Chapter 17
Temperature and Heat
by C.-R. Hu
1. Temperature
A (physical) system − A piece of solid, a liquid in an open or closed container, or a gas in a closed container.
Temperature is a physical property of a (physical) system. It measures how
much the atoms or molecules in the system are energetically excited. It is an
indicator of the common sense usually referred to as hot or cold.
Thermometer − A device to measure temperature.
Thermometry − Appropriate arrangements to measure temperature accurately. (That is, setting up an accurate thermometer and using it in the correct
way to measure the temperature of a physical system.)
Microscopic view − Looking at a system from the point of view of the large
number of atoms and molecules in it.
Macroscopic view − Looking at a system containing a large number of
atoms or molecules from the point of view of a small number of measurable
quantities, called the state variables. These variables include volume, number of atoms or molecules of each kind in the system, pressure, and temperature, and possibly more, if the system is more complicated. Each such variable is about the whole system rather than just a few atoms or molecules in
the system.
Brownian motion (or movement) − Each atom or molecule in a solid is
moving rapidly and randomly in a small region of space. In a liquid or gas
the atoms or molecules are also rapidly moving, but no longer limited to a
small region of space. But the movement is still random, is different even
between neighboring atoms or molecules, and changes direction frequently
due to interaction with other atoms or molecules. (Non-random movement
is called a flow for a liquid or gas, and translation for a solid.) The higher
is the temperature of the system, the more rapid is this random motion. It
can be seen by observing the behavior of a tiny speck of dust in a liquid or
gas using a microscope. Temperature is a measure of the energy associated
with this random motion (per atom or molecule).
Heat − energy transfer across a surface inside a material or across the contact surface between two materials, due the temperature of the two sides
being different. Heat is not a content, but is forever a flow, a transfer of
energy, when no movement of the surface or interface is involved, therefore
no apparent work is done by either side toward the other side. This energy
transfer is achieved with the Brownian motion ― explained in chapter 18
under the title “concept of diffusion” ― of the atoms or molecules. When
the temperatures of the two sides are different, the energies associated with
the Brownian motion on the two sides are different. Being associated with
random motion, the side with higher energy of this kind will clearly pass
energy to the side with lower energy of this kind. Thus heat energy will always flow from the higher temperature side across an interface to the lower
temperature side, and never the other way around!
Thermodynamics − The branch of physics to study heat flow and how to utilize it to get useful work. That is, how to make random motion of the
atoms or molecules do useful work for you, or to extract useful energy out of
this random motion. It is based on a macroscopic and phenomenological
approach, and so part of the logic is based on empirical facts, and the remaining logic is built on axioms (i.e., hypotheses). ( To see why the axioms
are correct, you have to study another branch of physics called statistical
mechanics. Within thermodynamics you rely on the agreements of its predictions with experimental measurements to justify the axioms. )
Celcius degree (ºC) − A temperature scale defined by setting the freezing
point of water at 1 atmospheric pressure (atm) to be 0ºC, and the boiling
point of water at 1 atm to be 100ºC. (Their difference is 100ºC.)
Fahrenheit degree (ºF) − Another temperature scale, defined by setting the
freezing point of water at 1 atm to be 32ºF, and the boiling point of water at
1 atm to be 212ºF. (Their difference is 180ºF.)
D
Relation between TC and TF : TC = (5 / 9)(TF − 32 ) ,
TF = (9 / 5)TC + 32D .
(Here TC , TF are temperature in Celcius and Fahrenheit scales, respectively.)
Phases − The term phase in thermodynamics has absolutely nothing to do
with the term phase in the discussion of waves. H2O can exist as ice (the
solid phase), water (the liquid phase) and water vapor (the gas phase).
Most materials can exist in all three phases, but helium can become a solid
only under high pressure (about 25 atm).
Freezing point of water or melting point of ice − This is the temperature of
a mixture of ice and water after a long wait, assuming that there are plenty of
both, so that at the end there are still both in coexistence. If the temperature
is initially above this point, then no ice can exist, and water can freeze only
after the temperature has dropped to or below this point. If the temperature is
initially below this point, then no water can exist, and ice can melt only after
the temperature has risen to or above this point. This is called phase transition, or phase change, change of phase. The coexistence of ice and water
at the melting/freezing point is called phase equilibrium. At P = 1 atm, this
melt-ing/freezing point for water is 0 °C. As P increases, this melting/freezing point for H2O actually decreases, but the opposite is often true for other
materials.
