(a) When ice floats in water at \(0^{\circ} \mathrm{C},\) what percent of its volume is submerged? (b) What is the specific gravity of ice?

According to Archimedes' principle, when an object is immersed in a liquid, the buoyant force acting on the object is equal to the weight of the liquid displaced by the object. Mathematically, this can be written as: \[F_{B} = \rho_{L} V_{L} g\] Where \(F_{B}\) is the buoyant force, \(\rho_{L}\) is the density of the liquid, \(V_{L}\) is the volume of the liquid displaced, and \(g\) is the acceleration due to gravity.

When the ice floats in water, the buoyant force is equal to the weight of the ice. Therefore, we can write the equation for the ice as: \[F_{B} = m_{ice}g\] Where \(m_{ice}\) is the mass of the ice. Since \(F_{B} = \rho_{L} V_{L} g\), we can also write the equation as: \[m_{ice}g = \rho_{L} V_{L} g\]

To find the percentage of ice volume submerged, we first need to find the volume of the ice (\(V_{ice}\)) using the mass and density of ice: \[V_{ice} = \frac{m_{ice}}{\rho_{ice}}\] The percentage of ice volume submerged (\(P_{submerged}\)) can be found using the volume of liquid displaced (\(V_{L}\)) and the volume of the ice (\(V_{ice}\)): \[P_{submerged} = \frac{V_{L}}{V_{ice}} \times 100\] Substitute the mass of the ice equation from Step 2: \[\frac{V_{L}g}{V_{ice}g} = \frac{\rho_{L}}{\rho_{ice}}\] Cancel out the acceleration due to gravity and rearrange for \(P_{submerged}\): \[P_{submerged} = \frac{\rho_{L}}{\rho_{ice}} \times 100\] The density of ice and water at \(0^{\circ} \mathrm{C}\) are \(\rho_{ice} = 917 kg/m^3\) and \(\rho_{L} = 1000 kg/m^3\). Substitute these values: \[P_{submerged} = \frac{1000}{917} \times 100 \approx 109.03\%\]

The specific gravity of a substance is the ratio of its density to the density of a reference substance, typically water. In this case, the specific gravity of ice can be found using its density and the density of water: \[Specific \; Gravity \; (ice) = \frac{\rho_{ice}}{\rho_{L}}\] Substitute the values of the densities: \[Specific \; Gravity \; (ice) = \frac{917}{1000} \approx 0.917\] #Final Answer# (a) The percentage of ice volume submerged in water at 0°C is approximately 109.03%. (b) The specific gravity of ice is approximately 0.917.