# Question Video: Determination of the Density of a Substance from the Difference between Its Mass and Submerged Apparent Mass

Bird bones have air pockets in them to reduce their weight. This also gives them an average density significantly less than that of the bones of other animals. An ornithologist weighs an impermeable bird bone in air and in water and finds its mass is 45.0 g and its apparent mass when submerged is 3.60 g. What mass of water is displaced? What is the volume of the bone? What is the average density of the bone?

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### Video Transcript

Bird bones have air pockets in them to reduce their weight. This also gives them an average density significantly less than that of the bones of other animals. An ornithologist weighs an impermeable bird bone in air and in water and finds its mass is 45.0 grams and its apparent mass when submerged is 3.60 grams. What mass of water is displaced? What is the volume of the bone? What is the average density of the bone?

Let’s start by highlighting some of the important information we’ve been given. We’re told that the mass of the bird bone is 45.0 grams and that its apparent mass when it’s under water is 3.60 grams. We want to know the mass of the water that the bone displaces; we’ll call that 𝑚 sub 𝑑𝑖𝑠𝑝. We want to know the volume of the bone; we’ll call that 𝑣 sub 𝑏. And we want to know the bone’s average density; we’ll call that 𝜌 sub 𝑏.

Let’s begin our solution by drawing a sketch of the two measurement methods for mass. Imagine that we have two identical scales: one in air and one submerged in water. When we weigh the bird bone on the first scale accounting for gravity, it reads out a mass of 45.0 grams. When we put the bird bone under water on top of the second submerge scale, it reads out an apparent mass of three 3.60 grams. Let’s call 𝑚 sub 𝑎 the mass of the bird bone in air and 𝑚 sub 𝑤 the apparent mass of the bone in water.

Since the true mass of the bird’s bone is constant, why does the second scale read less than 45.0 grams? It’s because of the buoyant force, where the water displaced by the bird bone pushes up on that bone so that its apparent mass as read by the scale is less. The mass of the water displaced is equal to the difference between the bird bone’s true mass, 𝑚 sub 𝑎, and its apparent mass under water, 𝑚 sub 𝑤. When we enter in the values for these two masses, we find that the mass of water displaced is 41.4 grams.

Now we move onto solving for the volume of the bone. As we do, let’s recall the definition for density. An object’s density 𝜌 is defined as its mass divided by its volume. Rearranging this equation to solve for volume, we see that the volume of an object is equal to its mass divided by its density. In our case, the volume of water displaced by the bone is equal to the mass of displaced water divided by the density of water.

In this problem, we’ll assume that the density of water is exactly one gram per cubic centimetre. We can now enter in the values for 𝑚 sub 𝑑𝑖𝑠𝑝 and 𝜌 sub 𝑤. This volume of displaced water because the bone is submerged is equal to the volume of the bone. So the volume of the bird’s bond is equal to 41.4 cubic centimetres. That’s how much space the bird’s bone takes up.

Now, we move onto solving for the density of the bone. From our density definition, we know that the density of the bone is equal to its mass in air divided by its volume. We’re given the bone’s mass in air and we’ve solved for its volume. We can substitute in those values now. 45.0 grams divided by 41.4 cubic centimetres is equal to 1.09 grams per cubic centimetre. That’s the density of this bird bone.