Boiling point of water or condensation point of water vapor − This is the
temperature of a mixture of water and water vapor after a long wait, assuming that there are plenty of both, so that at the end there are still both in coexistence. If the temperature is initially below this point, then no water vapor can exist, and water can evaporate only after temperature has risen to or
above this point. If the temperature is initially above this point, then no water can exist, and water vapor can condense into water only after the temperature has dropped to or below this point. This is also a phase transition, or
phase change. The coexistence of water and water vapor at the boiling/condensation point is another example of a phase equilibrium. At P = 1 atm,
this boiling/condensation point for water is 100 °C. As P increases, this
boiling/condensation point for H2O actually increases. This is like also true
with a lot of other materials.
Note 1: If you have just mixed some ice and water, the ice and water may be
at different temperatures. You must wait for a while before they reach the
same temperature, possibly after some ice melts or some water freezes. Similarly, if you have just mixed some water and water vapor, the water and
water vapor may be at different temperatures. You must wait for a while before they reach the same temperature, possibly after some water evaporates
or some water vapor condenses into water.
Note 2: When water freezes into ice, or when water vapor condenses into
water, heat is released. On the other hand, when ice melts, or when water
evaporates, heat must be absorbed. This is why after a long wait, a mixture
of water and ice will no longer change their ratio. If some more water
freezes, the heat released will melt exactly the same amount of ice, to keep
the amount of ice and water unchanged. This is not true if water and ice are
not at the same temperature. (For example, if ice is already at 0ºC, but water
is still at 10ºC, then some ice will melt, but no water will freeze. It will only
lower its temperature toward 0ºC. Whether it will actually reach 0ºC depends on how much ice is initially present.) Similar statements can be made
for a mixture of water and water vapor.
Note 3: Freezing point (= melting point) and boiling point (= condensation
point) are pressure dependent. This is why boiling water for a long time at
high altitude still may not kill all the germs, since at pressure much lower
than 1 atm, the boiling point of water can be much lower than 100ºC, and the
freezing point of water can be substantially higher than 0ºC. On the other
hand, if pressure is raised above 1 atm, then the boiling point of water can be
much higher than 100ºC, and the freezing point of water can be substantially
lower than 0ºC. A pressure cooker is designed to take advantage of this fact.
How to remember these facts? The easiest way is to remember the following picture:
1 atm
freezing
0°C
boiling
100 °C
T
Sublimation − The above picture is only for pressure above the triplepoint pressure. At the triple point (for water, it is at
Ptp = 0.00610×105 Pa = 0.00602 atm = 4.5765 torr , and Ttp = 273.16 K )
the solid, liquid, and gas phases of the same substance can coexist at thermal
equilibrium. For pressure at or below the triple point pressure, the solid phase exists below a (pressure-dependent) sublimation temperature, the gas
phase exists above this temperature, and the liquid phase does not exist at
any temperature! so as temperature rises above this (pressure-dependent)
particular temperature from below, the solid phase (such as ice) will directly
convert to the gas phase (such as water vapor), without going through the
liquid phase. This is called sublimation. We say that the solid phase has
sublimed to the gas phase. (This is another example of phase transition, or
phase change. The coexistence of the solid and gas phases at the sublimation
temperature for any pressure below Ptp is another example of phase equilibrium. Note that for water it happens at very low pressure only, but for
some other substances. such as carbon dioxide (CO2), it can happen at 1 atm.
This is just the behavior of the so-called “dry ice”, which is nothing but the
solid phase of CO2.
Superheating and supercooling − For very pure substances, and for
any pressure above the triple point pressure, it is possible to keep the solid
phase in a temperature range above the melting point, or to keep the liquid
phase in a temperature range above the boiling point. This is called superheating. On the other hand, it is also possible to keep the gas phase in a
temperature range below the condensation point, or to keep the liquid phase
in a temperature range below the freezing point. This is called supercooling.
That is, melting, boiling, condensation, or freezing should have happened,
but did not happen, because the system “does not know where to begin”.
For impure systems, or if the container wall is not smooth, then these conversion processes will begin to happen at an impurity site, or at a uneven
point on the container wall. The system will then avalanche, and the conversion will complete in a short time. This is what you often see in a boiling
process.
2. Thermal equilibrium and the zeroth law of thermodynamics
Thermal equilibrium − If a higher-temperature object is in contact with a
lower-temperature object, then heat will always flow from hot to cold, which
will raise the temperature of the cold object, and lower the temperature of
the hot object. This trend will stop only after the temperatures of the two
objects become equal. This final state is called thermal equilibrium. It
usually happens after a long wait.
