specific heat of water as our standard (as we practically do in
defining the units of heat), the specific heat of a substance is ex-
pressed by the number of heat units required to raise the unit
quantity of the substance one degree in temperature.
One of the simplest methods of determining specific heat is by
mixing the substance with water. Suppose that G pounds of mer-
cury at 100C are poured into 2 pounds of water at 0C, and
that the resulting temperature of the " mixture " is 9. The spe-
cific heat S of the mercury can then be found as follows :
In falling from 100 to 9 the C> pounds of mercury give out
G X (100 9) X S, or 546 S heat units. These have gone to heat
2 pounds of water from to 9, which requires 2 X 9, or 18 heat
units Hence we may write
546 S = 18
Therefore S = 0.033.
EXAflPLE FOR PRACTICE.
Half a pound of a metal at 212 F is dropped into one pound
of water at G8F. The temperature of the mixture is then
observed to be 7G F. What is the specific heat of the metal ?
TABLE OF SPECIFIC HEATS.
With the foregoing principles and the help of suitable tables,
many problems can be solved. For example, let us find how many
calories will be required to convert 10 grams of ice at 12 C
into steam at 100 C.
Solution. Required to raise the ice from 12 to 0,
10 X 12 X 0.504 = 60.48 calories.
Required to melt the ice,
10 X 80 = 800 calories.
Required to raise the water from to 100,
10 X 100 = 1,000 calories.
Required to vaporize the water,
10 X 536.5 = 5,365 calories.
Total number of calories required,
60.48 + 800 + 1,000 + 5,365 = 7,225.5 (nearly).
EXAMPLES FOR PRACTICE.
1. What weight of water at 75 C will just melt 15 pounds
of ice at ?
Ans. 16 pounds.
2. One kilogram of water at 40, 2 kilograms at 30, 3 kilo-
grams at 20, and 4 kilograms at 10 are mixed. Find the tem-
perature of the mixture.
3. How many heat units will be required to melt 5 grams
of ice at 20 C ? How many grams of water at 50 C would
Ans. 450.4 heat units ; 9.01 grams.
If we wish to use Fahrenheit degrees and British thermal
units in our calculations, it is necessary to remember that the
numbers representing the heats of fusion and of vaporization are
different, but that the specific he^t, which is a mere ratio, is the
same in both systems.
For example, let us find how many B. T. U. are required to
convert 12 Ibs. of ice at 10 F into steam at 212 F.
Solution. Required to raise the ice from 10 F to 32 F,
12 X 22 X 0.504 = 133.050 U. T. II.
Required to melt the ice,
12 X 144 = 1,728 B. T. U.
Required to raise the water from 32 to 212,
12 X 180 = 2,160 B. T. U.
Required to vaporize the water,
12 X 966 = 11,592 B. T. U.
Total number of B. T. U. required,
133.056+ 1,728 -f 2,160 -f 11,592 = 15013 (approx.).
EXArtPLE FOR PRACTICE.
How many B. T. U. are required to convert 10 Ibs. of ice at
15 F into steam at 212 F ?
Ans. 12,985 B. T. U.
For ordinary purposes we may proceed as above ; but as the
specific heat and latent heat of water vary for different tempera-
tures, we must, where great accuracy is necessary, employ a table
of the properties of steam and water.
THE PROPERTIES OF STEAM.
The relation between the external pressure and the boiling
point of water is a perfectly definite one, but cannot be exactly
expressed by any mathematical equation. In dealing with this
and other properties of steam and water, it is therefore customary
to refer to suitable tables where the values are given, as deter
mined by experiment. Such tables are called steam tables, and
are much used in engineering calculations.
In the following table are given (1) the pressure above abso-
lute vacuum, (2) the corresponding temperature, (3) the amount
of heat in B. T. U. required to raise a pound of water from 32 F
to the given temperature, (4) the amount of heat in B. T. U.
required to vaporize a pound of water at the given temperature ;
(5) equals the sum of (3) and (4).
A steam gage measures pressures above the atmospheric pres-
sure ; hence, when readings are taken from a steam gage, the baro-
metric pressure (averaging 14.7 Ibs. per sq. in., or in round numbers
15 Ibs.) must be added to obtain the " absolute " pressure.
With a steam table we can extend considerably the range of
problems like those on page 23. For example, let us find how
many pounds of steam at 65 Ibs. gage pressure will be needed to
raise the temperature of GO pounds of water from 50 F to 100 F.
