Thermometers are used to measure temperature according to welldefined scales of measurement, which use predefined reference points to help compare quantities. The three most common temperature scales are the Fahrenheit, Celsius, and Kelvin scales. A temperature scale can be created by identifying two easily reproducible temperatures. The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used.
The Celsius scale (which replaced the slightly different centigrade scale) has the freezing point of water at 0ºC0ºC size 12{0°C} {} and the boiling point at 100ºC100ºC size 12{"100"°C} {}. Its unit is the degree Celsius(ºC)(ºC) size 12{ \( °C \) } {}. On the Fahrenheit scale (still the most frequently used in the United States), the freezing point of water is at 32ºF32ºF size 12{"32"°F} {} and the boiling point is at 212ºF212ºF size 12{"212"°F} {}. The unit of temperature on this scale is the degree Fahrenheit(ºF)(ºF) size 12{ \( °F \) } {}. Note that a temperature difference of one degree Celsius is greater than a temperature difference of one degree Fahrenheit. Only 100 Celsius degrees span the same range as 180 Fahrenheit degrees, thus one degree on the Celsius scale is 1.8 times larger than one degree on the Fahrenheit scale 180/100=9/5.180/100=9/5. size 12{"180"/"100"=9/5 "." } {}
The Kelvin scale is the temperature scale that is commonly used in science. It is an absolute temperature scale defined to have 0 K at the lowest possible temperature, called absolute zero. The official temperature unit on this scale is the kelvin, which is abbreviated K, and is not accompanied by a degree sign. The freezing and boiling points of water are 273.15 K and 373.15 K, respectively. Thus, the magnitude of temperature differences is the same in units of kelvins and degrees Celsius. Unlike other temperature scales, the Kelvin scale is an absolute scale. It is used extensively in scientific work because a number of physical quantities, such as the volume of an ideal gas, are directly related to absolute temperature. The kelvin is the SI unit used in scientific work.
The relationships between the three common temperature scales is shown in Figure 4. Temperatures on these scales can be converted using the equations in Table 1.
Table 1: Temperature Conversions
To convert from . . . 
Use this equation . . . 
Also written as . . . 
Celsius to Fahrenheit 
T
º
F
=
9
5
T
º
C
+
32
T
º
F
=
9
5
T
º
C
+
32
size 12{T left (°F right )= { {9} over {5} } T left (°C right )+"32"} {}

T
º
F
=
9
5
T
º
C
+
32
T
º
F
=
9
5
T
º
C
+
32
size 12{T rSub { size 8{°F} } = { {9} over {5} } T rSub { size 8{°C} } +"32"} {}

Fahrenheit to Celsius 
T
º
C
=
5
9
T
º
F
−
32
T
º
C
=
5
9
T
º
F
−
32
size 12{T left (°C right )= { {5} over {9} } left [T left (°F right )  "32" right ]} {}

T
º
C
=
5
9
T
º
F
−
32
T
º
C
=
5
9
T
º
F
−
32
size 12{T rSub { size 8{°C} } = { {5} over {9} } left (T rSub { size 8{°F} }  "32" right )} {}

Celsius to Kelvin 
T
K
=
T
º
C
+
273
.
15
T
K
=
T
º
C
+
273
.
15
size 12{T left (K right )=T left (°C right )+"273" "." "15"} {}

T
K
=
T
º
C
+
273
.
15
T
K
=
T
º
C
+
273
.
15
size 12{T rSub { size 8{K} } =T rSub { size 8{°C} } +"273" "." "15"} {}

Kelvin to Celsius 
T
º
C
=
T
K
−
273
.
15
T
º
C
=
T
K
−
273
.
15
size 12{T left (°C right )=T left (K right )  "273" "." "15"} {}

T
º
C
=
T
K
−
273
.
15
T
º
C
=
T
K
−
273
.
15
size 12{T rSub { size 8{°C} } =T rSub { size 8{K} }  "273" "." "15"} {}

Fahrenheit to Kelvin 
T
K
=
5
9
T
º
F
−
32
+
273
.
15
T
K
=
5
9
T
º
F
−
32
+
273
.
15
size 12{T left (K right )= { {5} over {9} } left [T left (°F right )  "32" right ]+"273" "." "15"} {}

