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Chapter 9: Problem 82

The body of a 90.0 -kg person contains \(0.020 \mathrm{m}^{3}\) of body fat. If the density of fat is \(890 \mathrm{kg} / \mathrm{m}^{3},\) what percentage of the person's body weight is composed of fat?

Short Answer

Expert verified
Answer: The percentage of body weight composed of fat for this person is approximately 19.78%.

Step by step solution

01

Find the mass of body fat

To find the mass of body fat, use the mass density formula which is: mass_density = mass / volume. We need to find mass (m) in this case, so we rearrange the formula to mass = mass_density * volume. Given the density of fat (mass_density) = 890 kg/m^3 and volume of body fat (V) = 0.020 m^3, we'll calculate the mass of body fat. m = mass_density * V m = (890 kg/m³) * (0.020 m³)
02

Calculate the mass of body fat

Calculate the mass of body fat using the values from step 1. m = 17.8 kg This value represents the mass of the fat in the person's body.
03

Calculate the percentage of body fat

Now we calculate the percentage of body weight composed of fat. To do this, divide the mass of body fat by the total body mass and multiply by 100: Percentage of Body Fat = (mass of body fat / total body mass) * 100 Percentage of Body Fat = (17.8 kg / 90.0 kg) * 100
04

Calculate the final result

Now, solve for the percentage of body fat: Percentage of Body Fat = 19.78 % The person's body weight is composed of approximately 19.78% fat.

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