Question

A fish uses a swim bladder to change its density so it is equal to that of water, enabling it to remain suspended under water. If a fish has an average density of \(1080 \mathrm{kg} / \mathrm{m}^{3}\) and mass \(10.0 \mathrm{g}\) with the bladder completely deflated, to what volume must the fish inflate the swim bladder in order to remain suspended in seawater of density $1060 \mathrm{kg} / \mathrm{m}^{3} ?$

Short answer

Question: Calculate the volume to which the fish needs to inflate its swim bladder, given the fish's average density decreases to match that of seawater when the bladder is inflated. Answer: The fish needs to inflate its swim bladder to a volume of \(1.7 \times 10^{-7} \mathrm{m}^{3}\) to remain suspended in seawater of density \(1060 \mathrm{kg} / \mathrm{m}^{3}\).

Step-by-step solution

Calculate the initial volume of the fish

To calculate the initial volume of the fish when the swim bladder is deflated, we'll use the density formula: \(Density = \frac{Mass}{Volume}\) We'll rearrange the formula to find the initial volume (\(V_{1}\)): \(V_{1} = \frac{Mass}{Density}\) Now plug in the given fish's mass (\(10.0g\)) and density (\(1080 \mathrm{kg} / \mathrm{m}^{3}\)): Initial volume \(V_{1} = \frac{10.0 \times 10^{-3} \mathrm{kg}}{1080 \mathrm{kg} / \mathrm{m}^{3}} = 9.26 \times 10^{-6} \mathrm{m}^{3}\).

Calculate the new mass of the fish with swim bladder inflated

Since the fish inflates its swim bladder to match the density of seawater, their densities become equal when suspended: \(Density_{fish} = Density_{seawater}\) The mass of the fish doesn't change when the swim bladder is inflated, so we can find the mass of the fish (\(mass_{2}\)) by using the same mass as the initial mass (\(10.0g\)).

Calculate the new volume of the fish with the swim bladder inflated

Using the density formula, we will find the new volume (\(V_{2}\)) of the fish when the swim bladder is inflated: \(V_{2} = \frac{Mass_{2}}{Density_{seawater}}\) Plugging in the given mass and density of seawater: \(V_{2} = \frac{10.0 \times 10^{-3} \mathrm{kg}}{1060 \mathrm{kg} /\mathrm{m}^{3}} = 9.43 \times 10^{-6} \mathrm{m}^{3}\)

Calculate the volume of the inflated swim bladder

To find the volume of the inflated swim bladder (\(V_{bladder}\)), we will subtract the initial volume of the fish (\(V_{1}\)) from the new volume (\(V_{2}\)): \(V_{bladder} = V_{2} - V_{1}\) Now, we'll plug in the values we found for \(V_{1}\) and \(V_{2}\): \(V_{bladder} = 9.43 \times 10^{-6} \mathrm{m}^{3} - 9.26 \times 10^{-6} \mathrm{m}^{3} = 1.7 \times 10^{-7} \mathrm{m}^{3}\) So, the fish needs to inflate its swim bladder to a volume of \(1.7 \times 10^{-7} \mathrm{m}^{3}\) to remain suspended in seawater of density \(1060 \mathrm{kg} / \mathrm{m}^{3}\).