required to change a 40-g ice cube from ice at -10°C to steam at
110°C?
1.22x105 j
5-A 200-g aluminum cup contains 800 g of water in thermal equilibrium with the cup at
80°C. The combination of cup and water is cooled uniformly so that the temperature
decreases by 1.5°C per minute. At what rate is energy being removed? Express your
answer in watts.
6- How long can a man survive in an airtight room that has a volume of 27m3. Assume
that his surface area is 1.70m2. Use data provided in the text.
Ans t = 373 hours
7- Calculate the length of time that a person can survive without food but with adequate
water. Obtain a solution under the following assumptions:
(a) The initial weight and surface area of the person are 70 kg and 1.70m2, respectively.
(b) The survival limit is reached when the person loses one-half his or her body weight.
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(c) Initially the body contains 5 kg of fatty tissue.
(d)During the fast the person sleeps 8 hr/day and rests quietly the remainder of the
time. (e) As the person loses weight, his or her surface area decreases. However, here
we assume that the surface area remains unchanged.. t = 105 days
8- Suppose that a person of weight 60 kg and height 1.4m reduces her sleep by 1 hr/day
and spends this extra time reading while sitting upright. If her food intake remains
unchanged, how much weight will she lose in one year? Weight loss = 0.892 kg
9- Compute the heat loss per square meter of skin surface at −40oC in moderate wind
(about 0.5m/sec, Kc =10 Cal/m2- hr-oC). Assume that the skin temperature is 26oC.
Heat loss = 660 Cal/m2-h
10. A pair of eyeglass frames are made of epoxy plastic (coefficient of linear expansion
=1.30 x 10-4 oC-1). At room temperature (20.0oC), the frames have circular lens holes
2.20 cm in radius. To what temperature must the frames be heated if lenses 2.21 cm in
radius are to be inserted into them?
11. A gold ring has an inner diameter of 2.168 cm at a temperature of 15.00C.
Determine its inner diameter at 100oC ( gold = 1.42 x 10-5 oC-1).
12. The band in the Figure is stainless steel (coefficient of linear expansion _
17.3x10-6 oC-1; Young’s modulus 18 x 1010 N/m2). It is essentially circular with
an initial mean radius of 5.0 mm, a height of 4.0 mm, and a thickness of 0.50
mm. If the band just fits snugly over the tooth when heated to a temperature of
80oC, what is the tension in the band when it cools to a temperature of 37oC?
13-A popular brand of cola contains 6.50 g of carbon dioxide dissolved in 1.00 L of soft
drink. If the evaporating carbon dioxide is trapped in a cylinder at 1.00 atm and 20.0oC,
what volume does the gas occupy?
14. If 9.00 g of water is placed in a 2.00-L pressure cooker and heated to 500oC, what is
the pressure inside the container?
15. The human body must maintain its core temperature inside a rather narrow range
around 37°C. Metabolic processes (notably, muscular exertion) convert chemical
energy into internal energy deep in the interior. From the interior, energy must flow out
to the skin or lungs, to be lost by heat to the environment. During moderate exercise, an
80-kg man can metabolize food energy at the rate of 300 kcal/h, do 60 kcal/h of
mechanical work, and put out the remaining 240 kcal/h of energy by heat. Most of the
energy is carried from the interior of the body out to the skin by “forced convection” (as
a plumber would say): Blood is warmed in the interior and then cooled at the skin,
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which is a few degrees cooler than the body core. Without blood flow, living tissue is a
good thermal insulator, with a thermal conductivity about 0.210 W/m_ °C.
Show that blood flow is essential to keeping the body cool by calculating the rate of
energy conduction, in kcal/h, through the tissue layer under the skin. Assume that its
area is 1.40 m2, its thickness is 2.50 cm, and it is maintained at 37.0°C on one side and
at 34.0°C on the other side.
16. A “solar cooker” consists of a curved reflecting mirror that
focuses sunlight onto the object to be heated. The solar power per
unit area reaching the Earth at the location of a 0.50-m-diameter
solar cooker is 600 W/m2. Assuming that 50% of the incident
energy is converted to thermal energy, how long would it take to
boil away 1.0 L of water initially at 20°C? (Neglect the specific
heat of the container.)
17-The evaporation of perspiration is the primary mechanism for cooling the human
body. Estimate the amount of water you will lose when you bake in the sun on the beach
for an hour. Use a value of 1 000 W/m2 for the intensity of sunlight, and note that the
energy required to evaporate a liquid at a particular temperature is approximately
equal to the sum of the energy required to raise its temperature to the boiling point and
the latent heat of vaporization (determined at the boiling point).
18- (a) Find the number of moles in one cubic meter of an ideal gas at 20.0°C and
atmospheric pressure. (b) For air, Avogadro’s number of molecules has mass 28.9 g.
Calculate the mass of one cubic meter of air. Compare the result with the tabulated
density of air.
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