Solution. To raise one pound of water from 50 to 100
requires 50 B. T. U. ; and for GO pounds we need 50 X 60, or 3,000
B. T. U. At 65 Ibs. gage pressure (80 Ibs. absolute) the total
heat of one pound of steam is 1,177 B. T. U., and this amount
would all be available if we cooled it down to 32 F. But since
the cooling is not carried below 100 F, we cannot use 100 32
TABLE OF PROPERTIES OF SATURATED STEAM.
of cubic ft
1 '.13.2- 3
0.05023 : 1'..91
or 68 B. T. U., and the amount available is therefore 1,177 6'
= 1,109 B. T. U.
The quantity of steam therefore needed is
Ans. 2.705 -f- pounds,
The small quantity of stekm in this example well illustrates
the great heating power of steam.
EXAMPLES FOR PRACTICE.
1. How many pounds of steam at 100 Ibs. absolute pressure
will raise 250 pounds of water from 50 F to 150 F ?
Ans. 23.5 pounds.
2. How many pounds of steam at 35 pounds gage pressure
will just melt 1,000 pounds of snow at 32 F ? .
Ans. 123.3 pounds.
It will be seen that the values of the total heat column in the
table increase but slowly, while the pressure and temperature
increase rapidly. A pound of high-pressure steam thus contains
but little more heat than a pound of low-pressure steam, and con-
sequently requires but little more fuel to produce it. Since high-
pressure steam is more effective in steam engines, there is therefore
a decided thermodynamic advantage in using steam of the highest
Superheated Steam. From the table we see that there is a
perfectly definite temperature for steam at any given pressure.
Steam or other vapors in this condition are said to be saturated,
for if the temperature is lowered some of the vapor will condense
immediately into liquid. W, however, we pass steam through a
separately heated pipe or chamber it is easily possible to raise its
temperature by any desired amount above the given values.
Steam in this condition is called superheated. For steam engines
it has certain advantages over saturated steam, which are discussed
iu the Instruction Papers on the Steam Engine.
TRANSFER OF HEAT.
Heat may be transferred from one body to another by con-
duction, convection or radiation.
Conduction. When one end of a metal bar is heated in the
fire the other end gradually becomes warmer. The heated mole-
cules communicate their motion to their immediate neighbors ; and
the heat thus travels along the bar, and may be removed by a
cold body at the distant end. This process is called conduction.
In this way the heat of a boiler furnace is communicated to the
water in the boiler.
A brass pin held in a gas flame will burn the fingers almost in-
stantly, while a bit of glass may be melted at one end before -the
other becomes hot, and a match may be burned to the finger-tips
without discomfort. It is thus clear that substances differ greatly
in conductivity. There are great differences even among metals ;
a copper rod will conduct heat much more rapidly than an iron
If a piece of wire gauze is held over an unlighted gas jet, the
gas may be lighted on either side (Fig. 6), but the flame will not
pass through the meshes. The wires conduct the heat away so
rapidly that the gas on the other side does not get hot enough to
ignite. This is the principle of the safety lamp, used in coal
mines where inflammable gases collect. The lamp flame is sur-
rounded by a wire gauze, and thus kept from igniting the dan-
gerous gases outside.
The following table gives the relative conductivity of several
substances. The table well shows the great value of a layer of
snow as a protective blanket on the earth :
RELATIVE THERHAL CONDUCTIVITIES.
Convection. Kxcepting mercury, liquids and gases are poor
conductors of heat ; but when such bodies are heated from below,
the heated portion expands and rises through the mass, and is
replaced below by a colder portion, which is heated and rises in its
turn. In this way what are called convection currents arise, and
the heat is distributed throughout the fluid by actual motion
within the mass itself. The heating of houses by hot water is an
application of this principle.
Convection also takes place in gases ; the winds of the atmos-
phere illustrate it on a large scale.
Radiation. We have seen that the molecules of a hot body
are in very rapid vibration. Some of the energy of this vibration
is communicated in the form of waves to the space surrounding
the body. If the motion happens to lie within certain limits, the
waves affect our eyes and we call them light-waves. But all such
waves, visible or otherwise, represent energy which is sent out by
the hot body. "When tlicy fall upon any other body, they are
either reflected or absorbed and transformed into heat. Polished
silver reflects over 90 per cent of the waves falling on it ; charcoal
absorbs nearly all, and hence rises in temperature when exposed
to the radiations of a hot body.
Energy in this form is called radiant energy. By it the heat
of the sun is transmitted to the earth. Since it is in the form of
ether-waves, many of the experiments ordinarily performed with
light-waves may be repeated with the radiations from a hot body,
whether visible or not. The common burning-glass shows the
result of bringing such rays to a focus by refraction. If a pair of
concave mirrors be set facing each other, as shown in Fig. 7, and
a source of heat be placed in the focus of one, a thermometer in
the focus of the other will quickly show a rise in temperature
though the mirrors are many feet apart.
