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Bowling Balls

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It may come as a surprise that some bowling balls float! [1] [[Image:180px-BowlingBalls.jpg|left|thumb|Some bowling balls

Earlier we showed how to use unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will also how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, if bowling balls have a regulation volume what must their densities be to achieve the desired ball mass? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.

Suppose we have a spherical ball which measures 27 inches in circumference. Official Specifications Manual for the American Bowling Congress (ABC) reports circumferences in the range of 26.704–27.002 in. and diameters 8.500–8.595 in [2], but the World Tenpin Bowling Association(WTBA), specifies 27 in. as the maximum circumference for a bowling ball used in competitions and the maximum weight is 16 lb. Cite error: Invalid <ref> tag; refs with no name must have content. We can easily calculate that its volume (neglecting the drilled finger holes) is TAG Heuer Replica C = 2πr

V = \frac {4}{3} \pi r^3

= \frac {4}{3} \pi (\frac {C}{2 \pi})^3 = 5447 cm3

The construction of bowling balls is very technical. In the early 1900s they were made of wood, then rubber. But in the 1960s complex polymer chemistry was employed to create inner "cores" whose shape controls the rotational properties of the balls. The core is surrounded by a "filler" that determines the overall density of the ball, and the whole thing is coated with a "coverstock" that grips the alley, and allows the rotating ball to "hook". The density controls the weight (mass) of the ball. For example, we can calculate the mass of a ball whose density is 1.33 g/cm3 from the mathematical function which defines density:

\text{Density = }\frac{\text{mass}}{\text{volume}}\text{      or     }\rho \text{ = }\frac{\text{m}}{\text{V}}\text{                                                (1}\text{.1)}

If we multiply both sides by V, we obtain

\text{V}\times \rho =\frac{\text{m}}{\text{V}}\times \text{V}=\text{m}

m = V × ρ or mass = volume × density       (1.2)

So for a volume of 5447 cm3 and ρ = 1.33 g/cm3, we calculate a mass of

m = 5447 cm3 x 1.33 \frac {g}{cm^3} = 7245 g or in pounds 7245 g x \frac {1 lb}{453.59 gram} = 16.0 lb

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. (1.2) are multiplied by 1/ρ, we have

\begin{align}
  & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \\ 
 & \text{         V}=\text{m}\times \frac{\text{1}}{\rho } \\ 
\end{align}       (1.3)

The lowest density in modern bowling ball composition is 0.666 g/cm3 for 8 lb balls. What is the circumference of the ball?

V = m/D = 8 lb x \frac {453.59 g}{1 lb} / 0.666 \frac {g}{cm^3} = 5449 cm3

and the circumference is calculated from the formula above to be the regulation 27 in.

Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?

An Important Caveat

A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function

C = \frac{m}{V}.

In this case, V refers to the volume of a solution, which contains both a solute and solvent.

Given a concentration of an alloy is 10g gold in 100 cm3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of gold in 20 g of the alloy as follows:

20 g     x     \frac{100 cm^3}{10 g}     = 200 cm3

It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function.

The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.

EXAMPLE

A solution of ethanol with a concentration of 0.1754 g / cm3 has a density of 0.96923 g / cm3 and a freezing point of -9 ° F [3]. What is the volume of ethanol (D = 0.78522 g / cm3 at 25 °C) in 100 g of the solution?

Answer

The volume of 100 g of solution is

V = m / D = 100 g /0.96923 g cm-3 = 103.17 cm3.

The mass of ethanol in this volume is

m = V x C = 103.17 cm3 x 0.1754 g / cm3 = 18.097 g.

The volume of ethanol = m / D = 18.097 g / 0.78522 g / cm3 = 23.05 cm3.

Note that we cannot calculate the volume of ethanol by

 \frac {\frac{0.96923 g}{cm^3} x 100 cm^3}{\frac {0.78522 g}{cm^3}} = 123.4 cm3

even though this equation is dimensionally correct.

Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.


The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because of the mathematical formula, D = m/v, which relates density, mass, and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume, and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows:


EXAMPLE 1.11 A student weights 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?



We know that volume is related to mass through density.

Therefore

V = m × conversion factor


Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:


V=m\times \frac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \frac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3}


If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:


V=\text{98}\text{.0 g}\times \frac{\text{13,6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\; (no cancellation!)


It is clear that square grams per cubic centimeter are not the units we want.


Using a conversion factor is very similar to using a unity factor—we know the factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of that relationship [Eqs. (1.1), (1.2), and (1.3) in the case of density, mass, and volume], not because it is equal to one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.


A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:


\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume   or   }m\overset{\rho }{\longleftrightarrow}V\text{ }


This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.



EXAMPLE 1.12 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).


The road map

V\xrightarrow{\rho }m\text{ }


tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:


Mass = m = 47.3 cm3 × \frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}


Since the volume units are different, we need a unity factor to get them to cancel:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}


We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g}


In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.

Web Sources: TAG Heuer Replica

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