Calculating Lung Capacity in Liters: Freshwater Buoyancy Volume Question

[NY152]

Homework Statement

A 69.5 kg man floats in freshwater with 2.95% of his volume above water when his lungs are empty, and 4.75% of his volume above water when his lungs are full. Calculate the volume of air, in liters, that he inhales (this is called his lung capacity). Neglect the weight of air in his lungs.

Homework Equations

d=m/v
freshwater: d=1000 kg/m^3

The Attempt at a Solution

From the given information, there's a 1.8% increase in volume. I'm just not sure where to start given the information above.

 

[gneill]

You should be able to determine the man's volume for both cases. Start by considering the volume of water he displaces in order to float.

 

[NY152]

Okay so I did ((69.5 kg/1000 kg/m^3)*(4.75/100))-((69.5/1000 kg/m^3)(2.95/100))= .00125 m^3 then converted that to liters which is 1.25 Liters. The homework system says "it looks like you may have confused the denominator and the numerator, check your algebra" Not sure where I'm going wrong here though...

 

[gneill]

You've found the difference between 4.75% of the displaced water and 2.95% of the displaced water. That's not what you're looking for. The displaced water volume is constant and smaller than the man's volume. You need to find the two volumes of the man.

By the way, save yourself a heap of typing and just assign a variable name to repeated quantities. Call the volume of water displaced v_w, for example.

 

[NY152]

You've found the difference between 4.75% of the displaced water and 2.95% of the displaced water. That's not what you're looking for. The displaced water volume is constant and smaller than the man's volume. You need to find the two volumes of the man.

By the way, save yourself a heap of typing and just assign a variable name to repeated quantities. Call the volume of water displaced v_w, for example.

Oh okay, so I'd do the same sort of math setup but use 95.25% and 97.05% instead?

 

[gneill]

Oh okay, so I'd do the same sort of math setup but use 95.25% and 97.05% instead?

No. Let's concentrate on one of the volumes for the man first.

You've correctly determined that he displaces a volume of water $V_w = M/ρ$, where M is his mass and ρ the density of water. That is also the amount of his volume that is below water (since it's displacing the water). Let's call the man's total volume for the first case (the 2.95% above water case) $V_o$. What would be the volume above water (in symbols, no numbers yet)?