Significant digits (physical science)

[C_l_e_0..Q_c]

I always had problems with significant digits as a student, and it seems I still do as a teacher...

 

So...

If I were to cut three aluminium foil square of

1. 50mm

2. 100mm

3. 200mm

of length, how many significant numbers do I have for the length?

 

The solution manual tells me I have 3 significant numbers for the 50mm, and 4 for the other two. How come?

[Sandra in NC]

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/index.html

 

There seems to be some abiguity about significant figures for whole numbers as discussed in this tutorial.

 

 

(5) When a number ends in zeroes that are not to the right of a decimal point, the zeroes are not necessarily significant:

 

190 miles may be 2 or 3 significant figures,

 

50,600 calories may be 3, 4, or 5 significant figures.

 

 

 

The potential ambiguity in the last rule can be avoided by the use of standard exponential, or "scientific," notation. For example, depending on whether the number of significant figures is 3, 4, or 5, we would write 50,600 calories as:

 

5.06 × 104 calories (3 significant figures)

 

5.060 × 104 calories (4 significant figures), or

 

5.0600 × 104 calories (5 significant figures)

 

[C_l_e_0..Q_c]

I get that part, I think but it still doesn't tell me how 50 can be THREE significant digits.

There was no 50.0

it was clearly written as 50mm

[Kareni]

Cleo,

 

Is there more information available in the question? Does it say something to the effect that you can cut to the nearest tenth of a millimeter?

 

Regards,

Kareni

[C_l_e_0..Q_c]

Cleo,

 

Is there more information available in the question? Does it say something to the effect that you can cut to the nearest tenth of a millimeter?

 

Regards,

Kareni

 

Nope, nothing at all. And who would expect a child to cut aluminium foil into a perfect square to the nearest millimeter anyway? I was lucky to get two squares out of three that were square-shaped!

2.Cut out three squares of aluminum foil with sides of the following lengths: 50 mm, 100 mm, and 200 mm.

3.To determine the area of the 50-mm foil square, measure the length of one of its sides and then square it. Record the length and area in your data table

 

and the question:

How many significant figures were there in your measurement of the length of each square of aluminum foil?

[Caroline]

The answer is assuming you used a ruler with millimeters to measure your foil. The wat I read the question it is askining for the number of significant digits from your actual measurement of the foil, not from the number in the problem.

 

When measuring, you report one more digit than shows on your instrument. So you measured 50 mm, and then "guess" at the tenths digit. So you measured 50.5 mm. Or, say you didn't cut is perfectly, you measure 52.3 mm. I was taught in science classes from middle school through grad school to report one more digit than your device measures. That extra measurement digit gives you the third sig fig.

[MyThreeSons]

The answer is assuming you used a ruler with millimeters to measure your foil. The wat I read the question it is askining for the number of significant digits from your actual measurement of the foil, not from the number in the problem.

 

 

I'm pretty sure this is what they're trying to say.

[Maura in NY]

I was taught in science classes from middle school through grad school to report one more digit than your device measures. That extra measurement digit gives you the third sig fig.

 

The accuracy of your measurement is one place more than the markings on your device, and that determines the number of sig fig.

 

Maura

[C_l_e_0..Q_c]

Thanks everyone. I don't remember that part at all. Which is weird because I did study sciences in university (theoretical physics) for one year before becoming an engineer. I just never quite got the hang of significant digits though.

[StephanieZ]

I am pretty confident that significant digits indicate a level of "reality" of the measurement.

 

So, if your instrument of measurement only measures to the nearest whole number, then the significant digit would be the final whole number, not tenths. Now, if you feel you can estimate well to the tenths of the whole number, then you can decide that is significant and do it to the tenths. The idea of significant digits is to only report/use meaningful details.

 

So, if you can measure volume of a beaker to the nearest 10 mls . . .

 

And I say, OK there are 460 ml in the beaker based on my reading of the beaker markings (to nearest 10 mls), then there are only TWO sig. figs. (4 & 6).

 

So, now I say, OK, you have 1345 beakers full of lemonade -- each holding 460 ml of lemondae. . . So, how much lemonade do I have??