T
K
=
5
9
T
º
F
−
32
+
273
.
15
T
K
=
5
9
T
º
F
−
32
+
273
.
15
size 12{T rSub { size 8{K} } = { {5} over {9} } left (T rSub { size 8{°F} }  "32" right )+"273" "." "15"} {}

Kelvin to Fahrenheit 
T
(
º
F
)
=
9
5
T
K
−
273
.
15
+
32
T
(
º
F
)
=
9
5
T
K
−
273
.
15
+
32
size 12{T \( °F \) = { {9} over {5} } left [T left (K right )  "273" "." "15" right ]+"32"} {}

T
º
F
=
9
5
T
K
−
273
.
15
+
32
T
º
F
=
9
5
T
K
−
273
.
15
+
32
size 12{T rSub { size 8{°F} } = { {9} over {5} } left (T rSub { size 8{K} }  "273" "." "15" right )+"32"} {}

Notice that the conversions between Fahrenheit and Kelvin look quite complicated. In fact, they are simple combinations of the conversions between Fahrenheit and Celsius, and the conversions between Celsius and Kelvin.
“Room temperature” is generally defined to be 25ºC25ºC size 12{"25"°C} {}. (a) What is room temperature in ºFºF size 12{°F} {}? (b) What is it in K?
Strategy
To answer these questions, all we need to do is choose the correct conversion equations and plug in the known values.
Solution for (a)
1. Choose the right equation. To convert from ºCºC size 12{°C} {} to ºFºF size 12{°F} {}, use the equation
T
º
F
=
9
5
T
º
C
+
32
.
T
º
F
=
9
5
T
º
C
+
32
.
size 12{T rSub { size 8{°F} } = { {9} over {5} } T rSub { size 8{°C} } +"32" "." } {}
(1)2. Plug the known value into the equation and solve:
T
º
F
=
9
5
25
º
C
+
32
=
77
º
F
.
T
º
F
=
9
5
25
º
C
+
32
=
77
º
F
.
size 12{T rSub { size 8{°F} } = { {9} over {5} } "25"°C+"32"="77"°F "." } {}
(2)
Solution for (b)
1. Choose the right equation. To convert from ºCºC size 12{°C} {} to K, use the equation
T
K
=
T
º
C
+
273
.
15
.
T
K
=
T
º
C
+
273
.
15
.
size 12{T rSub { size 8{K} } =T rSub { size 8{°C} } +"273" "." "15" "." } {}
(3)2. Plug the known value into the equation and solve:
T
K
=
25
º
C
+
273
.
15
=
298
K
.
T
K
=
25
º
C
+
273
.
15
=
298
K
.
size 12{T rSub { size 8{K} } ="25"°C+"273" "." "15"="298"`K "." } {}
(4)The Reaumur scale is a temperature scale that was used widely in Europe in the 18th and 19th centuries. On the Reaumur temperature scale, the freezing point of water is 0ºR0ºR size 12{0°R} {} and the boiling temperature is 80ºR80ºR size 12{"80"°R} {}. If “room temperature” is 25ºC25ºC size 12{"25"°C} {} on the Celsius scale, what is it on the Reaumur scale?
Strategy
To answer this question, we must compare the Reaumur scale to the Celsius scale. The difference between the freezing point and boiling point of water on the Reaumur scale is 80ºR80ºR size 12{"80"°R} {}. On the Celsius scale it is 100ºC100ºC size 12{"100"°C} {}. Therefore 100º
C=80ºR100º
C=80ºR size 12{"100"°C="80"°R} {}. Both scales start at 0
º0
º size 12{0°} {} for freezing, so we can derive a simple formula to convert between temperatures on the two scales.
Solution
1. Derive a formula to convert from one scale to the other:
T
º
R
=
0
.
8
º
R
º
C
×
T
º
C
.
T
º
R
=
0
.
8
º
R
º
C
×
T
º
C
.
size 12{T rSub { size 8{°R} } = { {0 "." 8°R} over {°C} } times T rSub { size 8{°C} } "." } {}
(5)2. Plug the known value into the equation and solve:
T
º
R
=
0
.
8
º
R
º
C
×
25
º
C
=
20
º
R
.
T
º
R
=
0
.
8
º
R
º
C
×
25
º
C
=
20
º
R
.
size 12{T rSub { size 8{°R} } = { {0 "." 8°R} over {°C} } times "25"°C="20"°R "." } {}
(6)