Temperature Scales
Temperature Scales
Thermometers are used to measure temperature according to well-defined scales of measurement, which use pre-defined 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 \(0\text{º}\text{C}\) and the boiling point at \(\text{100}\text{º}\text{C}\). Its unit is the degree Celsius \((\text{º}\text{C})\). On the Fahrenheit scale (still the most frequently used in the United States), the freezing point of water is at \(\text{32}\text{º}\text{F}\) and the boiling point is at \(\text{212}\text{º}\text{F}\). The unit of temperature on this scale is the degree Fahrenheit \((\text{º}\text{F})\). 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 \(\text{180}/\text{100}=9/5\text{.}\)
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.
Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales, rounded to the nearest degree. The relative sizes of the scales are also shown.
The relationships between the three common temperature scales is shown in the figure above. Temperatures on these scales can be converted using the equations in the table below.
Temperature Conversions
| To convert from . . . | Use this equation . . . | Also written as . . . |
|---|---|---|
| Celsius to Fahrenheit | \(T(\text{º}\text{F})=\cfrac{9}{5}T(\text{º}\text{C})+\text{32}\) | \({T}_{\text{º}\text{F}}=\cfrac{9}{5}{T}_{\text{º}\text{C}}+\text{32}\) |
| Fahrenheit to Celsius | \(T(\text{º}\text{C})=\cfrac{5}{9}(T(\text{º}\text{F})-\text{32})\) | \({T}_{\text{º}\text{C}}=\cfrac{5}{9}({T}_{\text{º}\text{F}}-\text{32})\) |
| Celsius to Kelvin | \(T(\text{K})=T(\text{º}\text{C})+\text{273}\text{.}\text{15}\) | \({T}_{\text{K}}={T}_{\text{º}\text{C}}+\text{273}\text{.}\text{15}\) |
| Kelvin to Celsius | \(T(\text{º}\text{C})=T(\text{K})-\text{273}\text{.}\text{15}\) | \({T}_{\text{º}\text{C}}={T}_{\text{K}}-\text{273}\text{.}\text{15}\) |
| Fahrenheit to Kelvin | \(T(\text{K})=\cfrac{5}{9}(T(\text{º}\text{F})-\text{32})+\text{273}\text{.}\text{15}\) | \({T}_{\text{K}}=\cfrac{5}{9}({T}_{\text{º}\text{F}}-\text{32})+\text{273}\text{.}\text{15}\) |
| Kelvin to Fahrenheit | \(T(\text{º}\text{F})=\cfrac{9}{5}(T(\text{K})-\text{273}\text{.}\text{15})+\text{32}\) | \({T}_{\text{º}\text{F}}=\cfrac{9}{5}({T}_{\text{K}}-\text{273}\text{.}\text{15})+\text{32}\) |
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.
Example: Converting between Temperature Scales: Room Temperature
“Room temperature” is generally defined to be \(\text{25}\text{º}\text{C}\). (a) What is room temperature in \(\text{º}\text{F}\)? (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 \(\text{º}\text{C}\) to \(\text{º}\text{F}\), use the equation
\({T}_{\text{º}\text{F}}=\cfrac{9}{5}{T}_{\text{º}\text{C}}+\text{32}\text{.}\)
2. Plug the known value into the equation and solve:
\({T}_{\text{º}\text{F}}=\cfrac{9}{5}\text{25}\text{º}\text{C}+\text{32}=\text{77}\text{º}\text{F}\text{.}\)
Solution for (b)
1. Choose the right equation. To convert from \(\text{º}\text{C}\) to K, use the equation
\({T}_{\text{K}}={T}_{\text{º}\text{C}}+\text{273}\text{.}\text{15}\text{.}\)
2. Plug the known value into the equation and solve:
\({T}_{\text{K}}=\text{25}\text{º}\text{C}+\text{273}\text{.}\text{15}=\text{298}\phantom{\rule{0.15em}{0ex}}\text{K}\text{.}\)
Example: Converting between Temperature Scales: the Reaumur Scale
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 \(0\text{º}\text{R}\) and the boiling temperature is \(\text{80}\text{º}\text{R}\). If “room temperature” is \(\text{25}\text{º}\text{C}\) 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 \(\text{80}\text{º}\text{R}\). On the Celsius scale it is \(\text{100}\text{º}\text{C}\). Therefore \(\text{100}\text{º}\text{C}=\text{80}\text{º}\text{R}\). Both scales start at \(0\text{º}\) 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:
\({T}_{\text{º}\text{R}}=\cfrac{0\text{.}8\text{º}\text{R}}{\text{º}\text{C}}\phantom{\rule{0.15em}{0ex}}×\phantom{\rule{0.15em}{0ex}}{T}_{\text{º}\text{C}}\text{.}\)
2. Plug the known value into the equation and solve:
\({T}_{\text{º}\text{R}}=\cfrac{0\text{.}8\text{º}\text{R}}{\text{º}\text{C}}\phantom{\rule{0.15em}{0ex}}×\phantom{\rule{0.15em}{0ex}}\text{25}\text{º}\text{C}=\text{20}\text{º}\text{R}\text{.}\)
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws