THERMOMETRY
International Temperature Standards
B pt.
F pt.
B pt.
B pt.
F pt.
F pt.
O2 at 1 atm
H 2O
H2O at 1 atm
S
at 1 atm
Ag
Au
-182.97 ºC
0.0000 ºC
100.00 ºC
444.60 ºC
960.8 ºC
1063.0 ºC
Common Temperature Scales
Fahrenheit
B pt.
H2 O
F pt.
H2 O
absolute
zero
212 ºF
180
DIVISIONS
32 ºF
492
DIVISIONS
-459.7 ºF
Rankine
Celcius
100 ºC
373.15 K
100
DIVISIONS
100
DIVISIONS
0 ºC
273.15 K
671.7 ºR
180
DIVISIONS
491.7 ºR
492
DIVISIONS
Kelvin
273
DIVISIONS
273
DIVISIONS
0.00 K
-273.15 ºC
0.00 ºR
absolute
scale
absolute
scale
Note that between the F pt. and B pt. of water there are 180 divisions on the Fahrenheit
and Rankine scales but only 100 divisions on the Celsius and Kelvin scales, i.e.,
100 Cº = 180 Fº
or
5 Cº = 9 Fº
or
1 Cº = 1.8 Fº
Conversions
Fahrenheit & Rankine
ºF + 459.7 = ºR
or
ºR - 459.7 = ºF
Celsius &Kelvin
ºC + 273.2 = K
or
K - 273.2 = ºC
Celsius &Fahrenheit
(ºC  1.8) + 32 = ºF or
(ºF - 32)  1.8 = ºC
Kelvin &Rankine
K  1.8 = °R
°R  1.8 = K
or
Fahrenheit & Kelvin
or
Celsius &Rankine
or
Thermometry
1
1400
y = 1.0x + 459.4
R2 = 1.0
1200
°R
1000
800
y = 0.56x + 255.20
R2 = 1.00
K
°
600
400
200
y = 0.56x - 17.78
R2 = 1.00
°C
0
°C vs. °F
K vs. °F
-200
°R vs. °F
-400
Linear (K vs. °F)
-460
-260
-60
140
340
540
740
940
Linear (°C vs. °F)
°F
Linear (°R vs. °F)

Note that K is always (simply) 273.2° above °C

Note that °R is always (simply) 1.8 times greater than K

Note that °R is always > °C (starts higher and is a more sensitive temp. scale)
1000
y = 1.0x - 458.8
R2 = 1.0
800
600
°F
y = 0.56x
R2 = 1.00
400
°
°F
°C
K
200
K
°C
°F vs. °R
0
y = 0.55x - 272.69
R2 = 1.00
-200
°C vs. °R
Linear (K)
-400
-600
0
500
1000
1500
°R

Note that °F > °C at higher T but °C > °F at lower T

Note that °F > K at higher T but K > °F at lower T
Thermometry
2
1300
1100
y = 1.8x + 0.0 °R
R2 = 1.0
900
°F
700
y = 1.8x - 458.8
R2 = 1.0
°
500
°C
300
y = 1.0x - 272.7
R2 = 1.0
100
°C vs. K
°F vs. K
-100
°R vs. K
Linear (°F vs. K)
-300
Linear (°C vs. K)
Linear (°R vs. K)
-500
0
200
400
600
800
K
°

Note that °R is always (simply) 459.7° above °F
1400
1200
1000
800 y = 1.8x + 491.4 °R
R2 = 1.0
600
400
200
0
y = 1.8x + 32.0
-200
R2 = 1.0
°F
-400
-600
-273
-73
127
K
°R
K
°F
y = 1.0x + 273.0
R2 = 1.0
°R vs. °C
°F vs. °C
K vs. °C
327
527
°C

Note that °F > K at higher T but K > °F at lower T
Thermometry
3

Which is colder; 50°C or 50°F? Circle the correct answer.

Which is warmer; -30°C or -30°F? Circle the correct answer.

Which is warmer; 30°C or 30°F? Circle the correct answer.

Calculate the conversion for Fahrenheit to Kelvin.

Calculate the conversion for Celsius to Rankine.

Calculate the temperatures at which °C and °F are equal.

Calculate the temperatures at which °F and K are equal.
Temperature Conversions using Reference Points (rather than memorized equations)
One method of temperature conversion is to compare reference pts. such as Absolute zero,
B pt., or F pt. of H2O. This allows direct conversion between ºR and K scales.
Examples
1. Convert -20.5 ºF to ºC
-20.5 ºF is 20.5 + 32 Fº below F pt. = 52.5 Fº below F pt. of water
52.5 Fº  (5 Cº/ 9 Fº) = 29.2 Cº below F pt.
-29.2 ºC
2. Convert 165 ºC to ºF
165 ºC is 65 Cº above B pt. of water
65 Cº  (9 Fº / 5 Cº) = 117 Fº above B pt.
212 + 117 = 329 ºF
3. Convert 300 K to ºR
(recall 9 Rº = 5 Kº)
300 K is 300 Kº above Absolute zero
300 Kº  (9 Kº/ 5 Rº) = 540 Rº above Absolute zero
540 ºR

Do the 4 temperature conversion problems at the end of the notes for this unit.
Thermometry
4
thermometer
Ice cubes
room temperature
Thermometer
(sensor) can add
or absorb heat.
0° at ice-water
interface
Difficulties in Temperature Measurement
Temperature Gradient: e.g., An ice bath is 0 ºC at the ice-water interface but the measured
temperature is somewhere on a gradient from 0 ºC to the temperature of the container walls.
Location of a temperature sensor is very important.
Steady State (of a sensor): The temperature of a sensor (e.g., a thermometer bulb) is constant
only when the rate of heat conduction away = rate of heat conduction to sensor. Glass is a poor
thermal conductor so it will conduct very little heat to or from the sample. Temperature sensors
constructed of metal, however, are rapid thermal conductors and can affect the sample
temperature.
Lag: If temperature of a sample changes, the temperature of a sensor will not change instantly.
Glass equilibrates temperature slowly whereas metals equilibrate quickly.
Disturbance of Sample Temperature: A sensor with large mass will absorb or emit more heat
than a sensor with small mass. Heat capacity is also a factor. Materials with large heat capacity
require larger amounts of heat transfer to change their temperature than materials with low heat
capacities.
Heat capacity (specific heat) is the amount of heat (Btu, calories or Joules) required to raise the
temperature of a unit mass of substance by 1 degree. Water has one of the highest heat capacities
of all materials.
Substance
Specific Heat
Substance
Specific Heat
water
1.0
glass
0.21
ice
0.5
aluminum
0.22
Steam (100°C)
0.48
iron
0.12
ethanol
0.60
zinc
0.09
wood
0.42
mercury
Specific heat values are the same in units of cal/gC, kcal/kgC, and Btu/lbF°
0.03
Convert 1 cal/gC to Btu/lbF given that 1 Btu = 252 cal. and 1 lb = 454 g.
Thermometry
5
Ideally to minimize the disturbance of a sample the following should be true:
1.
2.
3.
4.
5.
sensor mass is small relative to the sample mass
the sensor material has a small heat capacity
the sensor will be insulated from its surroundings
the sample conducts heat readily to the sensor (high transfer coefficient)
the sensor has a large surface area
Temperature Measuring Devices
Any property that varies consistently with temperature can be used for temperature
measurement.
One of the simplest temperature measuring devices used in industry is a thermal strip,
which contains polymeric liquid crystals that change colour reversibly with temperature
changes. These are used as temperature sensors for electrical panels in industry and
domestically as fever strips, aquarium thermometers and mood rings.
Omega LCD reversible temperature strips
Continuous Temperature Measuring Devices
The main types of industrial temperature sensors include:
1. liquid-in-glass thermometers
2. bimetallic devices
3. filled thermal systems
4. resistance temperature detectors (RTD’s), e.g., Cu coils
5. thermistors (metal oxide mixtures)
6. thermocouples (dissimilar metal junctions)
7. optical and radiation pyrometers
Thermometry
6
Liquid-in-Glass Thermometers (Principle: volumetric expansion of a fluid)
Traditionally, these have been the most widely used in industry and laboratory
applications. Mercury or ethanol is the usual fill liquids.
overload capacity
scale etched on glass
stem
small bore
partial immersion line
maximum indication
constriction
range supression cavity
bulb
bulb
Liquid thermometers have a
large volume, thin-walled bulb
and a stem with a small bore.
Those designed for partial
immersion have an etched line
indicating correct immersion
depth.
A range suppression cavity is used
by the manufacturer to change the
range of the thermometer (by
varying its size). It provides a
reservoir that must be filled before
the bore fills.
The maximum indication
constriction allows fluid to
move up the thermometer but
restricts movement down.
The fluid must be shaken back into the bulb. This is more common in medical
thermometers where only the maximum temperature is of interest.
Industrial style thermometers are constructed the same as the lab type but due
to the fragile nature of glass, are protected by a metal case or metal thermowell.
The case may be calibrated rather than the thermometer itself.
Three types of liquid-in-glass thermometers are available: bulb
immersion, stem immersion (partial immersion) and total immersion.
Bulb immersion thermometers are calibrated by the manufacturer to give correct
temperature readings when only the bulb (not the rest of the thermometer) is placed in the
medium to be measured. Stem immersion thermometers are calibrated to give correct
temperature readings when they are immersed to a specified depth in the medium to be
measured. Stem immersion thermometers are easily recognized because the manufacturer
scores a mark, or ‘immersion ring’ below any of the thermometer’s divisions (usually
about 3 inches above the bulb). Total immersion thermometers are calibrated with the
entire thermometer immersed in the medium to be measured. The three types are often
identified on the back (opposite the graduations) by the words bulb, immersion, or total.
Thermometry
7
Accuracy Requirements for Liquid-in-Glass Thermometers
a) a large constant coefficient of thermal expansion of the fluid
b) adhesion of the fluid to the glass must be very low (cohesion >> adhesion); thus water
is not suitable.
c) the fluid level must be clearly visible (red dye is added to alcohol)
d) the fluids freezing and boiling points must satisfy the desired range of use.
Hg: (F pt. = -39 ºC, B pt. = +357 ºC)*
Range: -40 ºC  +500 ºC
Ethanol: (F pt. = -115 ºC, B pt. = 78 ºC)
Range: -100 ºC  + 110 ºC
Accuracy: Industrial = 1% of span
Laboratory = 0.1% of span
* Some Hg thermometers can read up to 500C. They are filled with gas under pressure
to prevent the mercury in the bulb and stem from boiling and condensing at the top of the
thermometer.
Disadvantages: fragile, slow response (lag) due to thermal resistance of glass,
readability is good only at close range, possibility of immersion error and parallax error.
Liquid-in-glass thermometers can’t be used in automatic controllers. They produce no
signal (electrical, pressure, mechanical or otherwise).
Accuracy of the Mercury-in-Glass Thermometer
The mercury thermometer is the simplest and most widely used laboratory thermometer.
Mercury is particularly suitable because it has a very uniform coefficient of expansion, it
does not wet the glass, it is easily purified and the thermometer is easily read.
As the temperature of the thermometer increases, the mercury rises in the capillary
because mercury expands more than glass. A thermometer can be calibrated using the
ice point (fp of H2O) and the steam point (bp of H2O). The distance between the mercury
lengths at 0 and 100C can be marked off into 100 intervals, each of which is 1 degree
Celsius. This assumes that there is a constant difference between the coefficients of cubic
expansion of glass and mercury. The error arising from this assumption is less than 5
parts per 1000 (0.5%) and is negligible, except for highly accurate work that requires
calibration of the thermometer at each temperature used.
High precision thermometers (e.g., for combustion calorimetry) have scale divisions of
0.02 C (0.05 F) and with a magnifier, can be read to  0.002 C ( 0.005 F)!
Thermometry
8
Exposed-Stem Correction for Mercury Thermometers
The depth of a bulb immersion thermometer can be readily adjusted by the technician to
avoid errors due to improper immersion depth.
Partial immersion thermometer readings require no correction for immersion depth
provided they are immersed to a depth approximately near the immersion line and
provided that the air temperature at the exposed stem is near 20C.
