3

# Measurement

Measurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison, and an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use SI (International System) units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and g/cm3 (for density). In many cases, it is convenient to use prefixes that yield fractional and multiple units, such as microseconds (10-6 seconds) and megahertz (106 hertz), respectively. Quantities can be defined or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the quantity’s number. The uncertainty of a calculated quantity depends on the uncertainties in the quantities used in the calculation and is reflected in how the value is rounded. Quantities are characterized with regard to accuracy (closeness to a true or accepted value) and precision (variation among replicate measurement results). Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.

## 3.1 Measurements

### Learning Objectives

By the end of this section, you will be able to:

• Explain the process of measurement
• Identify the three basic parts of a quantity
• Describe the properties and units of length, mass, volume, density, temperature, and time
• Perform basic unit calculations and conversions in the metric and other unit systems

Measurements provide much of the information that informs the hypotheses, theories, and laws describing the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: the size or magnitude of the measurement (a number); a standard of comparison for the measurement (a unit); and an indication of the uncertainty of the measurement. While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.

The number in the measurement can be represented in different ways, including decimal form and scientific notation. For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as 2.98×105 kg. The mass of the average mosquito is about 0.0000025 kilograms, which can be written as 2.5×10−6 kg.

Units, such as liters, pounds, and centimeters, are standards of comparison for measurements. A 2-liter bottle of a soft drink contains a volume of beverage that is twice that of the accepted volume of 1 liter. The meat used to prepare a 0.25-pound hamburger weighs one-fourth as much as the accepted weight of 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.

The measurement units for seven fundamental properties (“base units”) are listed in Table 3.1. The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Système International d’Unités). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964. Units for other properties may be derived from these seven base units.

Table 3.1

Base Units of the SI System

Property MeasuredName of UnitSymbol of Unit
lengthmeterm
masskilogramkg
timeseconds
temperaturekelvinK
electric currentampereA
amount of substancemolemol
luminous intensitycandelacd

Everyday measurement units are often defined as fractions or multiples of other units. Milk is commonly packaged in containers of 1 gallon (4 quarts), 1 quart (0.25 gallon), and one pint (0.5 quart). This same approach is used with SI units, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means “one thousand,” which in scientific notation is 103 (1 kilometer = 1000 m = 103 m). The prefixes used and the powers to which 10 are raised are listed in Table 3.2

Table 3.2

Common Unit Prefixes

PrefixSymbolFactorExample
femtof10−151 femtosecond (fs) = 1 $×$ 10−15 s (0.000000000000001 s)
picop10−121 picometer (pm) = 1 $×$ 10−12 m (0.000000000001 m)
nanon10−94 nanograms (ng) = 4 $×$ 10−9 g (0.000000004 g)
microµ10−61 microliter (μL) = 1 $×$ 10−6 L (0.000001 L)
millim10−32 millimoles (mmol) = 2 $×$ 10−3 mol (0.002 mol)
centic10−27 centimeters (cm) = 7 $×$ 10−2 m (0.07 m)
decid10−11 deciliter (dL) = 1 $×$ 10−1 L (0.1 L )
kilok1031 kilometer (km) = 1 $×$ 103 m (1000 m)
megaM1063 megahertz (MHz) = 3 $×$ 106 Hz (3,000,000 Hz)
gigaG1098 gigayears (Gyr) = 8 $×$ 109 yr (8,000,000,000 yr)
teraT10125 terawatts (TW) = 5 $×$ 1012 W (5,000,000,000,000 W)

### SI Base Units

The initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters.

### Length

The standard unit of length in both the SI and original metric systems is the meter (m). A meter was originally specified as 1/10,000,000 of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in 1/299,792,458 of a second. A meter is about 3 inches longer than a yard (Figure 3.1); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers (1 km = 1000 m = 103 m), whereas shorter distances can be reported in centimeters (1 cm = 0.01 m = 10−2 m) or millimeters (1 mm = 0.001 m = 10−3 m).

Figure 3.1

The relative lengths of 1 m, 1 yd, 1 cm, and 1 in. are shown (not actual size), as well as comparisons of 2.54 cm and 1 in., and of 1 m and 1.094 yd.

