Answer to Question #181528 in Physics for Kise

Question

A 60-kg man floats in freshwater with 4.6% of his volume above water when his lungs are empty, and 5.5% of his volume above water when his lungs are full. Calculate the volume of air he inhales—called his lung capacity—in liters.

Round your answer to 2 decimal places

Answer 1

Given:

$\rho_\text w$ - density of water

$\rho_\text b$ - density of body,

$\rho_\text a=1.225\text{ kg/m}^3$ - density of air,

$V$ - volume of body with no air in lungs

$V_\text l$ - volume of lungs.

Solution:

A body floats when the buoyancy force is equal to the force of gravity:

$$\rho_\text w gV(1-\kappa_1)=\rho_\text{b} Vg,\\ \rho_\text w g(V+V_\text l)(1-\kappa_2)=g\rho_\text{b} V+g\rho_\text aV_\text l.$$

From the first equation find density of the body:

$$\rho_\text{b} =\rho_\text w (1-\kappa_1)=1000(1-0.046)=954\text{ kg/m}^3.$$

Express volume of the body in terms of its mass and volume:

$$V=\frac{m}{\rho_\text b}.$$

Substitute and find volume of lungs:

$$\rho_\text w \bigg(\frac{m}{\rho_\text b}+V_\text l\bigg)(1-\kappa_2)=m+\rho_\text aV_\text l,\\\space\\ V_\text l=\frac{m(\rho_\text b-(1-\kappa_2)\rho_\text w)}{\rho_\text b((1-\kappa_2)\rho_\text w-\rho_a)}=5.998·10^{-4}\text{ m}^3,$$

or 0.60 L.