Total immersion thermometers require a stem correction when they are not immersed to
the full depth of the mercury column. Unless this thermometer is immersed to the level of
the mercury meniscus, there is a temperature gradient from the bulb to the meniscus and a
stem correction must be applied. This may occur, for example, when such a thermometer
is used for mp or bp determinations in the lab. The part of the mercury column exposed to
the cooler air of the laboratory is obviously not expanded as much as the mercury in the
immersed section and hence the reading will be lower than the true temperature. The error
introduced becomes significant at elevated temperatures reaching 3 to 5 C at 200C.
The temperature correction for an exposed stem with a total immersion thermometer is
given by the following equation
Stem correction ()  K  N  (Tmeasured - Texposed )
where
K = net expansion coefficient of Hg in glass
for C, K = 0.000164 @ 100C and 0.000174 @ 300C (for Pyrex glass)
for F, divide K by 1.8
N = length, measured in degrees, of the exposed section of the Hg column
Tmeasured = observed temperature
Texposed = mean temperature of the exposed section of the Hg column
(determined using an auxiliary thermometer placed along side with
its bulb at the middle of the exposed section of the Hg column)
Problem Determine the corrected temperature reading for a total immersion Pyrex
thermometer that is immersed to the 0C mark and reads 200 C. Ambient temperature at
the midpoint of the exposed Hg column is 35 C. (ans. = 206C)
In cases where the thermometer is used for low temperature measurements, the corrections
would be subtracted rather than added. It should be emphasized that stem corrections are
not accurate for very high or very low temperatures. It is better to avoid the exposed stem
by ensuring correct immersion depth.
In addition to error due to the exposed stem, ordinary thermometers of low cost are
subject to errors due to irregularities in the bore and/or inaccurate scale graduation. For
accurate work, it is therefore essential to check the thermometer at several temperatures
against the melting points of pure solids or the boiling points of pure liquids. Application
of an exposed stem correction will then be unnecessary.
Thermometry
9
Thermometer Calibration Curve
Calibration is accomplished by recording the melting point of five or six very pure
compounds, chosen to melt at a variety of temperatures. From these melting points, a
graph similar to the one shown is plotted. The graph shows the correction factor versus
the observed temperature (using the upper value of the melting point range). For
example, pure benzoic acid melts at 122.4C. If your thermometer records 120.6C, your
correction factor at approximately 120C would be + 1.8 C. Any time you record a
melting point near 120C, you would add 1.8 C to the observed temperature.
Corrected melting point = observed melting point + correction factor
Thermometer Calibration Graph
Correction Factor (C°)
4
3
2
1.8
1.0
1
0.6
0.2
0
0.5
-0.4
-1
-2
-3
80
100
120
140
160
180
200
Observed Temperature (°C)
Other Thermometer Errors
As with other instruments, the technician must be careful to take
readings at eye level to avoid parallax errors. Parallax error can be
eliminated by reading the thermometer with a reading lens that is
attached to the thermometer by a spring clamp so that the lens can
be slid up or down to line up with the meniscus.
For accurate work, standardization must be repeated periodically. Glass exhibits ‘cold
flow’, i.e., flow above the melting point. Slow permanent changes in the glass result in
changes in the volume of the bulb. Readings change as a glass thermometer ages.
Hysteresis: A thermometer should be read whenever possible with a rising Hg thread
rather than a falling Hg thread, and in either case it is good practice to tap the
thermometer gently before reading to minimize sticking.
If a mercury thermometer is kept at 100C for a few minutes and then put into an ice bath,
it will read below 0C for several hours. This effect is due to the slow contraction rate of
glass.
Thermometry
10
Bimetallic Thermometers
The sensing element is a firmly bonded sandwich of 2 metals having different coefficients
of thermal expansion. The low expansion metal is usually "Invar", an iron-36% nickel
alloy. The high expansion metal is usually brass for a low temperature range and a nickel
alloy for a high
FLAT
temperature range.
HELICAL
BIMETALLIC STRIP
BIMETALLIC STRIP
If the bimetallic strip
is flat, differential
Free end
expansion of the two
high expansion
metals causes a
coefficient
bending motion.
Free end
If the strip is wound
connected to
low expansion
into a helix and the
rotating dial
coefficient
lower end of the helix
is fixed while the
bulb
upper end is attached
to the dial of a scale,
the differential
Fixed End
Fixed End
expansion of the two
metals causes a twisting motion of the dial, which indicates temperature on the calibrated
circular scale.
The helix form is normally used as a dial thermometer that is rugged and requires no
maintenance. With age and repeated flexing some fatigue occurs and the thermometer
must be replaced.
Principle of Operation: motion caused by differential expansion of metals of different
coefficients of thermal expansion.
Range: -40 to +425 ºC
Accuracy: 1% of span
Speed of Response: faster than liquid in glass thermometers, i.e., the metal housing and
components have high thermal conductivity
Advantages: a) reasonable accuracy, b) simple, rugged construction, c) large, easy-toread scale, d) small diameter bulb, e) the mechanical action of the bimetallic strip is used
in temperature controllers, for example, when connected to a mercroid switch in a home
heating thermostat.
Thermometry
11
Filled Thermal Systems
The filled thermal system consists of a closed unit under pressure, comprising a bulb
connected by capillary tubing to a helical or C-shaped Bourdon tube.
The system is filled with
a) a liquid (Class I)
b) a liquid and vapour in equilibrium (Class II)
c) a gas (Class III).
In all three classes of instruments, an increase in temperature causes the internal pressure
to increase which in turn causes the Bourdon tube to uncoil. The movement of the
Bourdon tube causes the attached temperature-indicating dial to move across the face of
a temperature scale.
CLASS II FILLED THERMAL SYSTEM
Hotter at Sensor
Colder at Sensor
condensed
liquid
temperature
indicator
vapour
C-shaped
Bourdon tube
capillary
sensing
bulb
liquid
Class II systems, containing volatile fluids like benzene are most common and most
rugged. They need compensation (adjustment) depending upon whether the bulb is hotter
or colder than the ambient temperature. When the sensor is hotter than the Bourdon tube,
condensate held in the tube creates a static head (pressure), which requires compensation
usually by means of a small bimetallic strip attached to the end of the "C" or helix.
Capillary tubing can be up to 100 ft. long for Class I systems. Wide temperature ranges
are measurable, i.e., -400 to +1000 ºF.
Filled system temperature sensors are subject to the "dip effect", especially with mercuryfilled systems. With a sudden increase in temperature the bulb may expand faster than the
fluid causing the fluid to flow into the bulb, showing a temporary temperature dip. In
time, the correct reading is established.
Thermometry
12
Class I filling fluids (high-boiling liquids) include Hg (bp = +357C) and chlorobenzene
(bp = +132C).
Class II filling fluids (low-boiling liquids) include ethyl ether (bp = 35C), ethanol
(bp = +78C) and benzene (bp = +80C).
Class III filling fluids (gases) include ethane (bp = -89C), propane (bp = -42C),
dimethyl ether (bp = -23C), ethyl chloride (bp = +13C) and inert gases
Liquid (Hg) filled thermometer @ cncomco.com

