6516
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2011-02-01T00:00:01
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<li><a href="http://nrich.maths.org/6154">Warm-up</a></li>
<li><a href="http://nrich.maths.org/6349">Try this next</a></li>
<li><a href="http://nrich.maths.org/5830">Think higher</a></li>
<li><a href="http://en.wikipedia.org/wiki/Physical_constant">Read: mathematics</a></li>
<li><a href="http://en.wikipedia.org/wiki/Speed_of_light">Read: science</a></li>
<li><a href="http://en.wikipedia.org/wiki/International_System_of_Units">Explore further</a></li>
</ul>
<div> </div>
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<p>Physical constants can only be determined by experiment and can never be known exactly, even if in principle an exact value does exists. As a result, physical quantities are given as a probable range of values with an uncertainty registered in the last two digits, as follows:<br></br>
<br></br>
$1.234\, 5678(32) \rightarrow 1.234\, 5678 \pm 0.000\, 0032$<br></br>
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$245.234\, 789\, 123(45) \rightarrow 245.234\, 789\, 123\pm 0.000\, 000\, 045$<br></br>
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The following table contains the best currently known measurements for various physical quantities:<br></br>
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<table style="" border="1">
<tbody>
<tr>
<td>Name</td>
<td>Value</td>
</tr>
<tr>
<td>Avogadro constant</td>
<td>$6.022\, 141\, 79(30) \times 10 ^{23}$ mol $^{-1}$</td>
</tr>
<tr>
<td>Atomic mass constant</td>
<td>$1.660 \,538\, 782(83) \times 10^{-27}$ kg</td>
</tr>
<tr>
<td>Electron mass</td>
<td>$9.109\, 382 \,15(45)\times 10^{-31}$ kg</td>
</tr>
<tr>
<td>Proton-electron mass ratio</td>
<td>$1836.152\, 672\, 4718(80)$</td>
</tr>
<tr>
<td>Proton mass</td>
<td>$1.672\, 621\, 637(83) \times 10^{-27}$ kg</td>
</tr>
<tr>
<td>Neutron mass</td>
<td>$1.674\, 927 \,211(84) \times 10{-27}$ kg</td>
</tr>
<tr>
<td>Speed of light in vacuum</td>
<td>$299\, 792\, 458$ m s$^{-1}$ exactly</td>
</tr>
</tbody>
</table>
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<div>Consider the relationship between the error bounds for the proton-electron mass ratio and those for the electron mass and the proton mass. Are they consistent? Which appears to be known to best experimental accuracy?</div>
<br></br>
<div>Using this data, can you work out an upper limit on the mass of a mole of water? What is a lower limit?</div>
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<div>How much uncertainty is there is the energy is contained within the mass of a mole of water, according to Einstein's energy-mass equation $E=mc^2$?</div>
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<div>If the specific heat capacity of liquid water is about $4.1813$kJ kg$^{-1}$ K$^{-1}$ make an estimate of the number of cups of tea that you could make with this uncertain amount of energy.</div>
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<div class="framework">NOTES AND BACKGROUND<br></br>
<br></br>
National Institute of Standards and Technology Reference on Constants, Units and Uncertainty provides detailed information on the bounds of measurements of physical constants.<br></br>
<br></br>
See <a href="http://physics.nist.gov/cuu/Constants/index.html">http://physics.nist.gov/cuu/Constants/index.html</a> for more details .<br></br>
<br></br>
For a list of all constants, see <a href="http://physics.nist.gov/cuu/Constants/Table/allascii.txt">http://physics.nist.gov/cuu/Constants/Table/allascii.txt</a><br></br>
<br></br>
Interestingly, there is a strong element of statistics used to determine the probable values of constants. Key to this idea are the concepts of error and uncertainty in measurement. Cleverly designed experiments based on a strong understanding of statistics can be used to minimise this uncertainty.<br></br>
<br></br>
To read about the essentials of expressing measurement uncertainty see <a href="http://physics.nist.gov/cuu/Uncertainty/index.html">http://physics.nist.gov/cuu/Uncertainty/index.html</a><br></br>
<br></br>
Note that the speed of light given is a numerically exact quantity because the length of a metre has now been defined in terms of the speed of light!</div>
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<mdoxml version="1.0">It is not envisaged that <a href="https://nrich.maths.org/6516">this problem</a> would be used as a class problem. It is more appropriate for an enthusiastic student or small group of students looking for a challenge to work on independently.</mdoxml>
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<mdoxml version="1.0"><p>The error $\Delta Z$ of the quantity $Z=\frac{A}{B}$ where $A$ and $B$ are independent satisfies $\left(\frac{\Delta Z}{Z}\right)^2 = \left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2$.</p>
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<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 1ex; padding-right: 0px; padding-bottom: 0px; padding-left: 0px;">The error $\Delta Z$ of the quantity $Z=A+B$ where $A$ and $B$ are independent satisfies $(\Delta Z)^2 = (\Delta A)^2 + (\Delta B)^2$.</p>
<div>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; padding-top: 1ex; padding-right: 0px; padding-bottom: 0px; padding-left: 0px;">The error $\Delta Z$ of the quantity $Z=kA$ satisfies $(\Delta Z)^2 = (|k|\Delta A)^2 $.</p>
<div> </div>
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<div> </div>
<div>The error of $r' = \frac{m_p}{m_e} = 1836.15267...$ obeys:</div>
<p>$$\begin{align*}\left(\frac{\Delta r'}{r'}\right)^2 &= \left(\frac{8.3\times10^{-35}}{1.67...\times10^{-27}}\right)^2 + \left(\frac{4.5\times10^{-38}}{9.10...\times10^{-31}}\right)^2 \\&= 4.90\times^{-15}.\\\Rightarrow \Delta r' &= 1.29\times10^{-4}\end{align*}$$</p>
<p>Therefore, $r' = 1836.15267(13)$.</p>
<p>The proton/electron mass ratio, $r$, is $1836.152\, 672\, 4718(80)$. These values are consistent, as the given value for $r$ is within the error of $r'$. It appears the mass ratio is known to much greater accuracy than the individual masses. </p>
<p> </p>
<p>The atomic mass of oxygen-16 is <a href="http://en.wikipedia.org/wiki/Isotopes_of_oxygen">15.99491461956(16)</a>u, and the atomic mass of hydrogen-1 is <a href="http://en.wikipedia.org/wiki/Isotopes_of_hydrogen">1.007<span style="margin-left: 0.25em;">8</span>25<span style="margin-left: 0.25em;">0</span>32<span style="margin-left: 0.25em;">0</span>7(10)u</a>. The atomic mass of
a water molecule is therefore $18.0104536837(26)u = 2.99072411(15)\times10^{-26}kg$. Therefore 1 mole of water weighs (mulitplying by Avogadro's constant) $1.80105647(13)\times10^{-2}kg$.</p>
<div> </div>
<div>The energy in the mass of a mole of water is therefore $1.61870882(11)\times10^{15}J$.</div>
<div> </div>
<div>Suppose a cup of tea has a volume of $200$ml and we need to raise its temperature from $20^{\circ}\mathrm{C}$ to $100^{\circ}\mathrm{C}$. Supposing there's no loss of energy in our heating system, the amount of energy in a mole of water could make $2.4\times10^{10}$ cups of tea!</div>
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Constantly changing
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
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