Thermal insulator − A material which can be put between two other materials in order to reduce dramatically the heat flow between them which otherwise will occur. Note that a thermal insulator will take a long time to
reach thermal equilibrium with anything else, but eventually it still will.
Thermal conductor − A material which can conduct heat (i.e., let heat flow
through it) very effectively. A thermal conductor can reach thermal equilibrium with another thermal conductor very quickly.
The zeroth law of thermodynamics − If object A is in thermal equilibrium
with object B, and object B is in thermal equilibrium with object C, then object A is necessarily also in thermal equilibrium with object C. That is, if
heat has stopped flowing between A and B in contact, and between B and C
in contact, then no heat transfer will occur between A and C, when they are
brought into direct contact. ― These systems are said to be all at the same
temperature. Thus without this law, the concept of temperature would not
exist. But this law appears so elementary that it is called the zeroth law.
3. Gas thermometers and the Kelvin temperature scale
D
Absolute temperature scale (K) ― TK = TC + 273.15
(Also known as the Kelvin temperature scale. The unit K stands for
Kalvin.)
The absolute temperature of a physical system can be measured using a gas
in a closed container (so that its volume is fixed), with a pressure gauge attached to it. It is called a gas thermometer. First bring the system in
thermal contact with the gas container, and wait until they reach thermal
equilibrium. Measure the pressure of the gas and call it P. Then bring the
gas container in thermal contact with a container containing a mixture of
water, ice, and water vapor. (This mixture at thermal equilibrium will be
automatically at the absolute temperature Ttp = 273.16 K, and pressure Ptp =
4.58 torr.) Wait until they all reach thermal equilibrium, and measure the
pressure of the gas (not the mixture) and call it Ptp. If the gas obeys the ideal
gas law, then at a fixed volume its absolute temperature is proportional to its
pressure. That is,
T
P
=
,
Ptp
273.16 K
or
T = 273.16 K
P
,
Ptp
which would give the absolute temperature of the physical system. (This
proportionality is called Gay-Lussac's law.) Unfortunately, A real gas does
not obey the ideal gas law, until it has a very low density. That is, when it is
very dilute. But it is not convenient to work directly with a very dilute gas,
because it would take very long time to reach thermal equilibrium, which is
achieved through molecular collisions. So one rather uses a real gas of not
so low density (in a closed container of some fixed volume V ) to measure
the absolute temperature T of any physical system, by first measuring P (the
pressure of the gas when it is in thermal equilibrium with the system) and Ptp
(the pressure of the gas when it is in thermal equilibrium with an equilibrium
mixture of water, water vapor, and ice, which can occur at the triple point
of H2O only), for several densities of the gas, in order to obtain many (P, Ptp)
data pairs. One can then use these data pairs to generate the plot shown below, where each filled circle corresponds to a pair of values for (273.16K) ×
(P/Ptp) and Ptp, generated using a pair of measured (P, Ptp).
When this set of filled circles can be fitted with a straight line, the line can
then be extrapolated to vanishing Ptp, corresponding to an infinitely dilute
gas. The intercept of the straight line with the vertical axis is then the measured absolute temperature T (in K)
meassured
of the system. This way of measuring
(273.16K) (P/Ptp)
the absolute temperature of a system
is called the ideal gas temperature
scale, and corresponds to the formula:
true T (K)
 P 
T (K) = (273.16 K) lim 
.
Ptp → 0  P 
 tp 
Here Plim
simply means extrapolating to
→0
0
0
measured
P tp
tp
vanishing Ptp.
Triple point of water − ice, water, and water vapor can coexist at thermal
equilibrium only at a special pressure (P = 4.58 torr) and temperature (T =
0.01 ºC = 273.16 K). This special point in a P-T diagram is called the triple
point of water. It refers to this special pressure and temperature. We need
only the temperature part, since the pressure inside a closed container does
not affect anything outside. Mix some ice, water, and water vapor, and keep
it in a closed container for a long time, until they all reach the same temperature – they are said to have reached thermal equilibrium – then the mixture
will automatically be at this special temperature and pressure. Any other
system in thermal equilibrium with this mixture will also be at this temperature, but not at this pressure!
[Note: 1 torr =
1
atm = 1 mm Hg. That is, 1 atm = 760 torr = 760 mm Hg.]
