At room temperature, an iron horseshoe, when dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm})\) causes the water level to rise \(0.25 \mathrm{~cm}\) above the level without the horseshoe in the tank. When heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K}\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

First, we need to find out the initial volume of the horseshoe at room temperature. To do so, we can find the volume of water corresponding to the rise in water level when the horseshoe is dunked into the tank: \(V_{water} = \pi R^2 h_{rise}\) where \(R = 10.0 \mathrm{~cm}\) is the radius of the cylindrical tank, and \(h_{rise} = 0.25 \mathrm{~cm}\) is the rise in water level. Plug in the values to find the initial volume of the horseshoe: \(V_{1} = V_{water} = \pi (10.0)^2 (0.25)\)

Next, we need to find how much the horseshoe has expanded due to the increase in temperature. The linear expansion coefficient (\(\alpha\)) is given as \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\). The initial temperature of the horseshoe is room temperature, which is \(20^{\circ}\mathrm{C}\). The final temperature is \(7.00\cdot 10^{2} \mathrm{~K}\). Since the temperatures are in different units, we need to convert the final temperature to Celsius: \(T_{2} = 7.00\cdot 10^{2} \mathrm{~K} - 273.15 = 426.85^{\circ}\mathrm{C}\) Now, we can compute the change in temperature: \(\Delta T = T_{2} - T_{1} = 426.85 - 20 = 406.85^{\circ}\mathrm{C}\) Since the volume expansion for a solid is given by the following formula: \(\Delta V = V_{1} \cdot 3 \alpha \Delta T\) Plug in the values to find the volume expansion: \(\Delta V = V_{1} \cdot 3 \cdot (11 \cdot 10^{-6}) \cdot 406.85\)

Now that we've determined the volume expansion due to the increase in temperature, we can find the expanded volume of the horseshoe: \(V_{2} = V_{1} + \Delta V\)

Having found the final volume of the heated horseshoe, we can now determine the new rise in water level: \(h_{new} = \frac{V_{2}}{\pi R^2}\) Finally, we will have the rise in water level when the heated horseshoe is dunked back into the cylindrical tank.