### Mass

The standard unit of mass in the SI system is the kilogram (kg). The kilogram was previously defined by the International Union of Pure and Applied Chemistry (IUPAC) as the mass of a specific reference object. This object was originally one liter of pure water, and more recently it was a metal cylinder made from a platinum-iridium alloy with a height and diameter of 39 mm (Figure 3.2). In May 2019, this definition was changed to one that is based instead on precisely measured values of several fundamental physical constants1. One kilogram is about 2.2 pounds. The gram (g) is exactly equal to 1/1000 of the mass of the kilogram (10−3 kg).

Figure 3.2

This replica prototype kilogram as previously defined is housed at the National Institute of Standards and Technology (NIST) in Maryland. (credit: National Institutes of Standards and Technology)

### Temperature

Temperature is an intensive property. The SI unit of temperature is the kelvin (K). The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word “degree” nor the degree symbol (°). The degree Celsius (°C) is also allowed in the SI system, with both the word “degree” and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at 273.15 K (0 °C) and boils at 373.15 K (100 °C) by definition, and normal human body temperature is approximately 310 K (37 °C). The conversion between these two units and the Fahrenheit scale will be discussed later in this chapter.

### Time

The SI base unit of time is the second (s). Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds = 0.000003 s = 3×10−6 and 5 megaseconds = 5,000,000 s = 5×106 s. Alternatively, hours, days, and years can be used.

### Derived SI Units

We can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.

### Volume

Volume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure 3.3). The standard volume is a cubic meter (m3), a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.

A more commonly used unit of volume is derived from the decimeter (0.1 m, or 10 cm). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter (dm3). A liter (L) is the more common name for the cubic decimeter. One liter is about 1.06 quarts.

A cubic centimeter (cm3) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for cubic centimeter) is often used by health professionals. A cubic centimeter is equivalent to a milliliter (mL) and is 1/1000 of a liter.

Figure 3.3

(a) The relative volumes are shown for cubes of 1 m3, 1 dm3 (1 L), and 1 cm3 (1 mL) (not to scale). (b) The diameter of a dime is compared relative to the edge length of a 1-cm3 (1-mL) cube.

### Density

We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.

The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m3). For many situations, however, this is an inconvenient unit, and we often use grams per cubic centimeter (g/cm3) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm3 (the density of gasoline) to 19 g/cm3 (the density of gold). The density of air is about 1.2 g/L. Table 3.3 shows the densities of some common substances.

Table 3.3

Densities of Common Substances

SolidsLiquidsGases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3water 1.0 g/cm3dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3ethanol 0.79 g/cm3oxygen 1.31 g/L
iron 7.9 g/cm3acetone 0.79 g/cm3nitrogen 1.14 g/L
copper 9.0 g/cm3glycerin 1.26 g/cm3carbon dioxide 1.80 g/L
lead 11.3 g/cm3olive oil 0.92 g/cm3helium 0.16 g/L
silver 10.5 g/cm3gasoline 0.70–0.77 g/cm3neon 0.83 g/L
gold 19.3 g/cm3mercury 13.6 g/cm3radon 9.1 g/L

While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}$

### Example 3.1

#### Calculation of Density

Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm3. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?

#### Solution

The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.
$\text{volume of lead cube}=2.00 cm\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2.00 cm\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2.00 cm={\text{8.00 cm}}^{3}$
$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{90.7 g}}{{\text{8.00 cm}}^{3}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{\text{11.3 g/cm}}^{3}$

(We will discuss the reason for rounding to the first decimal place in the next section.)

(a) To three decimal places, what is the volume of a cube (cm3) with an edge length of 0.843 cm?

(b) If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?

(a) 0.599 cm3; (b) 8.91 g/cm3

### Example 3.2

#### Using Displacement of Water to Determine Density

This exercise uses a simulation to illustrate an alternative approach to the determination of density that involves measuring the object’s volume via displacement of water. Use the simulator to determine the densities iron and wood.

#### Solution

Click the “turn fluid into water” button in the simulator to adjust the density of liquid in the beaker to 1.00 g/mL. Remove the red block from the beaker and note the volume of water is 25.5 mL. Select the iron sample by clicking “iron” in the table of materials at the bottom of the screen, place the iron block on the balance pan, and observe its mass is 31.48 g. Transfer the iron block to the beaker and notice that it sinks, displacing a volume of water equal to its own volume and causing the water level to rise to 29.5 mL. The volume of the iron block is therefore:
${v}_{\text{iron}}=29.5 mL–25.5 mL=4.0 mL$

The density of the iron is then calculated to be:

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{31.48 g}}{4.0 mL}\phantom{\rule{0.2em}{0ex}}=\text{7.9 g/mL}$

Remove the iron block from the beaker, change the block material to wood, and then repeat the mass and volume measurements. Unlike iron, the wood block does not sink in the water but instead floats on the water’s surface. To measure its volume, drag it beneath the water’s surface so that it is fully submerged.

