Chapter 1: Problem 105
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
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
percentage calculation
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
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
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
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.