760
4. Thermal expansion
Changing temperature will cause a physical system to change its size. If it is
a long object, then mainly its length will change (called linear expansion).
The formula is
L = L0 (1+ α ∆T ),
where L0 is the length of the object at some reference temperature T0, and ∆T
is the change of temperature from T0, positive if the change is to increase the
temperature, and negative if the change is to decrease the temperature. α is
called the coefficient of thermal expansion.
Volume expansion − The formula is
V = V0 (1+ β ∆T ) ,
with
β = 3α ,
because the linear expansion formula applies to the length L, width W, and
height H of an object, and the volume of the object is the product of the three.
α and β are usually positive for most substances, but they are negative for
water between 0ºC and 4ºC. It means that the density of water drops as its
temperature drops further after reaching 4ºC. It can then stay on the top of a
deep-water lake or river, and freezes first, rather than sinking to the bottom,
which would allow more water to be chilled by the cold air above in a freezing weather. After ice forms on the surface of a lake or river, its density becomes even lower, so it continues to float on top. Ice is a poor thermal
conductor, which traps the heat of the warm water below, and not allowing
it to escape into the cold air above the ice layer. Thus deep water will not
freeze solid in harsh weather, and fishes and other under-water living things
can survive through severe winter. This fact is important for life to exist on
earth. (If fishes could not live through winter, higher forms of living things
probably would never be able to appear through evolution.)
[Footnote: Those of you who refuse to accept the concept of evolution,
think about why men and all kinds of male animals, including male cats,
male dogs, male lions, male tigers, male elephants, male monkeys, male
apes, male gorillas, etc., all have the same number of breasts and nipples as
their females, which not only cannot feed babies, but can give some of the
males breast cancer ― a medical fact, about one in 100,000 in man, mostly
occurring in older men. For more information, do a web-search on “male
breast cancer”. The situation is kind of like a strange model of automobile,
with two little flattened wheels on the roof, which not only cannot help the
car move, but can actually cause some cars of this model to leak in the rain!
Think why any designer would design that, unless this designer has a weird
mentality, and just wants to do strange things for the fun of it!? Evolution
can easily account for this fact, since evolution is not controlled by any
intelligence, but just lets things happen blindly. Some strange characteristics in a living species, once appeared through chance, can continue to
exist, as long as the species can survive. The species can even evolve into
higher species, while carrying that strange characteristics, as long as they do
not interfere with the survival of those species. Most men and male animals
surely do survive with breasts and nipples. Only a very small percentage of
them get breast cancer. On the other hand, roughly one in every 7 women
will get breast cancer in their lives, and even a radical mastectomy may or
may not save their lives. The problem is likely due to some genetic defect.
That is, they are destined to get this disease even before they were born. But
the problem is not bad enough to wipe out the whole human species. So
breast cancer continues to plague women, until one day a cure can be found
through scientific research. Also, if only you have noticed that mad cow
disease and avian (bird) flu can be passed to humans (as well as other animals), you would see that animals, birds, and humans must have the same
biological origin. The mere possibility of cloning of humans as well as all
kinds of animals is another strong indication. Whether cloning of humans
should happen or not is a moral decision, and is irrelevant here. What is
relevant is the fact that it can be done, just like the animals. This is already
an undeniable scientific truth. It proves that humans and animals have the
same biological origin. Humankind is merely a species of animal with a
better evolved brain. Animal brains are not as well evolved. Nevertheless,
they do function as brains in, for example, coordinating motion, or making
certain kinds of decisions, etc. If only you own a pet dog or cat, or even
some birds, you would have noticed that some of them are actually quite
intelligent. And yet some humans are actually quite dumb, such as those
who would use narcotic drugs, or cigarette, or alcohol, to the extent of destroying their health or life. They are actually dumber than some animals.]
The above footnote in square brackets has nothing to do with physics. So
you are not responsible for it in any exam or quiz. However, the interesting
behavior of water is physics, and its relevance to the survival of fish and
other water-dwelling living things is a direct consequence of physics.
Another important consequence of thermal expansion is that cement pavements and metal bridge structures must have gaps at certain distances, so
that when the temperature gets hot, pavements and bridges will not buckle.
Buckling will be very difficult to stop if no room is allowed for expansion,
since a tremendously large force is actually involved. To actually discuss
the force involved, you need the concepts in Sec. 11.4 of Young and Freeman, especially its equation (11.10), and the data in its Table 11.1. Basically,
the compressive stress (force per unit area) must be so large that the resultant
strain (∆l/l0, which is a percentage reduction of length) must be able to cancel out the percentage extension of length due to thermal expansion. When
the needed stress is larger than a certain limit value, the material will fracture or break into pieces.
