TEMPERATURE SCALES Table 13.1

13.2 Temperature Scales 455
13.2 TEMPERATURE SCALES
Thermometers measure temperature by exploiting some property of matter that is temperature-dependent.
The familiar liquid-in-glass thermometer relies on thermal expansion:
the mercury or alcohol expands as its temperature rises (or contracts as its
temperature drops) and we read the temperature on a calibrated scale. Since some materials
expand more than others, these thermometers must be calibrated on a scale using
some easily reproducible phenomenon, such as the melting point of ice or the boiling
point of water. The assignment of temperatures to these phenomena is arbitrary.
The most commonly used temperature scale in the world is the Celsius scale. On
the Celsius scale, 0°C is the freezing temperature of water at P = 1 atm (the ice point)
and 100°C is the boiling temperature of water at P = 1 atm (the steam point).
In the United States, the Fahrenheit scale is still commonly used (Fig. 13.2). At
1 atm, the ice point is 32°F and the steam point is 212°F, so the difference between the
steam and ice points is 180°F. The size of the Fahrenheit degree interval is therefore
smaller than the Celsius degree interval: a temperature difference of 1°C is equivalent to
a difference of 1.8°F:
° F
∆T F = ∆T C × 1.8 (13-1)
° C
Since the two scales also have an offset (0°C is not the same temperature as 0°F) conversion
between the two is:
T F = (1.8°F/°C) T C + 32°F
(13-2a)
T C = T F – 32°
F
(13-2b)
1.8°F/
° C
The SI unit of temperature is the kelvin (symbol K, without a degree sign). The
kelvin has the same degree size as the Celsius scale; that is, a temperature difference of
1°C is the same as a difference of 1 K. However, 0 K represents absolute zero—there
are no temperatures below 0 K. The ice point is 273.15 K, so temperature in °C (T C ) and
temperature in kelvins (T) are related.
T C = T – 273.15 (13-3)
Equation (13-3) is the definition of the Celsius scale in terms of the kelvin. Table 13.1
shows some temperatures in kelvins, °C, and °F.
Celsius
100°C
50°C
0°C
–40°C
The freezing and boiling
temperatures of water
depend on the pressure.
Fahrenheit
212°F (Steam
200°F point)
150°F
100°F
50°F
32°F (Ice point)
0°F
–40°F
Figure 13.2 The Fahrenheit
and Celsius temperature scales.
Table 13.1
Some Reference Temperatures in K, °C, and °F
K °C °F K °C °F
Absolute zero 0 –273.15 –459.67
Lowest transient temperature
achieved (laser cooling) 10 –9
Intergalactic space 3 –270 –454
Helium boils 4.2 –269 –452
Nitrogen boils 77 –196 –321
Carbon dioxide freezes
(“dry ice”) 195 –78 –108
Mercury freezes 234 –39 –38
Ice melts/water freezes 273.15 0 32.0
Human body temperature 310 37 98.6
Water boils 373.15 100.00 212.0
Campfire 1,000 700 1,300
Gold melts 1,337 1,064 1,947
Lightbulb filament 3,000 2,700 4,900
Surface of Sun; iron
welding arc 6,300 6,000 11,000
Center of Earth 16,000 15,700 28,300
Lightning channel 30,000 30,000 50,000
Center of Sun 10 7 10 7 10 7
Interior of neutron star 10 9 10 9 10 9

456 Chapter 13 Temperature and the Ideal Gas
Example 13.1
A Sick Friend
A friend suffering from the flu has a fever; her
body temperature is 38.6°C. What is her temperature
in (a) K and (b) °F?
Strategy (a) Kelvins and °C differ only by a shift of
the zero point. Converting from °C to K requires only the
addition of 273.15 K since 0°C (the ice point) corresponds
to 273.15 K. (b) The °F is a different size than the
°C, as well as having a different zero. In the Celsius
scale, the zero is at the ice point. First multiply by
1.8°F/°C to find how many °F above the ice point. Then
add 32°F (the Fahrenheit temperature of the ice point).
Solution (a) The temperature is 38.6 K above the ice
point of 273.15 K. Therefore, the kelvin temperature is
T = 38.6 K + 273.15 K = 311.8 K
(b) First find how many °F above the ice point:
∆T F = 38.6°C × (1.8°F/°C) = 69.5°F
The ice point is 32°F, so
T F = 32.0°F + 69.5°F = 101.5°F
Discussion The answer is reasonable since 98.6°F is
normal body temperature.
Practice Problem 13.1 Normal Body
Temperatures with Two Scales
Convert the normal human body temperature (98.6°F) to
degrees Celsius and kelvins.
13.3 THERMAL EXPANSION OF SOLIDS AND LIQUIDS
Most objects expand as their temperature increases. Long before the cause of thermal
expansion was understood, the phenomenon was put to practical use. For example, the
cooper (barrel maker) heated iron hoops red hot to make them expand before fitting
them around the wooden staves of a barrel. The iron hoops contracted as they cooled,
pulling the staves tightly together to make a leak-tight barrel.
Recall that the fractional length change (strain) caused by a tensile or compressive
stress is proportional to the stress that caused it [Eq. (10-4)]. Similarly, the fractional
length change caused by a temperature change is proportional to the temperature change,
as long as the temperature change is not too great. If the length of a wire, rod, or pipe is
L 0 at temperature T 0 (Fig. 13.3), then
∆ L
= a ∆T (13-4)
L
0
T 0
T > T 0
L 0
Figure 13.3 Expansion of a
solid rod with increasing
temperature.
L
∆L
where ∆L = L – L 0 and ∆T = T – T 0 . The length at temperature T is
L = L 0 + ∆L = (1 + a ∆T) L 0 (13-5)
The constant of proportionality a is called the coefficient of thermal expansion of
the substance. It plays a role in thermal expansion similar to that of the elastic modulus
in tensile stress. If T is measured in kelvins or in degrees Celsius, then a has units of K –1
or °C –1 . Since only the change in temperature is involved in Eq. (13-4), either Celsius or
Kelvin temperatures can be used to find ∆T; a temperature change of 1 K is the same as
a temperature change of 1°C.
As is true for the elastic modulus, the coefficient of thermal expansion has different
values for different solids and also depends to some extent on the starting temperature
of the object. Table 13.2 lists the coefficients of thermal expansion for various solids at
room temperature (20°C).