Do the 3 thermometer calibration problems.
Thermometry
13
Electrical Temperature Measuring Devices
A number of substances exhibit electrical properties that vary reproducibly with
temperature. Thermocouples, resistance circuits, and thermistors are preferred in industry
since they can be connected to electronic control circuits.
R1 = 100 
R2 = 200 
A galvanometer
G
is a very sensitive
ammeter measuring very
small electrical currents,
e.g., A
G
R4 = 3 
R3 = 6 
An undeflected
galvanometer needle is in
the null condition. The
bridge is said to be
balanced. I = 0 A.
6V
BALANCED WHEATSTONE BRIDGE
Resistance Temperature Circuits (The Wheatstone Bridge Circuit)
The Wheatstone bridge circuit (named in honor of Sir Charles Wheatstone) is used to
accurately measure resistance changes due to varying temperature. If three of the four
resistors are known, the fourth can be determined by altering the three known resistors
until the bridge is balanced (null point), i.e., no current flows across the galvanometer.
In this condition:
Ohm’s law: V=IR thus
dividing these gives:
V1 = V4
I1R1 = I4R4
I 1  R1 I 4  R 4

I1  R 2 I 4  R 3

and
and
V2 = V3
I1R2 = I4R3
R1 R 4

R2 R3
R 
 R1 = R 2   4 
 R3 
L 
R1 = R 2   4 
 L3 
and where R3 and R4 are replaced by a “slide wire”
where L4 and L3 are lengths of the slide wire measured to the position of the sliding contact
R1 = 100 
R2 = 200 
G
L4 = 25 cm
L3 = 50 cm
wire of uniform resistance
6V
SLIDE WIRE WHEATSTONE BRIDGE
Thermometry
14
Wheatstone bridges work with AC as well as DC by replacing the galvanometer with an
AC milliammeter. Wheatstone bridges can be modified to produce a self-balancing
bridge circuit. When the unknown resistance is temperature dependent, the instrument
scale can be calibrated to read directly in ºC.
A familiar laboratory example is the use of a Wheatstone bridge in the TCD (thermal
conductivity detector) of a gas chromatograph. Two of the resistors in the Wheatstone
bridge are tungsten ‘filaments’ located in the sample and carrier gas flow paths,
respectively, in the detector. The resistances of the tungsten filaments vary directly with
their temperatures. The temperature (and resistance) of the filaments in the sample path
will change as different gases (with different heat capacities) move through the heated
detector. At these times the bridge is unbalanced and causes a deflection of a
plotter/integrator. The magnitude of deflection is directly related to analyte concentration.
The wheatstone bridge is the most sensitive temperature-measuring device currently in
use.

Do the wheatstone bridge problems at the end of this unit.
Thermometry
15
Resistance Temperature Detectors (RTD's)
leads
resistance
wire
glass core
thermowell
RTD's have a resistance that is directly proportional to
temperature as per the following equation
RT = R0 ( 1 + aT + bT2 + cT3 + .... nTn )
where RT = R at T and R0 = R at 0 ºC
and a, b, & c are constants which depend on the material
used.
A simpler form of this equation, which is adequate for most
metals over moderate temperature ranges is ….
R = R0T or R = R0(1+ T)
[ approximates all constants (a, b, c … n) in one term].
Ni
6
RT/R0 5
4
W
3
Pt
2
1
0
100 200
300 400
T(°C)
500
The metals chosen for RTD's should be
corrosion resistant, high melting, and have
linear R vs. T relationships. Platinum is
accepted by the International Committee on
Weights and Measures for temperature
accuracy, is non-corroding and high
melting and linear in the range of -200 to
+1850 ºC.
Tungsten, nickel, and copper are also used
and are less expensive. Nickel, however, is
not linear at high temperatures. Copper is
only useful for low temp. measurements
owing to oxidation at higher temp.
A disadvantage of RTD's is that they required an applied voltage to operate and this
generates heat ("self-heating effect"), requiring compensation and adjustments to the
bridge circuit. One or two additional leads are sometimes added to compensate for the
resistance of the 2 basic leads between the RTD's and the bridge circuit that may be some
distance away. The accuracy of RTD's ranges from 0.01 to 0.1 ºF.
Note that R/R0 = T + 1 is the equation of a straight line with 1 as the y-intercept. The
slope of the line is  (called resistivity or specific resistance). It is unique to each
material.


Mathematically manipulate the equation, R = R0T
equivalent to R = R0(1+ T)
and show that it is
Do the RTD problems for this unit.
Thermometry
16
Thermistors
A thermistor is a thermally sensitive resistor formed from complex metal oxides. It is a
semiconductor whose resistance is a function of its temperature. The R vs. T relationship
is not linear but is logarithmic, as follows.
R  R0  e
 1 1
 
 T0 T 
 
where R  R at T and R 0  R at 0º C, T0  273 K,
 1 1
ln R  ln R0      
 is constant over a small temp. range
 T0 T 
 1 1
 R 
 1 1
ln R  ln R0        ln        
 T0 T 
 R0 
 T0 T 
All T values must be in Kelvin!
semiconducting
material
Thermistors produce relatively large changes in R with T so that high
precision temperature measurements are possible. When thermistors are
leads
connected into bridge circuits like RTD's, very small temperature changes
(0.0005 ºC) can be detected! Long leads can be used since their
resistance is negligible compared to the resistance of the thermistors.
Thermistors as Flow Indicators
In "self-heating" applications, a current is passed through the thermistor so that it heats up
and its R drops. The thermistor is suspended in a flowing fluid, which cools the
thermistor causing its resistance to rise. Eventually a steady state condition is reached
depending upon power supply to the circuit. Anything that changes the rate at which heat
is conducted away from the thermistor will be detected by a change in resistance.
Thermistors thus act as flow indicators (anemometers). Factors can include flow rate,
nature of the fluid (Cp ), and temperature of the fluid.

Sketch and label the plots for
o R vs. T
o lnR vs. (1/T0 – 1/T)
o lnR/R0 vs. (1/T0 – 1/T)

Do the thermistor problem
Omega thermistors
Thermometry
17
Pyrometers
A pyrometer is simply a thermometer. The term is sometimes used as a synonym for
thermometer, but more commonly the term refers to an instrument for measuring
temperatures beyond a thermocouple range. Pyrometers are based on electromagnetic
radiation. All hot bodies emit radiant energy whose intensity depends upon the absolute
temperature of the emitting surface. All objects above absolute zero T emit radiant
energy
Three Classes of Pyrometers
The temperature measuring instruments that respond to all wavelengths are referred to as
total-radiation pyrometers. They are based on the Stefan-Boltzman Law, i.e., the rate, R,
(in W/m2) at which an object of surface area A and absolute temperature T emits
P
radiation is given by
R =  e    T4 (T in Kelvins), P = Power (Watts)
A
-8
2
4
where  = 5.67  10 W/m K
and e = emissivity which depends on the nature of the emitting surface and ranges from
0, for a perfect reflector that does not radiate at all, to 1, for a "black body".
A total-radiation pyrometer is non-selective and measures all wavelengths of light in
determining the temperature.
A second class of pyrometer, called an optical pyrometer, measures only a narrow band of
wavelengths of radiation in the vicinity of the visible and infrared spectra.
The third class of pyrometer, called a partial-radiation pyrometer, is partially selective in
the wavelengths of light used for measurement. It operates on the photo-electric
principle, i.e., a photo sensitive material emits a current in proportion to the intensity and
wavelength of incident light. It can be adjusted to eliminate interference by ambient
radiation from such species as CO2, H2O, and smoke.
Temperature-Colour Relationship
Colour
Temp. ( ºC)
dull red
500-550
dark red
650-750
orange
850-950
yellowish-red
1050-1150
yellowish-white
1250-1350
white
1450-1550
Thermometry
Spectrodyne hand-held optical pyrometer
18
Thermometry
19
Optical Pyrometer
The radiation is focused by a convex lens onto the filament of a lamp. Current is supplied
to the lamp by a battery and can be adjusted by a variable resistor. There is a red optical
filter between the eyepiece and the lamp. This only allows red light to reach the
observer’s eye, which largely eliminates interference by other radiation, i.e., both nonvisible radiation and blue light are absorbed by the red filter. Radiant energy in the visible
range, which is due to heat, is red to white in colour and will pass through the red filter to
the viewer. The resistance of the circuit is varied which in turn varies the current and
hence the temperature of the lamp. When the color of the filament appears the same as the
incident radiation from the hot object, their temperatures must be the same. As with the
total-radiation pyrometer, the scale of the internal milliammeter is calibrated in ºC.
schematic of optical pyrometer from Photonics Directory (www.photonics.com)
Pyrometers measure the temperature of an object based exclusively on the intensity of radiation it
emits without physical contact.
Thermometers, thermocouples, RTD’s and resistance circuits usually mesure temperature by
physical contact.
Thermometry
20
RADIATION:
Radiant energy is transmitted in the form of electromagnetic (EM) waves. Visible light is
just a small portion of the electromagnetic spectrum. All forms of EM waves share
similar properties ...




