An iron cannonball of radius \(0.08 \mathrm{m}\) has a cavity of radius $0.05 \mathrm{m}$ that is to be filled with gunpowder. If the measurements were made at a temperature of \(22^{\circ} \mathrm{C}\) how much extra volume of gunpowder, if any, will be required to fill 500 cannonballs when the temperature is \(30^{\circ} \mathrm{C} ?\)

Let's find the volume of the cavity in a single cannonball at \(22^{\circ} \mathrm{C}\). The cavity is a sphere, so we can use the formula for the volume of a sphere: \(V = \dfrac{4}{3}\pi r^3\) where \(r\) is the radius of the cavity. Given radius \(r=0.05\ \mathrm{m}\), we can find the initial volume: \(V = \dfrac{4}{3}\pi(0.05\ \mathrm{m})^3 = 5.24 \times10^{-4}\ \mathrm{m^3}\)

To find the volume change of the cavity, we will use the formula for volume expansion: \(\Delta V = \beta V_i \Delta T\) where \(\Delta V\) is the change in volume, \(\beta\) is the volume expansion coefficient for iron (which is \(36 \times 10^{-6}\ \mathrm{^\circ C^{-1}}\)), \(V_i\) is the initial volume, and \(\Delta T\) is the temperature change. Given the temperature change from \(22^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\): \(\Delta T = 30^{\circ} \mathrm{C} - 22^{\circ} \mathrm{C} = 8^{\circ} \mathrm{C}\) Now we can find the volume change: \(\Delta V = (36 \times 10^{-6}\ \mathrm{^\circ C^{-1}})(5.24\times10^{-4}\ \mathrm{m^3})(8^{\circ} \mathrm{C}) =7.56\times 10^{-5}\ \mathrm{m^3}\)

To find the new volume of the cavity at \(30^{\circ} \mathrm{C}\), we add the change in volume to the initial volume: \(V' = V + \Delta V = 5.24\times10^{-4}\ \mathrm{m^3} + 7.56\times 10^{-5}\ \mathrm{m^3} = 5.996\times 10^{-4}\ \mathrm{m^3}\)

Now that we have the new volume of the cavity, we can find the additional volume needed to fill 500 cannonballs. First, we find the total volume needed to fill 500 cannonballs at \(30^{\circ} \mathrm{C}\): \(V_{total} = 500 \times 5.996\times 10^{-4}\ \mathrm{m^3} = 0.2998\ \mathrm{m^3}\) Next, we find the initial total volume for 500 cannonballs at \(22^{\circ} \mathrm{C}\): \(V_{initial} = 500 \times 5.24\times10^{-4}\ \mathrm{m^3} = 0.262\ \mathrm{m^3}\) Now we can calculate the additional volume of gunpowder required: \(V_{additional} = V_{total} - V_{initial} = 0.2998\ \mathrm{m^3} - 0.262\ \mathrm{m^3} = 0.0378\ \mathrm{m^3}\) Thus, an extra \(0.0378\ \mathrm{m^3}\) of gunpowder will be required to fill 500 cannonballs when the temperature is \(30^{\circ} \mathrm{C}\).