A copper calorimeter of \(200\,\text{g}\) behaves, for heating, like a small mass of water. The chapter on calorimetry finds that mass as \(80/4200\,\text{kg}\). A calculator shows \(0.019047619\). The masses and the specific heats were known to two or three figures at most. So which of these eight digits mean anything?
The last digit is a guess
Lay a rod along a rule and read its length. Near the end of the rod you must judge the position between two marks. That last digit is an estimate. The digits before it are sure, because the marks fix them. Fig 2.1 shows the same rod against two rules, one marked in centimetres and one in millimetres.
The sure digits and the one estimated digit together are the significant figures of a reading. The centimetre rule gives \(4.3\,\text{cm}\), two significant figures. The millimetre rule gives \(4.27\,\text{cm}\), three. Writing \(4.270\,\text{cm}\) would claim a fourth digit that neither rule can give. So the number of significant figures tells the reader how fine the measurement was.
The significant figures of a measured number are all its digits that are known reliably, together with the first digit that is uncertain. Zeros that only fix the position of the decimal point are not significant.
Counting the significant figures
Most digits are easy to judge. The zeros need four rules.
- Every non-zero digit is significant: \(4.27\) has three.
- A zero between non-zero digits is significant: \(4.07\) has three.
- Zeros before the first non-zero digit only place the point, so they are not significant. \(0.0250\) has three: the 2, the 5 and the final 0.
- A zero at the end of a number with a decimal point is significant: \(4.270\) has four.
The rule for a result
A result can be no more reliable than the data it comes from. For a product or a quotient, count the significant figures of each factor. The result keeps the smallest of these counts. For a sum or a difference, look at the decimal places instead. The result keeps as many decimal places as the term with the fewest.
Why should the product keep so few figures? A rectangle measured as \(3.61\,\text{cm}\) by \(2.1\,\text{cm}\) could really be a little smaller or a little larger. The breadth \(2.1\,\text{cm}\) means a value between about 2.05 and 2.15. Fig 2.2 draws the rectangles that the readings allow.
The possible areas run from \(7.39\) to \(7.77\,\text{cm}^2\). The digit 7 is sure, and the first decimal is already uncertain. The calculator's \(7.581\) claims two decimals that the rule could never give. So the area is written \(7.6\,\text{cm}^2\), with two significant figures, the same number as the breadth. The weakest datum sets the limit.
How many significant figures does the length \(0.02050\,\text{m}\) have?
The length and the breadth of a rectangular sheet are measured as \(3.61\,\text{cm}\) and \(2.1\,\text{cm}\). What are its area and its perimeter, written to the correct significant figures?
Rounding off
To round a number, keep the digits you need and look at the first digit to be dropped. If it is less than 5, the kept digits stay as they are, so \(7.54\) to two figures is \(7.5\). If the dropped part is more than 5, raise the last kept digit by one. So \(7.581\) becomes \(7.6\), because the dropped 8 is more than 5. In the same way, \(0.019047\) to two figures is \(0.019\).
A dropped part of exactly 5 needs one more rule. If we always raised the digit, rounding would push many results upward. So the rule used in NCERT books is: raise the kept digit if it is odd, and leave it if it is even. Then \(2.45\) becomes \(2.4\) and \(2.35\) also becomes \(2.4\). Over many roundings, half of such numbers are raised and half are not.
Rounding is like reporting a cricket score to the nearest ten runs. A score of 243 is "about 240" and 247 is "about 250". A score of 245 lies exactly between them, and a fixed habit decides it. The even rule is that habit, chosen so that the rounded scores of a whole season do not drift upward.
Round only at the end. Every intermediate result should keep one digit more than the answer will need. Rounding too early can shift the last digit of the answer. The example below shows this with the calorimeter of the phenomenon.
A \(200\,\text{g}\) copper calorimeter (\(c = 400\,\text{J/(kg K)}\)) holds \(100\,\text{g}\) of water (\(c_w = 4200\,\text{J/(kg K)}\)). How much heat warms both by \(10\,\text{K}\)? Work it once through the water equivalent rounded to \(19\,\text{g}\), and once without rounding.
The heat capacity of the copper is \(0.2 \times 400 = 80\,\text{J/K}\). Its water equivalent is \(80/4200 = 0.019047\,\text{kg}\), which the physics chapter rounds to \(19\,\text{g}\).
Rounded early: \((0.100 + 0.019) \times 4200 \times 10 = 0.119 \times 42\,000 = 4998\,\text{J}\).
Without rounding: \((0.1 \times 4200 + 80) \times 10 = (420 + 80) \times 10 = 5000\,\text{J}\).
The early rounding moved the fourth figure. The data carry about two significant figures, so the answer is \(5.0 \times 10^3\,\text{J}\), the 5000 J of the physics chapter. The water takes 4200 J and the copper 800 J.
The phenomenon is answered. Of the calculator's \(0.019047619\), only the first two significant figures, \(0.019\), are worth writing. The rest describe the calculator, not the calorimeter.
Round \(2.45\) and \(2.35\) to two significant figures by the rule of this lesson.
A child's fever is \(102\,°\text{F}\). The conversion gives \(C = (102 - 32) \times 5/9 = 38.888\ldots\). The reading has three significant figures. How should the Celsius temperature be written?
Zeros at the end of a whole number
The specific heat of water is often written \(4200\,\text{J/(kg K)}\). Are its two zeros measured, or do they only fill places? The number alone cannot say. Scientific notation removes the doubt. \(4.2 \times 10^3\) has two significant figures, and \(4.200 \times 10^3\) has four. By convention, a whole number such as 4200 written without a point is read with the zeros not significant.
A change of unit moves the decimal point but does not change the measurement. So it never changes the number of significant figures. A guitar string of \(2.0\,\text{g}\) is \(0.0020\,\text{kg}\) and \(2.0 \times 10^{-3}\,\text{kg}\), two significant figures each time. Exact numbers are different again. The 2 in \(2\pi r\), the \(\tfrac12\) in \(\tfrac12 mv^2\) and a count of 20 swings have no uncertainty, so they never limit a result.
Which of these measured values have exactly three significant figures?
Significant figures are only a rough way to show uncertainty. The value \(9.9\) has two figures and \(10.1\) has three, yet both are known to about \(0.1\). Subtraction is the worst case. Two lengths of \(10.25\) and \(10.21\,\text{cm}\) have four figures each, but their difference, \(0.04\,\text{cm}\), has only one. When the uncertainty truly matters, estimate it directly, as the chapter Measurement and Error does.
Significant figures decide how every worked answer of the book is written. The calorimetry chapter rounds a water equivalent to 19 g and reports the heat as 5000 J. The temperature chapter writes a fever of \(102\,°\text{F}\) as \(38.9\,°\text{C}\). The next chapter, Measurement and Error, replaces these rules by the actual uncertainty of each reading.
- Copying every digit of the calculator into the answer.Keep the fewest significant figures of the data for a product or quotient, and the fewest decimal places for a sum or difference.
- Counting the leading zeros of \(0.0250\).Leading zeros only place the point. Write \(2.50 \times 10^{-2}\) and count three.
- Rounding each intermediate step.Carry one extra digit through the working and round only the final answer.
- Letting the 2 of \(2\pi r\) limit the answer to one figure.Exact numbers and counts have unlimited precision. Only measured values limit a result.