How to convert density ratio of grams/ mm^3 into metric tonnes/ m^3

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Discussion Overview

The discussion revolves around converting density from grams per cubic millimeter to metric tonnes per cubic meter, specifically in the context of a right cylinder with given dimensions and mass. Participants explore the implications of significant digits, volume calculations, and unit conversions, while addressing misunderstandings related to metric conversions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the volume of a right cylinder and derives a density ratio, questioning the impact of significant digits from the mass of 49 grams on the final result.
  • Another participant challenges the assumption that 1,000 cubic millimeters equals 1 cubic meter, clarifying that 1,000 cubic millimeters equals 1 cubic centimeter, and that 1 cubic meter equals 1,000,000,000 cubic millimeters.
  • A mathematical explanation is provided showing the correct conversion factor from grams per cubic millimeter to tonnes per cubic meter, emphasizing the importance of treating units as mathematical entities.
  • Participants discuss the significance of the number of significant digits in the mass measurement and how it affects the final density calculation.
  • One participant expresses confusion over the conversion process and the results obtained, leading to a correction in their approach that ultimately yields a correct answer.

Areas of Agreement / Disagreement

Participants generally agree on the correct conversion factors and the significance of significant digits, but there is some initial disagreement regarding the assumptions about volume conversions and the implications of those assumptions on the calculations.

Contextual Notes

Limitations include potential misunderstandings about unit conversions and the handling of significant digits, which may affect the accuracy of the calculations presented.

RayDonaldPratt
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For the dimensions of a right cylinder, I am given three significant digits for the diameter (17.4 mm) and the height (50.3 mm). The formula for the volume of a right cylinder is V = Pi x r^2 x h, which would lead here to Pi x (17.4 mm / 2)^2 x 50.3 mm = 11,960.69354 mm^3 before rounding to 3 significant digits (i.e., 12,000 mm^3).

I am also told that this right cylinder has a mass of 49 grams, and I am told that the "density" is the mass (49 g) divided by the volume (11,960.69354 mm^3), which is .004096752402 g / mm^3 (before rounding).

My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?

I am then asked to express (convert) this density ratio into metric tonnes per cubic meter (Tn / m^3). And, I am told that a metric ton is 1,000 kilograms, or, as it were, 1,000 x 1,000 grams. That would be a Mg (Mega-gram). However, some online calculators that I have looked at for converting grams per cubic millimeter to metric tonnes per cubic meter would simply multiply 49 grams from above by 1,000, i.e., 49,000 tonnes per cubic meter. Right or wrong, the math does not make sense to me.

I assume that 1,000 cubic millimeters is 1 cubic meter. And, raising 49 grams proportionately would give me 49,000 grams, or, 49 kilograms. So, raising 49 kilograms to a metric tonne without proportionately changing from cubic meters would move the decimal place back three spaces to .049 metric tonnes per cubic meter.

With two significant digits, the answer after rounding would be .05 metric tonnes per m^3, but that answer has been marked wrong. I will try again with the answer of .049, but if that fails (it will be my fourth attempt), I will post this question to the Internet and ask how to correctly solve this problem. I am taking a Coursera MOOC about physics solely for my personal use. (I want to actually use math and physics to solve a more complex problem that I am personally interested in solving, and I am therefore more interested in understanding all the math and physics than in getting a passing grade with missed answers.)

Where and why does my thinking go wrong?
 
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Welcome to PF.
RayDonaldPratt said:
My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?
Or is it 49.00 gram accurate to 4 digits ?
RayDonaldPratt said:
I assume that 1,000 cubic millimeters is 1 cubic meter.
A bad assumption.
1000 cubic mm are one cubic cm.
1000 cubic cm are one litre.
1000 litres are 1 cubic metre.
 
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RayDonaldPratt said:
My first question, or request for affirmation, is that the "49" grams reduces the number of significant digits in the final result to two digits, true?
Correct.

As for the conversion factor, consider the units as mathematical entities, along with the fact that multiplying by 1 doesn't alter any equality. Therefore,
$$
\begin{align*}
1 \frac{\mathrm{g}}{\mathrm{mm}^3} &= 1 \frac{\mathrm{g}}{\mathrm{mm}^3} \times \frac{1\, \mathrm{Tn}}{1000\, \mathrm{kg}} \times \frac{1\, \mathrm{kg}}{1000\, \mathrm{g}} \times \left( \frac{1000\, \textrm{mm}}{1\, \textrm{m}} \right)^3 \\
&= 1000 \frac{\mathrm{Tn}}{\mathrm{m}^3}
\end{align*}
$$
So the conversion factor you found online is indeed correct. The mistake you made is
RayDonaldPratt said:
I assume that 1,000 cubic millimeters is 1 cubic meter.
1,000 mm is 1 m, so you get 1,000,000,000 mm3 for 1 m3.
 
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Just to add a little to what has already been said...

Imagine a cube 1m x 1m x 1m.

In mm, its dimensions are 1000mm x 1000mm x 1000mm. So in mm³ its volume is 1000x1000x1000 mm³ = 10⁹ mm³.

You might like to ask yourself what an area of 1m² is in units of mm².
 
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Baluncore said:
Welcome to PF.

Or is it 49.00 gram accurate to 4 digits ?

A bad assumption.
1000 cubic mm are one cubic cm.
1000 cubic cm are one litre.
1000 litres are 1 cubic metre.
Thank you, it was given just as 49, so it was just 2 significant digits.
And, I was able to visualize 1,000 cubic millimeters as only creating the length (in cubic millimeters) of one edge of a cubic meter, but not the full volume of a cubic meter. Thank you.
 
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DrClaude said:
Correct.

As for the conversion factor, consider the units as mathematical entities, along with the fact that multiplying by 1 doesn't alter any equality. Therefore,
$$
\begin{align*}
1 \frac{\mathrm{g}}{\mathrm{mm}^3} &= 1 \frac{\mathrm{g}}{\mathrm{mm}^3} \times \frac{1\, \mathrm{Tn}}{1000\, \mathrm{kg}} \times \frac{1\, \mathrm{kg}}{1000\, \mathrm{g}} \times \left( \frac{1000\, \textrm{mm}}{1\, \textrm{m}} \right)^3 \\
&= 1000 \frac{\mathrm{Tn}}{\mathrm{m}^3}
\end{align*}
$$
So the conversion factor you found online is indeed correct. The mistake you made is

1,000 mm is 1 m, so you get 1,000,000,000 mm3 for 1 m3.
Thank you. I tried to do what you did above, but because my assumptions were wrong, my results were wrong. When I corrected my approach to yours above, I got 4.09 x 10^-3 x 10^9 Tn in the numerator, and I got 10^6 m^3 in the denominator. The net result after rounding to two significant digits was 4.1 Tn/m^3 and it was marked correct. Thank you!
 
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