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{1.95 g}}{3.0 mL}\phantom{\rule{0.2em}{0ex}}=\text{0.65 g/mL}$

Note: The sink versus float behavior illustrated in this example demonstrates the property of “buoyancy” (see Supplemental Exercise 42 and Supplemental Exercise 43).

Following the water displacement approach, use the simulator to measure the density of the foam sample.

0.230 g/mL

Supplemental exercises are available if you would like more practice with these concepts.

## 3.2 Measurement Uncertainty, Accuracy, and Precision

### Learning Objectives

By the end of this section, you will be able to:

• Define accuracy and precision
• Distinguish exact and uncertain numbers
• Correctly represent uncertainty in quantities using significant figures
• Apply proper rounding rules to computed quantities

Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. By counting the eggs in a carton, one can determine exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.

### Significant Figures in Measurement

The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.

Figure 3.4

To measure the volume of liquid in this graduated cylinder, you must mentally subdivide the distance between the 21 and 22 mL marks into tenths of a milliliter, and then make a reading (estimate) at the bottom of the meniscus.

Refer to the illustration in Figure 3.4. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.

This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If the coin is weighed on a more sensitive balance, the mass might be 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.

A measurement result is properly reported when its significant digits accurately represent the certainty of the measurement process. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them. Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point. Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located. The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407×10−3; then the number 8.32407 contains all of the significant figures, and 10−3 locates the decimal point.

The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3×103 (two significant figures), 1.30×103 (three significant figures, if the tens place was measured), or 1.300×103 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant. When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17×108 people.

### Significant Figures in Calculations

A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. Take the uncertainty in measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:

1. When adding or subtracting numbers, round the result to the same number of decimal places as the number with the least number of decimal places (the least certain value in terms of addition and subtraction).
2. When multiplying or dividing numbers, round the result to the same number of digits as the number with the least number of significant figures (the least certain value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, “round down” and leave the retained digit unchanged; if it is more than 5, “round up” and increase the retained digit by 1. If the dropped digit is 5, and it’s either the last digit in the number or it’s followed only by zeros, round up or down, whichever yields an even value for the retained digit. If any nonzero digits follow the dropped 5, round up. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)

The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:

• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and a nonzero digit follows it)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)

Let’s work through these rules with a few examples.

### Example 3.3

#### Rounding Numbers

Round the following to the indicated number of significant figures:

(a) 31.57 (to two significant figures)

(b) 8.1649 (to three significant figures)

(c) 0.051065 (to four significant figures)

(d) 0.90275 (to four significant figures)

#### Solution

(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)

(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)

(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)

(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)

Round the following to the indicated number of significant figures:

(a) 0.424 (to two significant figures)

(b) 0.0038661 (to three significant figures)

(c) 421.25 (to four significant figures)

(d) 28,683.5 (to five significant figures)

(a) 0.42; (b) 0.00387; (c) 421.2; (d) 28,684

### Example 3.4

#### Addition and Subtraction with Significant Figures

Rule: When adding or subtracting numbers, round the result to the same number of decimal places as the number with the fewest decimal places (i.e., the least certain value in terms of addition and subtraction).

(a) Add 1.0023 g and 4.383 g.

(b) Subtract 421.23 g from 486 g.

#### Solution

(a) $\begin{array}{l}\\ \begin{array}{l}\hfill \\ \frac{\begin{array}{c}\phantom{\rule{1.4em}{0ex}}1.0023 g\\ \text{+ 4.383 g}\end{array}}{\phantom{\rule{1.5em}{0ex}}5.3853 g}\hfill \end{array}\end{array}$

Answer is 5.385 g (round to the thousandths place; three decimal places)

(b) $\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\\ \frac{\begin{array}{l}\text{}\phantom{\rule{0.8em}{0ex}}486 g\hfill \\ -421.23 g\hfill \end{array}}{\phantom{\rule{1.3em}{0ex}}64.77 g}\end{array}$

Answer is 65 g (round to the ones place; no decimal places) (a) Add 2.334 mL and 0.31 mL.

(b) Subtract 55.8752 m from 56.533 m.