Equation (11.10) of Young and Freeman reads: F / A = Y (∆L / L0 ) . If this
∆ L / L0 is to cancel the thermally induced ∆ L / L0 = α ∆T , the former must be
the negative of the latter. That is, we must have
F
= −α ∆T .
AY
Here A is the cross sectional area, and Y is the Young’s modulus. F / A is
called the stress, and ∆L / L 0 is called the strain. So this equation can be said
as the stress-induced strain canceling out the thermally-induced strain. That
is, one strain is the negative of the other strain. So if ∆T is positive (negative), then the thermally-induced strain is positive (negative), and then the
stress-induced strain must be negative (positive), meaning that it is a compression (expansion). You would then know what must be the direction of the
force needed.
5. Quantity of heat
Heat (Q) is a type of energy flow, so its unit is that of energy. However, we
usually give it a new unit caller cal, which stands for calorie. We have:
1 cal = 4.186 J,
1 kcal = 1000 cal = 4186 J.
There is another British unit called Btu, for British thermal unit. We have:
1 Btu = 778 ftÿlb = 252 cal = 1055 J.
When heat is flowing into a system, we regard Q as positive. It should normally raise the temperature of the system. That is, we will have a positive
∆T. When the two are both small, they are linearly proportional to each
other. That is, we have
Q = mc∆T ,
where m is the total mass of the substance involved, and c is the specific
heat of the material. (It is the amount of heat needed to raise the temperature
of one gram of the substance by one Celcius degree.) When the change of
temperature is infinitesimal, we change the notation ∆T to d T. The
corresponding very small amount of heat needed can be denoted as d Q .
Thus we have
mc =
dQ
.
dT
The right hand side is read as “d-bar Q over d T ”. It should not be written as
simply d Q / d T , because this is not a derivative. That is, there is no function Q( T ) that can let you take its derivative to get d Q / d T. so d Q / d T does
not exist. Note that d-bar Q is just an infinitesimal quantity. It is not a differential of any quantity.
The specific heat of water is:
cwater = 1 cal/g ⋅ C D = 4186 J/kg ⋅ K = 1 Btu/lb ⋅ F D .
you should remember that c water = 1 cal/gÿC°, because it is so special. (Is it
accidental that it is exactly 1? No. The heat unit “cal” is defined so that this
is exactly true for water at 15 C°, and the error is less than 1 % for water in
the whole temperature range between 0 C° and 100 C°.) Note also that
c ice = 0.50 cal/gÿC°, and c steam = 0.48 cal/gÿC°. They are not equal to c water !
Molar heat capacity: If n is the number of moles of the substance, and M
is the molar mass of the substance (or mass per mole), then m = nM, and we
have
Q = nMc ∆T = nC ∆T
where C ≡ M c is called the molar heat capacity of the substance. That is,
it is the amount of heat needed to raise the temperature of one mole of the
substance by 1 Celcius degree.
For a solid, the specific heat (or molar heat capacity) is usually measured at
a constant pressure (mostly at 1 atm). So they are called the specific heat
(or molar heat capacity) at constant pressure. For a gas, the measurement can be made at constant pressure or volume, and the result is larger
for the constant-pressure measurement than for the constant-volume measurement, because when raising the temperature of a gas at a constant pressure, its volume will expand, doing work to the wall of the container, which
requires extra heat input. The notation for specific heat at constant pressure
(or volume) is c p (or c V ) , and the notation for molar heat capacity at constant pressure (or volume) is C p (or C V ) . It is important that you pay attention to the difference between the capital letter C and the small-case letter c
here.
The above equations for Q are to be used when no phase transitions are involved, so that all heat input is used to raise the temperature of the system.
If phase transitions are also involved, you need to also use the equation for
the latent heat given below.
Latent heat (L) − Heat needed to convert 1 gm of a substance in a phase
transition from the solid phase to the liquid phase, or from the liquid phase
to the gas phase, or directly from the solid phase to the gas phase, if the pressure is below the triple-point pressure so that this can happen. ( Exactly this
amount of heat is released if the phase transition is in the opposite direction.)
To calculate the total amount of heat involved in the transition, we use
Q = mL ,
where m is the total mass of the substance in gm or kg, depending on the unit
of L. Note that the temperature does not change during a phase transition. If
the transition is from the solid phase to the liquid phase, L is called the heat
of fusion ( L f ). If the transi-tion is from the liquid phase to the gas phase, L
is called the heat of vapor-ization ( L v ). If the transition is from the solid
phase directly to the gas phase, L is called the heat of sublimation ( L s ).
For H2O at 1 atm, the heat of fusion is 79.6 cal/g = 3.34 × 105 J/kg, and the
heat of vaporization is 539 cal/g = 2.256 × 106 J/kg. Note that these latent
heats are pressure dependent, so they don’t have these values at P other than
1 atm.
6. Heat calculations (calorimetry problems)
The term calorimetry refers to the use of a set-up called a calorimeter to
quantitatively measure an amount of heat exchange. One application of it is
to determine the specific heat of a given material, or the latent heat of a certain phase transition of a given material. Another application is to predict the
final equilibrium state of a system containing two or more kinds of substances at different initial temperatures and/or phases.
Example 1: If 5g of steam at 120 °C are mixed with 4g of ice at -10 °C,
what is the final equilibrium state and temperature? (Assume that the pressure is at 1 atm, and there is no heat input from outside this mixture and
there is no heat leaked to the outside of this system.)
Answer: In this problem, there is no heat input into the system or heat loss
to the environment. Whatever amount of heat released by the 5 g of steam
for it to reach the final equilibrium state must equal to the amount of heat
absorbed by the 4 g of ice so it will reach the same final equilibrium state.
We however do not know whether the final phase is ice or water or steam,
or even some mixture of these phases, and what is the final temperature. So
we must do this problem by trial and error. That is, we first postulate a final
state, and then see in what direction we must correct this postulate. Let us
first postulate that the final state is 0 °C water. That is, we postulate that
both the 5g of steam at 120 °C and the 4g of ice at -10 °C will become 0 °C
water at the end when thermal equilibrium has been reached. Then the 5 g of
steam at 120 °C will release 0.48 cal/g·C° × 5g × (120 °C – 100 °C) + 539
cal × 5g + 1 cal/g·C° × 5g × (100 °C – 0 °C) = 48 cal + 2695 cal + 500 cal
= 3243 cal of heat, for the steam to first cool down to 100 °C, then condense
to 100 °C water, then to cool to 0 °C water. On the other hand, the 4 g of ice
at -10 °C will absorb 0.50 cal/g·C° × 4g × (0 °C – (-10) °C) + 4 g × 79.6
cal/g = 20 cal + 318.4 cal = 338.4 cal of heat, in order to first warm up to 0
°C ice, and then to melt into 4 g of 0 °C water. The 3243 cal is released, and
the 338.4 cal is absorbed. So we see that there is more heat released than
absorbed. That is, there is 3243 cal − 338.4 cal = 2905 cal of heat available.
Is it enough to raise all water (9 g tatal) from 0 °C to 100 °C? It would need
1 cal/g·C° × 9 g × (100 °C − 0 °C) = 900 cal. So, yes. There is enough heat
to raise all water to 100 °C, and there is still 2905 cal − 900 cal = 2005 cal of
heat left. Thus some water will be converted to steam. How much? 2005
cal / 539 cal/g = 3.72 g. We thus conclude that the final equilibrium state is
made of 3.72 g of steam and 9 g – 3.72 g = 5.28 g of water all at 100 °C.
(Steam and water can only coexist at 100 °C if P = 1 atm.)
Example 2: If 2g of steam at 120 °C is mixed with 10g of ice at -10 °C,
what is the final equilibrium temperature? (Assume that the pressure is at 1
atm, and no heat input from the environment or heat loss to the environment.)
Answer: Let us again first postulate that the final state is all water at 0 K.
Then the heat absorbed by the 10 g of ice is 0.50 cal/g·C° × 10g × (0 °C –
(-10) °C) + 79.6 cal × 10g = 846 cal. The heat released by the 2 g of steam
is 0.48 cal/g·C° × 2g × (120 °C – 100 °C) + 539 cal/g × 2g + 1 cal/ g·C° ×
2 g × (100 K – 0 K) = 1297 cal. We see that the heat is not balanced.