travel at the speed of light (3.0  108 m/s in a vacuum)
CO2 and H2O absorb large quantities of electromagnetic radiation (EMR)
air is relatively transparent to EMR
all substances which absorb radiation are also capable of emitting EMR
An important example of heat transfer by radiation is global heating. Radiation emitted
by the sun is absorbed by H2O vapor and CO2 in the atmosphere and by the earth itself.
These in turn re-emit radiation usually at a different (longer) wavelength than that of the
radiation absorbed. At least some of the radiation emitted is in the infrared (IR) region
which causes objects, including the atmosphere, to heat up.
Radiant energy striking a body may a) be absorbed, b) be reflected, c) pass through a
body. Bodies that absorb all radiant energy are called “black bodies”, i.e., perfect
absorbers. There is no such a thing as a perfect absorber but porous black surfaces, e.g., a
handful of black soil, approach this condition. A cavity or hollow box with a small
pinhole and a rough, black, interior surface is the best approximation of a black body.
If a body is placed in a container and thermal equilibrium exists between the body and the
container, then the areas receiving energy must emit energy at the same rate, i.e., equal
and opposite rates is the definition of equilibrium.
If a body is a black body then it must emit energy at the maximum possible rate for its
temperature or equilibrium would shift to a higher temperature. Thus a perfect absorber
must also be a perfect emitter.
If the body is non-black, it absorbs only a fraction of the radiant energy and must emit the
same fraction, e.g., polished metal surfaces. This concept, that at thermal equilibrium, the
emissivity of a body equals the absorptivity is known as Kirchoff’s Law.
Emissivity, e, is the ratio of the rate of energy emission by a body to the rate of energy
emission by a perfect radiator (a black body) of the same area and temperature.
Absorptivity, a, is the ratio of the rate of energy absorption by a body to the rate of
energy absorption by a black body of the same area and temperature.
Emissive power, E, of a surface is the total energy emitted per unit time and per unit
surface area ....
E = Q/A
Intensity, i, of radiation is the amount of energy emitted per unit time per solid angle
(steradian).
Thermometry
21
From thermodynamic relationships, the Stephan Boltzman law states that the emissive
power of radiation emitted from a black body per unit area depends solely upon the 4th
power of its absolute temperature ...
E = eT4
e = emissivity (no units)
T = absolute temperature (K)
E = intensity (W/m2)
 = Stephan Boltzman constant = 5.67  10-8 Wm-2K-4 = 1.36  10-11 kcals-1m-2K-4
When the area of the emitting surface is considered, the rate of emitted energy is called
“radiant power”, Q. Q = q/t = AeT4
A = surface area of emitting object (m2)
Q = radiant power (Watts)
The emissivities of various surfaces are listed below ...
surface
e
surface
e
black body
1
glossy white paint
0.91
flat black paint
0.97
polished copper
0.3
human body
0.96
polished steel
0.07
common brick
0.93
perfect reflector
0
The color of a heated object varies with its temperature approximately as shown ...
T (C)
color
T (C)
color
500
just visible
1080
yellow
600
dark red
1250
light yellow
750
cherry red
1400 and above
white
880
orange-red
Since all objects above Absolute zero constantly emit radiation one might wonder why
objects do not cool to Absolute zero. Bear in mind that objects are continuously
absorbing radiant energy at the same time as they are emitting it. A body at the same
temperature as its surroundings emits and absorbs energy at the same rate. The net rate of
transfer of radiant energy between an object and its surroundings at different temperatures
is readily calculated ..
Consider an hot body at T1 surrounded by another cooler body at T2 ...
Net rate of radiation = [rate of radiant emission - rate of radiant absorption]
Q = Q1 - Q2 = [(AeT14) - (AaT24)] = A(eT14 - aT24)
Thermometry
22
Since a  e for a given substance we can further simplify this calculation ...
Q = Ae( T14 - T24)
Problem: Determine the power radiated from the surface of a spherical black body if it
has a diameter of 30.0 cm and is maintained at a temperature of 1500C.
ans. = 1.5  105 W
Problem: The tungsten filament of a light bulb has a length of 15 cm, a diameter of
0.15mm, and an emissivity of 0.31. Neglecting radiative gains, what power does it
radiate at 3.0  103 K?
ans. = 101 W
Problem: The filament of a light bulb maintains a temperature of 1.50  103 K when it is
supplied with 1.50  102 watts. What temperature would it maintain if it were supplied
with 2.00  102 watts? ans. = 1610K

The absorptivity (and emissivity) of an object is a function of the wavelength of
radiation, e.g., you cannot get a sun tan through a glass window, because glass absorbs
UV light (but is transparent to visible light).

The wavelength of emitted (and absorbed)
radiation is a function of temperature, i.e., the
hotter a surface, the shorter the wavelength of
radiation emitted by it. For example, at 37 C,
your body emits IR, while at 600 to 1000 C,
heated metals emit visible light, and the sun
(surface temperature  6200 K) emits short 
radiation including visible, UV, gamma, etc.



relative
intensity
A ‘gray body’ is a surface that absorbs all ‘s
with the same absorptivity and emissivity.
Study the graph of relative intensity amounts of
wavelengths of light emitted at various
temperatures. Recall that the wavelength IR 0
radiation is from 0.8 to 25 m.
1400 oC
1200 oC
1000 oC
600 oC
1
2
3
wavelength (microns)
Do the pyrometer problem at the end of this unit.
Thermometry
23
4
Thermocouples:
The thermocouple is one of the most important temperature measuring devices. They are
used in thousands of temperature measurement applications.
Copper
T2
measuring
junction
T1
reference
junction
Thermoelectricity was discovered in 1821 by
Thomas Seebeck (‘The Seebeck Effect’).
He observed that an emf is generated causing
an electric current to flow continuously in a
closed circuit of two dissimilar metals when
their junctions are maintained at different
temperatures.
A thermocouple consists of two dissimilar
Iron
metals, such as iron and constantan alloy
wires joined at two points (junctions). The
measuring junction or ‘hot junction’ is
inserted into the medium where the temperature is to be measured. The reference junction
or ‘cold junction’ is normally connected to a sensitive potentiometer, which converts the
emf signal into a temperature difference between junctions.
The voltage is nonlinear with respect to
temperature. However, for small changes in
temperature, the voltage is approximately linear, or
V = ST
where V is the change in voltage, S is the Seebeck
coefficient, and T is the change in temperature.
Accurate equations are complicated polynomial
expression. As a result tables of emf vs. T are
generally used to determine temperatures.
Omega thermocouple
Thermometry
24
Law of Intermediate Metals
Additional junctions and additional metals inserted into the thermocouple will not change
the emf provided that both junctions of the new material are at the same temperature
(since T = 0 , emf = 0 for the inserted materials). As a result junctions can be made using
solders of different (third, fourth, etc.) metals.
Law of Intermediate Temperatures:
The emf developed between junctions at different temperatures, say T1 and T3, is the
algebraic sum of the emfs developed by that thermocouple with junction temperatures of
T1 and T2 plus the emf developed by that thermocouple with junction temperatures at T2
and T3.
Advantages:
Thermocouples are rugged and show fast response. They can read very wide temperatures
ranges (from -350C up to +1800C). They are constructed from a variety of metals
including temperature and chemical resistant materials such as platinum. As a result they
can be used in corrosive and high temperature applications where other temperature
sensors would fail, e.g., in molten steel or in liquid HF. Since thermocouples generate an
electrical signal they are well suited to use in automatic controllers.
Reference Junction Temperature and Tables:
The emf measured with a thermocouple can be converted to a temperature using tables
supplied by various manufacturers. The tables are based on a reference temperature of
32F or 0C. However, in practice, the instrument and its reference junction are rarely
that cold. In such cases, if the instrument does not have built-in compensation, one must
apply a correction factor.
For example, suppose the emf measured using a type J thermocouple with its reference
junction at 100F is 41.61 mV. Simply reading the table, without correction, gives a
measuring junction temperature of 1363F. This is not correct because the actual
reference temperature, 100F, is higher than 32F. This results is a decrease in the emf
being generated by the thermocouple compared to what it would generate if the reference
temperature were 32F. Therefore it is necessary to add a corrective emf value to the
observed emf reading. From the appropriate tables the emf between 32 and 100F is
1.942 mV. This should be added to the 41.61 mV emf measured with the 100F reference
temperature. The total of these two values is 43.552 mV, which is what the thermocouple
would generate if its reference junction temperature were 32F. The corrected emf is
43.552 mV and the true temperature is found to be 1418F.
To ensure accurate readings, most thermocouples are now installed with instruments that
provide automatic reference junction compensation. Shielding is often used over
thermocouple cables to ground static charges that might cause interference. Thermocouple
wires are often twisted to cancel out electromagnetic interference.
Thermometry
25
Average and Differential Temperatures:
-
.
+
+
-
+
.
-
Thermocouples connected in series read T
+
.
-
.
+
.
-
+
.
Differential temperatures can be measured by
connecting thermocouples in series.
Connections are made such that the emfs
developed oppose each other. Thus if the
temperatures of both measuring junctions are
equal, regardless of the magnitude, the net
emf will be zero.
Average temperatures can be measured by
connecting thermocouples in parallel. The
voltage measured will be the average of that
developed by all the measuring junctions.
-
Thermocouples connected in parallel read average temp.
Some Standard Thermocouples:
The common types are the Types E, J, K, N and T. The better ones (types B, S, and R) all
contain platinum at varying percentages. The exotic class includes several tungsten alloy
thermocouples usually designated as Type W (something).
Many types of thermocouples are in use.
THERMOCOUPLE TYPES
TYPE
MATERIAL
RANGE °C
Continuous
OUTPUT µV/C°
B
PtRh6%/PtRh30%
100~1600
6
E
Chromel/Constantan
0~1000
56
J
Iron/Constantan
0~700
52
K
Chromel/Alumel
0~1100
40
N
Nicrosil/Nisil
0~1100
38
R
Platinum/PtRh13%
0~1600
10
S
Platinum/PtRh10%
0~1600
10
T
Copper/Constantan
-185~300
38