(a) 2.64 mL; (b) 0.658 m

### Example 3.5

#### Multiplication and Division with Significant Figures

Rule: When multiplying or dividing numbers, round the result to the same number of digits as the number with the fewest significant figures (the least certain value in terms of multiplication and division).

(a) Multiply 0.6238 cm by 6.6 cm.

(b) Divide 421.23 g by 486 mL.

#### Solution

(a) $\begin{array}{l}\begin{array}{l}\text{0.6238 cm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}6.6\phantom{\rule{0.2em}{0ex}}\text{cm}=4.11708\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is}\phantom{\rule{0.2em}{0ex}}4.1\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}\left(\text{round to two significant figures}\right)\hfill \\ \text{four significant figures}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{two significant figures}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{two significant figures answer}\hfill \end{array}\hfill \end{array}$

(b) $\begin{array}{l}\frac{\text{421.23 g}}{\text{486 mL}}\phantom{\rule{0.2em}{0ex}}=\text{0.866728... g/mL}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is 0.867 g/mL}\phantom{\rule{0.2em}{0ex}}\left(\text{round to three significant figures}\right)\\ \frac{\text{five significant figures}}{\text{three significant figures}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{three significant figures answer}\end{array}$

(a) Multiply 2.334 cm and 0.320 cm.

(b) Divide 55.8752 m by 56.53 s.

(a) 0.747 cm2 (b) 0.9884 m/s

In the midst of all these technicalities, it is important to keep in mind the reason for these rules about significant figures and rounding—to correctly represent the certainty of the values reported and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.

### Example 3.6

#### Calculation with Significant Figures

One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.

#### Solution

$\begin{array}{lll}V\hfill & =\hfill & l\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}w\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}d\hfill \\ & =\hfill & \text{13.44 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{5.920 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{2.54 dm}\hfill \\ & =\hfill & \text{202.09459}...\phantom{\rule{0.2em}{0ex}}{\text{dm}}^{3}\left(\text{value from calculator}\right)\hfill \\ & =\hfill & {\text{202 dm}}^{3}\text{, or 202 L}\phantom{\rule{0.2em}{0ex}}\left(\text{answer rounded to three significant figures}\right)\hfill \end{array}$

What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm3?

1.034 g/mL

### Example 3.7

#### Experimental Determination of Density Using Water Displacement

A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown. (a) Use these values to determine the density of this piece of rebar.

(b) Rebar is mostly iron. Does your result in (a) support this statement? How?

#### Solution

The volume of the piece of rebar is equal to the volume of the water displaced:
$\text{volume}=\text{22.4 mL}-\text{13.5 mL}=\text{8.9 mL}={\text{8.9 cm}}^{3}$

(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)

The density is the mass-to-volume ratio:

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{69.658 g}}{{\text{8.9 cm}}^{3}}={\text{7.8 g/cm}}^{3}$

(rounded to two significant figures, per the rule for multiplication and division)

From Table 3.3, the density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.

An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown. (a) Use these values to determine the density of this material.

(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.

(a) 19 g/cm3; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in Table 3.3.

### Accuracy and Precision

Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to evaluate both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 3.5).

Figure 3.5

(a) These arrows are close to both the bull’s eye and one another, so they are both accurate and precise. (b) These arrows are close to one another but not on target, so they are precise but not accurate. (c) These arrows are neither on target nor close to one another, so they are neither accurate nor precise.

Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table 3.4.

Table 3.4

Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers

Dispenser #1Dispenser #2Dispenser #3
283.3298.3296.1
284.1294.2295.9
283.9296.0296.1
284.0297.8296.0
284.1293.9296.1

Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).

Supplemental exercises are available if you would like more practice with these concepts.

## 3.3 Mathematical Treatment of Measurement Results

### Learning Objectives

By the end of this section, you will be able to:

• Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities
• Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties

It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:

$\text{speed}=\phantom{\rule{0.2em}{0ex}}\frac{\text{distance}}{\text{time}}$

An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of

$\frac{\text{100 m}}{\text{10 s}}\phantom{\rule{0.2em}{0ex}}=\text{10 m/s}$

Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation among the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:

$\text{time}=\phantom{\rule{0.2em}{0ex}}\frac{\text{distance}}{\text{speed}}$

The time can then be computed as:

$\frac{\text{25 m}}{\text{10 m/s}}\phantom{\rule{0.2em}{0ex}}=\text{2.5 s}$

Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/(m/s) = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”

These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.

### Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

$\frac{\text{2.54 cm}}{\text{1 in.}}\phantom{\rule{0.2em}{0ex}}\text{(2.54 cm}=\text{1 in.) or 2.54}\phantom{\rule{0.2em}{0ex}}\frac{\text{cm}}{\text{in.}}$

Several other commonly used conversion factors are given in Table 3.5.

Table 3.5

Common Conversion Factors

LengthVolumeMass
1 m = 1.0936 yd1 L = 1.0567 qt1 kg = 2.2046 lb
1 in. = 2.54 cm (exact)1 qt = 0.94635 L1 lb = 453.59 g
1 km = 0.62137 mi1 ft3 = 28.317 L1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m1 tbsp = 14.787 mL1 (troy) oz = 31.103 g

When a quantity (such as distance in inches) is multiplied by an appropriate unit conversion factor, the quantity is converted to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

$34\phantom{\rule{0.2em}{0ex}}\overline{)\text{in.}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{2.54 cm}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{in.}}}\phantom{\rule{0.2em}{0ex}}=\text{86 cm}$

Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield $\frac{\text{in.}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{cm}}{\text{in.}}$. Just as for numbers, a ratio of identical units is also numerically equal to one, $\frac{\text{in.}}{\text{in.}}\phantom{\rule{0.2em}{0ex}}=\text{1,}$ and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Dimensional analysis may be used to confirm the proper application of unit conversion factors as demonstrated in the following example.

### Example 3.8

#### Using a Unit Conversion Factor

The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table 3.5).

#### Solution

Given the conversion factor, the mass in ounces may be derived using an equation similar to the one used for converting length from inches to centimeters.
$x\phantom{\rule{0.2em}{0ex}}\text{oz}=\text{125 g}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{unit conversion factor}$

The unit conversion factor may be represented as:

$\frac{\text{1 oz}}{\text{28.349 g}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{\text{28.349 g}}{\text{1 oz}}$

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

$\begin{array}{ccc}x\phantom{\rule{0.2em}{0ex}}\text{oz}\hfill & =& 125\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 oz}}{\text{28.349}\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}\hfill \\ \hfill & =\hfill & \left(\frac{125}{\text{28.349}}\right)\phantom{\rule{0.2em}{0ex}}\text{oz}\hfill \\ \hfill & =\hfill & \text{4.41 oz (three significant figures)}\hfill \end{array}$

Convert a volume of 9.345 qt to liters.

8.844 L

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to ensure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

### Example 3.9

#### Computing Quantities from Measurement Results and Known Mathematical Relations

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.

#### Solution

Since $\text{density}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}$, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A $×$ unit conversion factor. The necessary conversion factors are given in Table 3.5: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. Mass may be converted from pounds to grams as follows:
$\text{9.26}\phantom{\rule{0.2em}{0ex}}\overline{)\text{lb}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{453.59 g}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{lb}}}\phantom{\rule{0.2em}{0ex}}=4.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{g}$

Volume may be converted from quarts to milliliters via two steps:

1. Step 1. Convert quarts to liters.
$\text{4.00}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 L}}{\text{1.0567}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}\phantom{\rule{0.2em}{0ex}}=\text{3.78 L}$
2. Step 2. Convert liters to milliliters.
$\text{3.78}\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1000 mL}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}=3.78\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}$

Then,

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{4.20}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{g}}{3.78\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}}\phantom{\rule{0.2em}{0ex}}=\text{1.11 g/mL}$

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

$\frac{\text{9.26}\phantom{\rule{0.2em}{0ex}}\overline{)\text{lb}}}{\text{4.00}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{453.59 g}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{lb}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1.0567}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}{\text{1000 mL}}\phantom{\rule{0.2em}{0ex}}=\text{1.11 g/mL}$

What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?

2.956 × 10−2 L

### Example 3.10

#### Computing Quantities from Measurement Results and Known Mathematical Relations

(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?