 

You would do 1345 x 460 ml = 618,700 ml rounded to 620,000 -- we rounded down to two sig fig b/c you have to go with the fewest sig figs (two in the 460, 4 in 1345).

 

The reason we use sig figs is b/c we are being clear that that number 620,000 is only PRECISE to the two digits. . . it could just as well be 618,000 or 624,000. . . any number in the range of 615,000 to 624,999. . . if you go back to your 460 that was only accurate to 10 ml. . . recall that since it is only accurate to 10 ml, that beaker coudl REALLY have, say 456 ml or 464 ml in it. . .not exactly 460. . . . If you substitute those numbers into your x1345 equation, you'll get a range of values in the ballpark of the 615,000 to 624,999. . .

 

The whole purpose of rounding to sig figs is to be honest about how precise our data is. That way, if you are using a beaker that measures to the nearest ml VS a beaker that measures the nearest 100 ml. .. well, that precision is reflected in your final answer. ..

 

HTH

[MyThreeSons]

The accuracy of your measurement is one place more than the markings on your device, and that determines the number of sig fig.

 

Maura

 

I majored in Mechanical Engineering, so I took LOTS of math and science classes, including plenty of labs. I don't remember being concerned with the number of significant figures when solving textbook problems, but it did come into play when we were doing our calculations for our lab experiments. I do remember that we would set our calculators to a set number of digits past the decimal point (usually 3 or 4, depending on the course) when doing homework or tests. And when we did take significant figures into account, I don't remember having to think about whether I was adding/subtracting or multiplying/dividing, as is taught in the Apologia texts. Our final answer simply couldn't indicate more precision than our least precise measurement.

 

Does this jive with anyone else's memories, or am I all wet?

[C_l_e_0..Q_c]

Does this jive with anyone else's memories, or am I all wet?

 

That's pretty much what I remember too, but I was also in engineering.

Significant digits rarely had any significance, because we would multiply any answer by 10% (or more) for some extra security. If a bridge needs to hold X tons, make sure it can hold 1.1X (or more!) but keep it within budget. Who cared about significant digits ? :)

[C_l_e_0..Q_c]

When measuring, you report one more digit than shows on your instrument. So you measured 50 mm, and then "guess" at the tenths digit. So you measured 50.5 mm. Or, say you didn't cut is perfectly, you measure 52.3 mm. I was taught in science classes from middle school through grad school to report one more digit than your device measures. That extra measurement digit gives you the third sig fig.

 

Thanks Caroline. I finally got the science book away from my son, and yes, it does say to report one more digit, and then proceeds with an example where they don't! Argh. No wonder I never got the hang of significant digits.

 

In their example, they have an analog clock where you can read in 5 minutes increments. They claim this clock has one significant digit. Then a digital clock that reads to the second, so you read 5 min 15 secs, or 5.25 min, for three significant figures. Where is that extra digit?

 

I'll be glad when we're passed that part!

[Caroline]

Thanks Caroline. I finally got the science book away from my son, and yes, it does say to report one more digit, and then proceeds with an example where they don't! Argh. No wonder I never got the hang of significant digits.

 

In their example, they have an analog clock where you can read in 5 minutes increments. They claim this clock has one significant digit. Then a digital clock that reads to the second, so you read 5 min 15 secs, or 5.25 min, for three significant figures. Where is that extra digit?

 

I'll be glad when we're passed that part!

 

The digital clock only has three because it is digital. You don't have a way of estimating the next digit. So the rules are different for a digital instrument. Sorry I didn't mention that before. Analog instruments have an extra sig fig. Digital ones do not. Does that make sense?

[C_l_e_0..Q_c]

The digital clock only has three because it is digital. You don't have a way of estimating the next digit. So the rules are different for a digital instrument. Sorry I didn't mention that before. Analog instruments have an extra sig fig. Digital ones do not. Does that make sense?

 

Thanks Caroline, that makes sense, and in fact, I got to that conclusion on my own. Except that the analog clock in their example is still only one digit, not two. I guess it would be hard to estimate seconds on a clock that is only marked every 5 minutes. Maybe that's screws up the rule? I could estimate 6 minutes, but not 5 and a quarter.

 

But then why did the authors use such examples, right after saying you need to report one more digit??? I swear, it's in the same paragraph. ARGH!