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a. Some athletes have as little as \(3.0 \%\) body fat. If such a person has a body mass of \(65 \mathrm{~kg}\), how many pounds of body fat does that person have? b. In liposuction, a doctor removes fat deposits from a person's body. If body fat has a density of \(0.94 \mathrm{~g} / \mathrm{mL}\) and \(3.0 \mathrm{~L}\) of fat is removed, how many pounds of fat were removed from the patient?

Short Answer

Expert verified
a. 4.29 pounds. b. 6.21 pounds.

Step by step solution

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01

- Understand the Percentage of Body Fat

For part (a), determine the amount of body fat by taking 3.0% of the total body mass. The body mass is given as 65 kg. Calculate 3.0% of 65 kg.
02

- Convert Kilograms to Pounds

Use the conversion factor that 1 kg = 2.20462 pounds to convert the body fat mass from kilograms to pounds.
03

- Calculate Volume to Mass of Fat

For part (b), use the given density of fat (0.94 g/mL) and the volume of fat removed (3.0 L). First, convert the volume from liters to milliliters: 3.0 L = 3000 mL. Then calculate the mass in grams by multiplying the volume by the density.
04

- Convert Grams to Pounds

Finally, convert the mass of fat from grams to pounds using the conversion factor that 1 pound = 453.592 grams.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

percentage calculation
Understanding percentage calculation is crucial for solving problems related to body fat. A percentage represents a part of a whole number in terms of 100.
For instance, 3.0% is the same as saying 3 out of every 100 parts.
In our exercise, the body mass is 65 kg. To find 3.0% body fat, we multiply the body mass by the percentage:
\(3.0\text{ of } 65 = \frac{3.0}{100} \times 65 = 1.95 \text{ kg}\text{ of fat}\)
This calculation is simple yet vital for determining the specific quantity, and it can be used in various applications.
unit conversion
Unit conversion helps translate measurements from one unit to another. This is essential when comparing or combining values measured in different units. In our problem, we need to convert kilograms to pounds:
The conversion factor is:\ 1 \text{ kg} = 2.20462 \text{ pounds}
So, converting 1.95 kg of fat to pounds involves multiplying by this factor:
\(1.95 \text{ kg} \times 2.20462 = 4.299 \text{ pounds}\text{ of fat}\text{ in the body} \)
Unit conversion is widely applicable and helps in making information universally understandable.
density and volume
Density is defined as mass per unit volume, which helps us relate the mass and volume of substances. In our example, the fat's density is given as 0.94 g/mL, and we need to find the mass from a known volume of 3.0 L:
First, convert liters to milliliters since the density is in g/mL:
3.0 \text{ L} = 3000 \text{ mL}
Next, use the density to find the mass: \( \text{Mass} = \text{Density} \times \text{Volume} \)
\( \text{Mass} = 0.94 \text{ g/mL} \times 3000 \text{ mL} = 2820 \text{ g of fat removed}\text{.}/ \)
This demonstrates how knowing the density and volume allows us to determine the mass of a substance.
mass to weight conversion
Converting mass to weight is another significant aspect of unit conversion, particularly when needing to express weight in pounds. The removed fat's mass is calculated in grams, which we must convert to pounds.
The conversion factor is: \( 1 \text{ pound} = 453.592 \text{ grams} \)
Convert the calculated mass from grams to pounds:
\( 2820 \text{ grams} \times \frac{1 \text{ pound}}{453.592 \text{ grams}} \text{} \text{ pounds } = 6.215 \text{ pounds}/ \)
This conversion allows for an easier understanding of the fat removed in commonly used weight units.
Mastering these conversions is essential for accurately interpreting and communicating measurements in science and everyday life.

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