Instead, we have 1297 cal – 846 cal = 451 cal left. Can this much heat raise
all water (12 g of it) from 0 K to 100 K? The answer is no, since it would
need 1 cal/ g·C° × 12 g × (100 K – 0 K) = 1200 cal. Thus we conclude that
the final state is 12 g of water at a temperature x K which should be above
0 K but below 100 K. To find this temperature, we solve 1 cal/ g·C° × 12 g
× x K = 451 cal, and find x K = 451 cal / (1 cal/ g·C° × 12 g) = 37.6 K.
This final temperature can also be obtained by setting up the equation
1097 cal + 1 cal/g· C° × 2g × (100 C° − Tf ) = 846 cal + 1 cal/g · C° × 10g × Tf ,
and solving it to obtain Tf = (1297 cal – 846 cal ) / (1 cal/g · C° × 12g) =
37.6 K. However, this approach is valid only if you know already that the
final temperature is somewhere between 0 K and 100 K. If you blindly do
this and obtain for the final temperature a negative temperature, or a positive
temperature above 100 K, then you better realize that you have reached a
contradiction, implying that your approach is incorrect. In that case, you
should abandon your approach, and switch to our first approach.
The following diagram illustrates what is happening: ( p = 1 atm assumed. )
The above two examples assume that the whole mixture is inside a wellinsulated container, and no heat is provided to the mixture or taken away
from the mixture. If a certain amount of heat is provided by an external
source (such as a burner), you should add it to the total heat released, which
is heat available (to raise temperature, or melt ice, or evaporate water, etc.).
If some amount of heat is lost to the environment due to poor heat insulation,
then you should add this amount to the total heat absorbed, which is also the
total heat needed. That is, it has to come from somewhere. Then the heat
balance equation is simply:
absorbs heat
releases heat
10g
2g
T
-10°C 0°C
ice
37.6°C
100°C
water
total heat released = total heat absorbed.
120°C
steam
(1st approach)
Note that if you use this equation to solve any calorimetry problem, then
every entry should be positive. So you must not use ∆T = Tfinal − Tinitial
when Tfinal is less than Tinitial , because it would give a negative quantity. On
the other hand, if you do have both positive and negative heat terms, then
you should use the equation:
sum of all heat terms = 0,
(2nd approach)
where a positive term on the left hand side corresponds to: (i) some
∆T = Tfinal − Tinitial > 0 ; or (ii) some solid melted to liquid, or some liquid evaporated to gas, or some solid sublimed to gas; or (iii) Some heat is lost to the
environment. (All these three cases correspond to some heat is needed to get
such changes to happen.) On the other hand, a negative term on the left hand
side corresponds to: (i) some ∆T = Tfinal − Tinitial < 0 ; or (ii) some liquid frozen
to solid, or some gas condensed to liquid, or some gas directly converted to
solid; or (3) Some heat is provided to the system. (All these three cases correspond to some heat is released when these changes happen.) That is, all
positive terms correspond to some heat absorbed, or heat needed, and all negative terms have their absolute values correspond to some heat released, or
heat available. You can use either approaches, but you should never mix up
the two approaches! That is the main reason that you cannot get the right
answer!
It is important to realize that the specific heat of ice, water, and steam are all
different. Use 1 cal/g·C° for water only (at 1 atm). For ice, it is 0.50 cal/g·C°
(near 0°C at 1 atm), and for steam, it is 0.48 cal/g·C° (near 100 °C at 1 atm).
If the unit “ J/ kg·K ” is used for the specific heats, then the specific heat of
water is 4186 J/ kg·K, the specific heat of ice is 2093 J/ kg·K, and the specific heat of steam is 2009 J/ kg·K. You do not need to remember these numbers, but must know when to use what when they are all given to you.
7. Mechanisms of heat transfer
a. Conduction
This is through molecular collisions and interactions to transfer the kinetic energy of the faster molecules to the slower molecules. It can occur in
a solid, a liquid, or a gas, but it is very weak in a gas, because the gas
molecules are separated by large distances, and collide or interact with
each other much less frequently. The formula for heat conduction is
H =
T − TC
dQ
=κ A H
,
dt
L
or
H =
dQ
dT
= −κ A
,
dt
dx
where H is called the heat current and is the rate of heat transfer across a
cross-sectional area A, through a distance L ), and κ is a material-dependent constant called the thermal conductivity, and TH and TC are the temperatures at the hot and cold ends of the distance L . The notation d Q / dt
(read as “d-bar Q over d t”), instead of simply d Q / d t as in Young and
Freeman, emphasizes that this is not a derivative of a function Q ( t ) that
doesn’t exist, but is simply the ratio of two very small quantities. That is,
a very small amount of heat flow d Q, divided by the very small amount
of time d t that it took to take place. You should notice that Young and
Freeman uses the notation k for κ , which is very confusing, since k has
already been used to denote three other quantities: (1) wave number
which obeys k = 2π/λ , (2) force constant which obeys F = − k x, and (3)
the Boltzmann constant which appears in R = NA k , the ideal gas constant.)
