## Kitchen Math: What is the density of paper towel sheets?

Kitchen math is important math for food service and catering. Gastronomy requires both math and measurement. The related kitchen duties also require some understanding of the physical properties of equipment and supplies.

In the kitchen, paper towels are useful for food preparation and essential for food safety. Like everything in the food service industry, paper towels are resources that must be managed to help control costs.

Food service managers use comparison shopping to find the best deals. For paper towels, density and absorbency are very important product features.

In this math practice activity, we we will focus on the density of paper towel sheets.

TL;DR
The density of the paper towel is 0.00784 pounds per cubic inch in this specific math solution. That is roughly equal to 0.2 grams per cubic centimeter.

## How do you find the density of paper towel sheets?

The answer depends on the quality of the paper towel sheets. Our test sample for this math activity is part of the package called Kirkland Signature Big Roll Paper Towels from Costco Wholesale.

Let us first review the governing equation for density.

Density = Weight / Volume

Density is a property that tells us how closely-packed molecules are within a unit volume. Higher densities tend to be heavier, and lower densities tend to be lighter.

Subways during rush hour are densely-packed with passengers and are heavier. During off-hours, the passenger load is much lighter. The same principle applies for molecules.

We already know that matter is anything that has mass and takes up space. Density helps us better understand how much matter is packed into a specified three-dimensional space.

To achieve that understanding, we need to know the amount of mass, which is affected by gravity (which causes the phenomenon called weight) and the amount of volume.

Units for density are weight per unit volume. For this example, we will specifically use the units pounds per cubic inch [lbs. / cu. in.]

The weight of the paper towel material is unknown, so we will use a digital kitchen scale to take some measurements.

We find that the total weight of an unopened, unused paper towel roll with its packaging is 12.4 ounces.

The cardboard roll that holds the paper towel is 0.2 ounces. The packaging that wraps the paper towel roll is also 0.2 ounces.

In other words, we can calculate the weight of the paper towel material itself by subtracting the weight of everything else (e.g., all that other stuff).

W_material = W_total โ W_stuff
W_material = 12.4 โ 0.4

The paper towel material weighs 12 ounces. There are 16 ounces in one pound. So, we can say that material weight is 12/16 pounds.

`"how many pounds is 12 ounces?"`

We have calculated the weight of the paper towel material. Now, let us calculate the volume of the paper towel material.

We know that the area of the entire paper towel roll is 85 square feet. We can calculate the area A (in square inches) to be:

A = 85 ft. ft. * 12 * 12 in./ft. in./ft. ==> 12,240 in. in.

In other words, we simply converted the area from square inches into square feet. All the voice assistants can readily convert common units.

`"85 square feet in square inches"`

CCSS.MATH.CONTENT.5.MD.A.1: Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems.

The paper has a thickness, but a ruler won’t be practical for this measurement. We need something that can measure very small fractions of an inch.

Let us use digital calipers.

Using digital calipers, we measure that the thickness t (in inches) of a paper towel sheet is 1/128″.

We now have enough information to calculate the volume of the paper towel material. The volume V (in cubic inches) is:

V = A * t
V = ( 85 * 12 * 12 ) * 1/128

The weight W = 12/16 lbs. Combining all our previous calculations, we find that the density must be:

ฯ = W / V = (12/16) / ( 85 * 12 * 12 * (1 / 128))

The density of the paper towel sheets is 0.00784 lbs. / cu. in.

All the calculations above can be organized into a computational solution that can be solved using Wolfram Alpha.

Sample Wolfram Alpha input
`W=(12/16);V=(85*12*12*(1/128));rho=W/V`
View Result

## Conclusion

This information can be recorded and used later for comparison shopping of paper towels that may enter the market at a future date (hint, hint). While price may be an important factor, the density of the paper towel sheets for cooking and cleaning is also a very important metric in food service.

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.

## CCSS.MATH.CONTENT.5.MD.A.1: Coffee Jar Flower Vase Activity

The following activity is an opportunity for math practice in situations when measurement requires math-based design solutions. If possible, read the problem statement aloud. After reading the problem statement, we will have a brief discussion.

## Problem Statement

An empty coffee jar has been converted into a flower vase. The 3D modeling of this coffee jar flower vase is described as two frusta stacked end-to-end to form an hourglass container. The top areas of both frusta share a common square cross-section, which forms a bottleneck at 3.5″ above the bottom of the jar (sitting on a table). The capacity of the container is 800 mL of water. The total height of the container is 5.5″. The base area for the lower frustum and upper frustum are 3.5″ x 3.5″ and 3″ x 3″, respectively. What are the dimensions of the cross-sectional area at the bottleneck?

## Discussion

Wow! ๐๐๐ Lots of words. That is the point.

This activity is designed to help you practice reading a lot of words in test items. You will gain experience organizing these words and ideas using computational thinking.

Let us take a moment to visualize the solid model described above with 3D file viewer on GitHub. Or you can simply view the video below:

Before we begin this problem, we need to recall some prior knowledge and best practices. First, we choose and confirm the units of the final answer. In this case, the units are [inches x inches] or [in. x in.].

Note that the question did not ask for the side length or the actual area. The call to action asked for the dimensions. In the workplace, it is very important to pay attention to what the request for proposal (RFP) needs.

Second, we convert all quantities into compatible units. We convert 800 millimeters into cubic inches. The result is about 49 cubic inches.

In Wolfram Alpha, type this:
`800 ml in cu in` (view)

For voice assistants, try this command:
`800 milliliters in cubic inches`

CCSS.MATH.CONTENT.5.MD.A.1: Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems.

Next, we look at the formula for the volume of a frustum.

There are two frusta in this problem, so we will be using the same formula twice to help solve for a system of equations.

Let us place the container on the table. The table will be our reference point. The total height of the container is the distance from the table to the top of the container (see Figure 1). The total height (in inches) of the container is H = 5.5.

This measurement excludes the the jar mouth section, where the screw top lid has been removed.

For the lower frustum volume Y, we suppose that the base area (in inches squared) is A = 3.5 * 3.5 and the height (in inches) from the table is h = 3.5.

The units of area are square inches. We will let Wolfram Alpha make the calculations for us by leaving the dimensions as multiplicands. This methods lets us make design changes to the dimensions as needed without re-building our solution.

For the upper frustum volume Z, we conclude that the base area is B = 3 * 3, with a height of H – h.

We know the exact height of the upper frustum, but we will still use variables to express the difference between the total height and the height from the table.

Both frusta share the common bottleneck area X. Their combined volume is 49 cubic inches; that is, Y + Z = 49.

Our analysis is complete. We now know enough information to solve for the unknown bottleneck area X.

In Wolfram Alpha, type this:
`Y + Z = 49; H = 5.5; A = 3.5 * 3.5; B = 3 * 3; h = 3.5; Y = (h/3) * (A + X + SQRT(A*X)); Z = ((H-h)/3) * (B + X + SQRT(B*X)); J = SQRT(X);`(view)

The dimensions of the bottleneck cross-section are 2.63″ x 2.63″.

Key vocabulary: request for proposal, units, unit conversion

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.