It does not follow, however, that bodies transparent to light
are equally transparent to other radiations. Glass, for example, is
quite opaque to the invisible radiations that are most effective in
producing heat. Also, a solution of alum in water will cut oft
most of these radiations, while allowing the light-waves to pass
freely ; and a glass tank of alum water is often used in stereopti-
cons to keep the heat of the lamp from the rest of the apparatus.
It is important to note that though the radiation from a hot
body is often called radiant heat, yet in the process of transmission
it is not heat at all, but a wave motion in the ether, which is
energy of a very different kind. A somewhat analogous case is
present in the incandescent lamp; the heat which appears at the
lamp does not come along the wires as heat, but as electrical
energy, which is altogether different. It is indeed transformed
into heat in the lamp, but is not itself heat. In like manner*,
radiant energy is transformed into beat only by falling on some
body wbicb absorbs it. It is thus possible to make a good burn-
ing-glass by shaping a piece of clear ice into the form of a lens,
a thing which would clearly be impossible if the energy passing
through the ice lens were in the form of heat.
This is the science which deals with the relations between
heat and mechanical energy. It rests on two fundamental propo-
sitions called the first ami second laws of thermodynamics.
The first law states that when heat is transformed into
mechanical energy, or the reverse, the quantity of heat is always
exactly equivalent to the quantity of mechanical energy.
The production of heat by friction is familiar to every one;
but ic is not so clear that there is an exact equivalence between
the energy lost by friction and the heat produced. Joule was the
first to establish the relation accurately. The principle of his
apparatus is shown in Fig. 8. The falling weights EE turned a
paddle-wheel stirrer inside the cylindrical vessel G, which was filled
with water and was much like the common ice-cream freezer.
The friction of the stirrer heated the water ; and when the distance
was measured through which the weights fell, it was possible to
calculate the relation between the work done by the falling weights
and the heat developed in the water.
Later experiments on a larger scale have given results which
are more accurate than was possible with this apparatus. The
values now accepted are the following :
427.3 kilogrammeters of work or energy are required to raise
the temperature of one kilogram of water from 15 to 16 C at
sea-level, in latitude 45.
In English units, 7T8.8 foot-pounds of work are required to
raise the temperature of one pound of water from 59 to 60 Fah-
renheit at sea-level, in latitude 45.
These values vary slightly for different places, because the
weight of a pound depends on the pull of gravity, and this varies
in different places ; but for most engineering purposes 779 foot-
pounds would be near enough.
This law states in effect that we cannot get energy for noth-
ing. Whenever we get work from heat, a definite quantity of heat
disappears ; and whenever we convert mechanical energy into heat,
we must expend 779 foot-pounds for every British thermal unit
The second law of thermodynamics asserts that heat cannot of
itself pass from a cold to a hot body. Since a hot body in cooling
gives out heat which, in part at least, may be converted into work,
it might seem that by cooling it indefinitely we could get an infi-
nite amount of work from it Bufe the second law declares that
the process stops as soon as the hot body has cooled to the tem-
perature of its surroundings ; and if we wish to cool it further we
must expend energy in so doing. It follows from this that no
heat engine can convert into work all the heat which it receives.
As soon as the steam (or other working fluid) has fallen to the
temperature of the exhaust, the remaining heat in it is no longer
available for doing useful work. If the heat is supplied at the
absolute temperature T l and the exhaust is at the temperature T 2 ,
the efficiency of the engine cannot be greater than , no
matter what is used as the working fluid of the engine.
THERHODYNAniCS OF PERFECT GASES.
The subject of thermodynamics cannot be fully treated by
elementary methods of analysis ; in the following brief discussion,
however, no advanced methods are used.
The thermodynamics of steam will be more readily under-
stood by first taking up the simpler case of a perfect gas. Air.
oxygen, nitrogen, hydrogen and some others, behave very nearly
as perfect gases ; others, as ammonia, carbon dioxide and sulphur
dioxide, do not.
Boyle's Law. The product of the pressure P and the volume
V of a perfect gas is constant if the temperature is constant; that
is, if at a pressure l\ a body of gas lias the volume V lf and at
some other pressure P 2 has the volume V 25 then
PiVj P 2 V 2 constant.
Example. If 12 cubic feet of air at 135 pounds absolute
pressure expand to 27 cubic feet at the same temperature, what
will be the pressure? What pressure would a gage indicate?
Solution. 1*135^135 = 1*27^27
135 X 12=P 27 x 27
Therefore P 27 = 60 pounds, absolute.
