Question

The record blue whale in Problem 68 had a mass of $1.9 \times 10^{5} \mathrm{kg} .\( Assuming that its average density was \)0.85 \mathrm{g} / \mathrm{cm}^{3},$ as has been measured for other blue whales, what was the volume of the whale in cubic meters \(\left(\mathrm{m}^{3}\right) ?\) (Average density is the ratio of mass to volume.)

Short answer

Answer: The volume of the whale is approximately \(223.53 \mathrm{m}^{3}\).

Step-by-step solution

Write down the given information

We have, - Mass (m) = \(1.9 \times 10^{5} \mathrm{kg}\) - Average density (D) = \(0.85 \mathrm{g}/ \mathrm{cm}^{3}\)

Convert the density to \(\mathrm{kg} / \mathrm{m}^{3}\).

To convert the density from \(\mathrm{g} / \mathrm{cm}^{3}\) to \(\mathrm{kg} / \mathrm{m}^{3}\), we will follow these steps: 1 g = 0.001 kg, 1 cm = 0.01 m. Then, \(0.85 \mathrm{g} / \mathrm{cm}^{3} = 0.85(0.001 \mathrm{kg}) / (0.01\mathrm{m})^{3} = 850 \mathrm{kg} / \mathrm{m}^{3}\)

Rearrange the formula for density

Density = Mass / Volume So, Volume = Mass / Density.

Find the volume of the whale

To find the whale's volume, we will use the formula derived in step 3 and the given information: Volume = \( \frac{1.9 \times 10^{5}\mathrm{kg}}{850 \mathrm{kg}/\mathrm{m}^3}\) Volume \(\approx \frac{1.9 \times 10^{5}}{850}\) \( \mathrm{m}^{3}\) Volume \(\approx 223.53 \mathrm{m}^{3}\) Hence, the volume of the whale is approximately \(223.53 \mathrm{m}^{3}\).