Do the thermocouple problems in this unit.
Thermometry
26
THERMOMETRY PROBLEMS
Temperature Conversions
1.
Convert the temperature difference of 180 C to a temperature difference in F
2.
Convert the temperature difference of 100 F to a temperature difference in K
3.
Convert the temperature difference of 273 R to a temperature difference in K
4.
Convert
a)
32.0C to F
…………………
b)
0.00F to C
…………………
c)
373.2R to C
…………………
d)
459.7 K to R
…………………
e)
273.2R to K
…………………
f)
32.0 K to F
…………………
g)
273R to C
…………………
h)
–32.0C to F
…………………
i)
-212F to R
…………………
j)
32R to F
…………………
Thermometer Calibration Problems
1.
A thermometer reads 2.0, 3.0 and 4.0 F too high at the respective temperatures of 60.0,
70.0 and 90.0F. At an actual temperature of 100.0F, what would be the most probable
temperature indicated by the thermometer? (hint: plot a graph or use linear regression)
2.
At the freezing point of water the temperature indicated on a thermometer is –3.6C. When
the thermometer reads 32.0C, the true temperature is 38.3C. What would be the most
probable temperature indicated by the thermometer when the true temperature is 15.0C?
3.
A student constructed a homemade thermometer that was marked to read between 0.0C
and 100.0C. When the thermometer was calibrated, the student found the following
relationship:
indicated Temp
actual Temp.
0.0
0.0
28.0
16.0
66.6
90.0
70.0
100.0
At what temperature other than 0.0C would the indicated and actual temperatures be the
same?
Thermometry
27
Wheatstone Bridge Problems
1.
2.
A wheatstone bridge is set up using a galvanometer and a 10 volt battery. The upper
branch consists of two fixed resistances, R1 of 760  and R 2 of 1771  (left to right,
respectively). The lower branch consists of a variable resistor, R3, and an unknown
resistance, Rx (left to right, respectively).
a)
Draw a labeled diagram of the circuit
b)
The variable resistor must be set to a value of 520  to balance the bridge. What is
the value of the unknown resistance, Rx?
A wheatstone bridge has two fixed resistances, R1 and R2, in its lower branch (left to right,
respectively) and in its upper branch a variable resistance, R3, and a temperature sensitive
resistance, R4 (left to right, respectively).
a)
Draw a diagram including a galvanometer and dc power source
b)
If R1 is 1000  and R2 is 100 , the variable resistance, R3, is 30  at 20C and
90  at 320C. Calculate the resistance of the R4 when both R3 and R4 are at 0C.
Assume a linear relationship between temperature and resistance. (hint: use linear
regression)
RTD, Thermistor and Pyrometer Problems
1.
A copper coil ( = 3.9010-3 C -1) has a resistance of 5.00  at 0.0C.
a)
Calculate its resistance at 85.0C
b)
What temperature difference will produce a resistance of 3.2 ?
2.
Calculate the resistance of a carbon conductor at 80C given that  = -0.50  10-3 C -1 and
R = 240  at 0C.
3.
A thermistor with  = -4012 has R = 7759  at 0C. Calculate R at 120C
4.
Calculate the temperature (C) that would cause an object (e = 0.75) to emit radiation at a
rate of 5.00  102 W/m2.
Thermocouple Problems (Calculate all answers to the nearest 0.1)
1.
If an iron-constantan thermocouple produces an emf of 2.00 mV with a cold junction of
79F, what temperature is indicated by the thermocouple?
2.
With a cold junction of -14C an iron-constantan thermocouple produces an emf of 4.82
mV. Calculate the temperature of the hot junction.
3.
The emf of an iron-constantan thermocouple was found to be 44.40 mV with a cold junction
of 0C. If the cold junction changes to 100C, what emf would the thermocouple produce?
4.
An iron-constantan thermocouple is placed in a known temperature of 400F. If the
thermocouple produces an emf of 10.81 mV, what is the temperature of the cold junction?
Thermometry
28
5.
At a temperature of 447C an iron-constantan thermocouple produces an emf of 8.11 mV.
Calculate the temperature of the cold junction.
6.
The emf of an iron-constantan thermocouple was found to be 5.90 mV with a cold junction
of 83F.
a)
Calculate the temperature of the not junction.
b)
If the temperature of the cold junction changes to 67F, what emf would be produced
by the thermocouple?
7.
The measuring junction of an iron-constantan thermocouple is placed in a mixture of ice,
water and salt that is known to have a temperature of 20F. If the room temperature is
75F, calculate the emf that would be produced by the thermocouple.
8.
A copper-constantan thermocouple is used to check the temperature inside a freezer. A
reading of –2.13 mV is obtained with an instrument temperature of 80F. Calculate the
temperature indicated by the thermocouple.
9.
Using a copper-constantan thermocouple with a room temperature of 27C a reading of
–0.91 mV is obtained. Calculate the indicated temperature.
Thermometry
29
Answers to ICS Problem Set on Thermometry
Temperature Conversions
1.
324 Fº
2.
55.6 Kº
3.
151.7 Kº
4.
a)
89.6 ºF
f)
-402.1 F
b)
-17.8 C
g)
-121.5 C
c)
-65.9 C
h)
-25.6 F
d)
827.5 R
i)
247.7 R
e)
151.8 K
j)
-427.7 F
Thermometer Calibration Problems
1.
104.7F
2.
10.3C
3.
ca. 50C
Wheatstone Bridge Problems
1.
Rx = 1212 
2.
R4 = 2.6  (R3 = 26  at 0C)
RTD, Thermistor and Pyrometer Problems
1.
2.
a)
R = 6.66  at 85C
b)
T = -92.3 C
3.
R = 87.3  at 120C
4.
T = 56C
R = 230  at 80C
Thermocouple Problems
1.
2.
3.
4.
5.
6.
7.
8.
147.4 F
78.8 C
39.13 mV
39.7 F
300.1 C
a) 280.5 F b) 6.359 mV
–1.55 mV
–19.5 F
Thermometry
30
9.
4.2 C
Thermometry
30
CH 614 THERMOMETRY TEST SUMMARY
Problems:
1. Convert between the four temperature scales (K, R, C and F) for both temperatures and
temperature differences
2. Do calculations using the Wheatstone bridge equation, RTD’s, thermistors,
thermocouples, thermometry calibration problems (without graph paper – using
algebraic or regression analysis), pyrometers.
Theory:
1. Liquid-in-Glass Thermometer: Draw and or label one. Be able to explain the purpose of
the following: overload capacity cavity, range suppression cavity, partial immersion line,
and bulb
2. Bimetallic Thermometer: Explain the principle of operation and identify its advantages
over the liquid-in-glass thermometer.
3. Filled Thermal Systems: Draw and or label one. Explain the principle of operation and
identify which type of fluid is used in each of the three classes, i.e., liquid, liquid + vapor,
or gas in class 1, or class II or class III.
4. RTD’s. Explain the principle of operation and give examples of typical composition and
describe the output signal as a function of temperature.
5. Thermistors: Explain the principle of operation and give an example of typical
composition and describe the output signal as a function of temperature. State the main
advantage of Thermistors over all other temperature measuring devices.
6. Thermocouples: Explain the principle of operation and give an example of typical
composition and describe the output signal as a function of temperature. Describe the
Seebeck effect, and show how thermocouples can be connected to read average
temperature and temperature differentials. State the main advantages of
thermocouples over other temperature measuring devices.
7. Radiation: Explain a black body, a gray body, absorptivity and emissivity and the type of
surfaces that exhibit these characteristics.
Lab #4: Density, viscosity & surface tension:
 Given the equation for relative viscosity determine relative viscosity, density or time of
flow of fluids in a Ostwalt viscometer
 Given the equation for kinematic viscosity determine kinematic viscosity, density or
absolute viscosity
 Given the equation for surface tension and appropriate data, calculate the surface
tension of a liquid including units in cgs or SI
Lab #3 Particle size distribution:
 given screen analysis data, complete a table like that in Figure 3 on page 4 of the
experiment including calculation of the average particle diameter of each fraction. Be
able to explain/interpret the notation used, i.e., +10-14 and explain/summarize the
purpose and method of quartering a sample.
Thermometry
31
CH 614 THERMOMETRY TEST SUMMARY
The test is closed book. You will be provided with appropriate thermocouple tables as well as
the following info and formulas.
°C
K
°F
°R
bp H2O
100
373.2
212
671.7
fp H2O
0
273.2
32
491.7
absolute
zero
-273.2
0
-459.7
0
R = R0T or R = R0(1+ T)
R  R0  e
 1 1
 
 T0 T 
 
where R  R at T and R 0  R at 0º C, T0  273 K,
 1 1
ln R  ln R0    
 
 T0 T 
P
(T in Kelvins)
 e    T4
A
Q = Ae( T14 - T24)
R=
1 1  t1

2 2  t2
Thermometry
=


where  = 5.67  10-8 W/m2K4
=
F
4   r
32