(b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip? #### Solution (a) First convert distance from kilometers to miles: $1250\phantom{\rule{0.2em}{0ex}}\overline{)\text{km}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{0.62137 mi}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{km}}}\phantom{\rule{0.2em}{0ex}}=\text{777 mi}$ and then convert volume from liters to gallons: $213\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1.0567}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{1 gal}}{4\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}\phantom{\rule{0.2em}{0ex}}=\text{56.3 gal}$ Finally, $\text{(average) mileage}=\phantom{\rule{0.2em}{0ex}}\frac{\text{777 mi}}{\text{56.3 gal}}\phantom{\rule{0.2em}{0ex}}=\text{13.8 miles/gallon}=\text{13.8 mpg}$ Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows: $\frac{1250\phantom{\rule{0.2em}{0ex}}\overline{)\text{km}}}{213\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{0.62137 mi}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{km}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}{\text{1.0567}\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{4\phantom{\rule{0.2em}{0ex}}\overline{)\text{qt}}}{\text{1 gal}}=\text{13.8 mpg}$ (b) Using the previously calculated volume in gallons, we find: $56.3\phantom{\rule{0.2em}{0ex}}\overline{)\text{gal}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{3.80}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{gal}}}=\text{214}$ #### Check Your Learning A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits). (a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip? (b) If gasoline costs$3.90 per gallon, what was the fuel cost for this trip?

(a) 51 mpg; (b) \$62

### Conversion of Temperature Units

We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The liquid in a common glass thermometer changes its volume as the temperature changes, and the position of the trapped liquid's surface along a printed scale may be used as a measure of temperature.

Temperature scales are defined relative to selected reference temperatures: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).

Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:

$\text{length in feet}=\left(\frac{\text{1 ft}}{\text{12 in.}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{length in inches}$

where y = length in feet, x = length in inches, and the proportionality constant, m, is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one (y = mx + b). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points (b).

The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as x and the Fahrenheit temperature as y, the slope, m, is computed to be:

$m=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}y}{\text{Δ}x}\phantom{\rule{0.2em}{0ex}}=\frac{\text{212 °F}-\text{32 °F}\phantom{\rule{0.2em}{0ex}}}{\text{100 °C}-\text{0 °C}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{180 °F}}{\text{100 °C}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{9 °F}}{\text{5 °C}}$

The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:

$b=y-mx=\text{32 °F}-\phantom{\rule{0.2em}{0ex}}\frac{\text{9 °F}}{\text{5 °C}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0 °C}=\text{32 °F}$

The equation relating the temperature (T) scales is then:

${T}_{\text{°F}}=\left(\frac{\text{9 °F}}{\text{5 °C}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{T}_{\text{°C}}\right)+\text{32 °F}$

An abbreviated form of this equation that omits the measurement units is:

${T}_{\text{°F}}=\phantom{\rule{0.2em}{0ex}}\left(\frac{9}{5\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{T}_{\text{°C}}\right)+32$

Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:

${T}_{\text{°C}}=\phantom{\rule{0.2em}{0ex}}\frac{5\phantom{\rule{0.2em}{0ex}}}{9}\left({T}_{\text{°F}}-32\right)$

As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. Since the kelvin temperature scale is absolute, a degree symbol is not included in the unit abbreviation, K. The early 19th-century discovery of the relationship between a gas’s volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).

The freezing temperature of water on this scale is 273.15 K and its boiling temperature is 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of $1\phantom{\rule{0.2em}{0ex}}\frac{\text{K}}{\text{°C}}$. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:

${T}_{\text{K}}={T}_{\text{°C}}+\text{273.15}$
${T}_{\text{°C}}={T}_{\text{K}}-\text{273.15}$

The 273.15 in these equations has been determined experimentally, so it is not exact. Figure 3.6 shows the relationship among the three temperature scales.

Figure 3.6

The Fahrenheit, Celsius, and kelvin temperature scales are compared.

Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.

### Example 3.11

#### Conversion from Celsius

Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?

#### Solution

$\text{K}=\text{°C}+273.15=37.0+273.2=\text{310.2 K}$
$\text{°F}=\frac{9}{5}\text{°C}+32.0=\left(\frac{9}{5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}37.0\right)+32.0=66.6+32.0=\text{98.6 °F}$

Convert 80.92 °C to K and °F.

354.07 K, 177.7 °F

### Example 3.12

#### Conversion from Fahrenheit

Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?

#### Solution

$\text{°C}=\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\text{(°F}-\text{32)}=\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\left(450-32\right)=\phantom{\rule{0.2em}{0ex}}\frac{5}{9}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}418=\text{232 °C}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{set oven to 230 °C}\phantom{\rule{2em}{0ex}}\left(\text{two significant figures}\right)$

$\text{K}=\text{°C}+273.15=230+273=\text{503 K}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\text{K}\phantom{\rule{2em}{0ex}}\left(\text{two significant figures}\right)$

Convert 50 °F to °C and K.

10 °C, 280 K