Also, x is measured along the direction of the heat conduction. Along
this direction, temperature should be a decreasing function of x, so that
heat is flowing from the hot side to the cold side. So dT/d x should be
negative if d Q / dt is positive, hence the minus sign in the second form
of the formula.
Metals are the best heat conductors, because there are freely moving electrons inside metals to carry the heat from the hot side to the cold side.
The R value of a building material is defined as R = L / κ , where L is
the thickness of the material, and κ is thermal conductivity of the material.
The higher is the R value, the better is the thermal insulation achieved by
this material. (Clearly, high thermal conductivity is bad for this purpose,
so κ is in the denominator. It is easy to see why L is in the numerator.)
b. Convection
In a liquid or gas, heat can be transferred much more effectively through
the circulation of the liquid or gas. For example, when you boil water in
a kettle, the water at the bottom is hotter than the water at the top. But
hotter water is less dense than colder water (if only T is higher than 4 °C),
so hotter water will rise to the top, carrying heat up, and colder water will
sink to the bottom, to be heated by the burner below. This is convection.
If a surface at temperature T is in contact with a fluid at a lower temperature Tf, then convection will carry heat away from the surface at a rate ap5/ 4
proximately proportional to (T − Tf ) . It is also proportional to the total
area of the surface.
c. Radiation
Conduction and convection can not occur in vacuum. But heat can still
be transferred through vacuum via radiation. This is because any object
with a temperature above absolute zero will emit electromagnetic waves
in all directions. The formula governing this form of heat flow is
H = Aeσ T 4 ,
(the Stefan-Boltzmann equation)
where the left hand side still denotes the rate of heat transfer (via radiation), and on the right hand side, e is called emissivity, and is a materialspecific constant indicating how effective the surface of the material can
emit radiation. It takes values between 0 and 1 only. It is closer to 1 for
a very black non-shiny surface, and it is close to 0 for a bright shiny surface. σ is called the Stefan-Boltzmann constant. It has a universal value σ = 5.670400(40) ×10−8 W / m 2 ⋅ K 4 . (If e is close to unity, the surface
is a good emitter. Then it is also a good absorber.) Again, A is the crosssectional area, and T is the absolute temperature in K. Between an object
at temperature T , and the surroundings at temperature Ts , the net heat
transfer via radiation from the system to the surroundings is given by the
formula (which can be negative if T < Ts ):
H net = Aeσ T 4 − Aeσ Ts 4 = Aeσ (T 4 − Ts 4 ),
where the substracted term is the radiation coming from the surroundings
and absorbed by the system. This net heat transfer is again from hot to
cold, and it drops to zero when T becomes equal to Ts. Note that e is the
emissivity of the system, with the emissivity of the surroundings not involved. (If the surroundings have emmisivity less than unity, they will
make up the reduced radiation by reflecting back more of the radiation
emitted by the system.)
The sun also radiates heat. By the time the radiation reaches the atmosphere of the earth it is at the rate of 1350 W/m2, which is called the solar
constant. A large fraction, 70%, may be absorbed by the atmosphere on a
cloudy day. On a clear day, about 1000 W/m2 reaches the surface of the
earth, if the surface is perpendicular to the sun rays. (If the surface is tilted at an angle θ from being perpendicular to the sun rays, then the amount
is reduced by a factor cos θ.) The amount actually absorbed is also reduced by the emissivity e of the surface.
A blackbody is a perfect emitter (e = 1). It is also a perfect absorber in
the sense that any radiation falling on it is absorbed by it with no part
reflected. On the other end of the range, an ideal reflector has e = 0. It
absolutely cannot absorb radiation nor can it emit radiation. A small hole
on the wall of a large enclosed cavity is the best example of a black body.
Thermograph, which measures the amount of radiation from various
parts of the surface of a human body, can determine the temperature distribution on the surface of a human body, from which a tumor can be detected, because the cells inside a tumor is more active, so they generate
more heat (which means that they convert more other forms of energy,
such as chemical energy, into heat).
The eardrum thermometer measures the radiation from the eardrum in
order to determine the body temperature!