Gage pressure = absolute pressure atmospheric pressure.
Pgage = 60 14.7 45.3 pounds. Ans.
EXAMPLE FOR PRACTICE.
Ten cubic feet of air at 2.3 Ibs. gage are compressed until the
gage pressure is 7.3 Ibs. Find the volume.
Ans. 7.727 cub. ft.
Since one pound of air at 32 F occupies 12.387 cubic feet,
we may calculate the product PV of a pound of air as follows:
V = 12.387
P = 14.7 X 144 = 2,117 (nearly).
Hence PV = 2,117 X 12.387 = 26,223.
Law of Boyle and Charles. For a perfect gas the product
PV is proportional to ths absolute temperature T. In the form
of an equation, this becomes -r- = constant, or PV= constant X T.
This is usually written PV = HT. For air we may easily calcu-
late R as follows : We have just seen that at 32 F, P V = 26,223,
and T - 32 -|- 461, or 493 absolute temperature. Therefore,
= K = 53.2.
Example. What volume will be occupied by a pound of air
at 50 F and 40 pounds pressure (absolute) per sq. in.?
Solution. P = 40 X 144 = 5,760 Ibs. per sq. ft.
T = 50 -f- 461 = 511 absolute temperature.
Therefore 5,760 X V = 53.2 X 511.
V = 4.72 cub. ft. (nearly).
Example. A quantity of air at 75 Ibs. gage pressure and 60
F is heated to 90 F. What is the pressure ?
Solution. Since the volume is unaltered, the pressure is pro-
portional to the absolute temperature. We have
60 F = 60+461 = 521 absolute.
90 F = 90+461 551 absolute
Therefore l:l:: T : T,.
P 2 = '^ X (75 -f 14.7) = 94.86 Ibs. Ans.
EXAflPLES FOR PRACTICE.
1. What is the weight of 6 cubic feet of air at 60 F and
25 pounds absolute pressure per square inch ?
Ans. .78 Ibs.
SUGGESTION. First find the volume of one pound under the given
2. A reservoir containing 4 cubic feet of air at a tempera-
ture of 40 F and a pressure of 100 pounds per square inch abs.,
is heated to 80 F. What will the pressure be, and how much
does the air weigh ?
A ( 107.98 +lbs.
8 - i 2.16 +lbs.
Isothermal and Adiabatic Expansion. When a gas expands
and does work, as by pushing a piston
in a cylinder, we see from the first law
of thermodynamics that the equivalent
in the form of heat must be supplied
from somewhere. If the temperature of
the gas is to be kept constant, heat must
be supplied to it from the outside, in
exact equivalent to the work done. In
such cases the expansion is said to be
isothermal, and the relation between
pressure and volume is as shown by
.ythe dotted curve I of Fig. 9. This
curve is an equilateral hyperbola. But
if no heat be allowed to enter the gas,
as would be the case if the cylinder and piston were perfect
non-conductors of heat, the work done in expansion will be at the
expense of the heat energy in the gas itself, and its temperature
will therefore fall during the expansion. We have seen that the
pressure is less as the temperature falls, other things being equal ;
hence under the conditions the pressure p
will fall faster than if the temperature
were kept up by the addition of heat from
outside. This is shown by the curve A
of Fig. 9. Curves of this kind, represent-
ing expansion or compression without
communication of heat to or from the gas,
are called adiabatics. It is evident that
adiabatic expansion along A fromv 2 tovi(
is accompanied by a greater fall of pressure
than isothermal expansion along I.
Both isothermal and adiabatic curves are of great importance
in thermodynamic studies, but they represent conditions that are
only imperfectly realized in practice. The general expression for
an adiabatic curve is PV U = constant. For air, n = 1.405.
Most problems involving adiabatic and isothermal curves cannot
be solved without the aid of higher mathematical processes than
are used in this Paper.
Work Done in Expansion. Suppose we have a piston whose
area is A square inches, which is acted upon by a pressure of p
pounds per square inch, and which moves through a distance of
m feet in consequence. Then the total pressure isj^A pounds,
and the work done is pA X in foot-pounds. But A X m is the
volume of the cylinder swept out by the piston in its stroke ; arid
calling this V, we have :
Work done = pressure X volume = PV.
This can be conveniently shown on the pressure-volume dia-
gram. Suppose B (Fig. 10) represents the pressure and volume
of a gas, which then expands a little to the condition A. Then
the average pressure during the expansion will he ^ (B6 -f- Aa),
and the work done will be J (Eb -f- Aa) X ab = the area BAao.
Since we may regard the whole change from C to A as made
up of portions like that from B to A, it follows that in changing