Consider a sphere of diameter \(5 \mathrm{~cm}\), a cube of side length \(5 \mathrm{~cm}\), and a rectangular prism of dimension \(4 \mathrm{~cm} \times\) \(5 \mathrm{~cm} \times 6 \mathrm{~cm}\), all initially at \(0^{\circ} \mathrm{C}\) and all made of silver \((k=\) \(\left.429 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=10,500 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.235 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). Now all three of these geometries are exposed to ambient air at \(33^{\circ} \mathrm{C}\) on all of their surfaces with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine how long it will take for the temperature of each geometry to rise to \(25^{\circ} \mathrm{C}\).

First, let's calculate the volume of each object: 1. Sphere: The formula for the volume of a sphere is: \(V_\text{sphere} = \frac{4}{3}\pi r_\text{sphere}^3\), where \(r_\text{sphere}\) is the radius of the sphere. Since the diameter is 5 cm, the radius is \(\frac{5}{2}\) cm. So, \(V_\text{sphere} = \frac{4}{3}\pi\left(\frac{5}{2}\right)^3 = \frac{125}{2}\pi\ \mathrm{cm^3}\). 2. Cube: The formula for the volume of a cube is: \(V_\text{cube} = a^3\), where \(a\) is the side length of the cube. In this case, \(a = 5\) cm. So, \(V_\text{cube} = 5^3 = 125\ \mathrm{cm^3}\). 3. Rectangular prism: The formula for the volume of a rectangular prism is: \(V_\text{rectangular\_prism} = lwh\), where \(l\), \(w\), and \(h\) are the length, width, and height of the prism, respectively. In this case, \(l = 4\ \mathrm{cm}, w = 5\ \mathrm{cm},\) and \(h = 6\ \mathrm{cm}\). So, \(V_\text{rectangular\_prism} = 4 \times 5 \times 6 = 120\ \mathrm{cm^3}\).

Next, we will calculate the mass of each object, given their volumes and the density of silver, which is \(\rho = 10,500\ \mathrm{kg/m^3}\). First, we need to convert the volumes to \(\mathrm{m^3}\): \(V_\text{sphere} = \frac{125}{2}\pi\ \mathrm{cm^3} \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^3 = \frac{125}{2}\pi\times 10^{-6}\ \mathrm{m^3}\), \(V_\text{cube} = 125\ \mathrm{cm^3} \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^3 = 125\times 10^{-6}\ \mathrm{m^3}\), and \(V_\text{rectangular\_prism}=120\ \mathrm{cm^3} \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^3 = 120\times 10^{-6}\ \mathrm{m^3}\). Now, we calculate the mass: \(M_\text{sphere} = \rho V_\text{sphere} = 10,500 \times \frac{125}{2}\pi \times 10^{-6}\ \mathrm{kg}\), \(M_\text{cube} = \rho V_\text{cube} = 10,500 \times 125 \times 10^{-6}\ \mathrm{kg}\), and \(M_\text{rectangular\_prism} = \rho V_\text{rectangular\_prism} = 10,500 \times 120 \times 10^{-6}\ \mathrm{kg}\).

We can now calculate the heat required to raise the temperature of each object to 25°C, using the specific heat capacity of silver, given by \(c_p = 0.235\ \mathrm{kJ/kg\cdot K}\), which is equal to 235 J/kg⋅K: First, calculate the temperature difference (\(\Delta T = 25 - 0 = 25\) K). Now, we find the heat required for each object: \(Q_\text{sphere} = M_\text{sphere} c_p \Delta T = 10,500\ \frac{125}{2}\pi \times 10^{-6} \times 235 \times 25\ \mathrm{J}\), \(Q_\text{cube} = M_\text{cube} c_p \Delta T = 10,500 \times 125 \times 10^{-6} \times 235 \times 25\ \mathrm{J}\), and \(Q_\text{rectangular\_prism} = M_\text{rectangular\_prism} c_p \Delta T = 10,500 \times 120 \times 10^{-6} \times 235 \times 25\ \mathrm{J}\).

Finally, we use the heat transfer equation for each object to find the time, given the heat transfer coefficient \(h = 12\ \mathrm{W/m^2\cdot K}\) and the surface area of each object. 1. Sphere: The surface area of a sphere is \(A_\text{sphere} = 4\pi r_\text{sphere}^2\). So, \(A_\text{sphere} = 4\pi\left(\frac{5}{2}\right)^2 = 25\pi\ \mathrm{cm^2}\). Now, convert it to \(\mathrm{m^2}\): \(A_\text{sphere} = 25\pi \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^2 = 0.025\pi\ \mathrm{m^2}\). Next, apply the heat transfer equation: \(Q_\text{sphere} = h A_\text{sphere} (\Delta T) t_\text{sphere}\), where \(t_\text{sphere}\) is the time taken. Rearrange for time: \(t_\text{sphere} = \frac{Q_\text{sphere}}{h A_\text{sphere} (\Delta T)}\), And substitute the values: \(t_\text{sphere} = \frac{10,500\ \frac{125}{2}\pi \times 10^{-6} \times 235 \times 25}{12 \times 0.025\pi \times 25}\ \mathrm{s}\). 2. Cube: The surface area of a cube is \(A_\text{cube} = 6a^2\). So, \(A_\text{cube} = 6 \times 5^2 = 150\ \mathrm{cm^2}\). Converting to meters: \(A_\text{cube} = 150 \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^2 = 0.015\ \mathrm{m^2}\). Apply the heat transfer equation: \(Q_\text{cube} = h A_\text{cube} (\Delta T) t_\text{cube}\), and rearrange for time: \(t_\text{cube} = \frac{Q_\text{cube}}{h A_\text{cube} (\Delta T)}\), Substitute the values: \(t_\text{cube} = \frac{10,500 \times 125 \times 10^{-6} \times 235 \times 25}{12 \times 0.015 \times 25}\ \mathrm{s}\). 3. Rectangular prism: The surface area of a rectangular prism is \(A_\text{rectangular\_prism} = 2lw + 2lh + 2wh\). So, \(A_\text{rectangular\_prism} = 2 \times 4 \times 5 + 2 \times 4 \times 6 + 2 \times 5 \times 6 = 148\ \mathrm{cm^2}\). Convert to meters: \(A_\text{rectangular\_prism} = 148 \times \left(\frac{\mathrm{m}}{100\ \mathrm{cm}}\right)^2 = 0.0148\ \mathrm{m^2}\). Apply the heat transfer equation: \(Q_\text{rectangular\_prism} = h A_\text{rectangular\_prism} (\Delta T) t_\text{rectangular\_prism}\), and rearrange for time: \(t_\text{rectangular\_prism} = \frac{Q_\text{rectangular\_prism}}{hA_\text{rectangular\_prism} (\Delta T)}\), Substitute the values: \(t_\text{rectangular\_prism} = \frac{10,500 \times 120 \times 10^{-6} \times 235 \times 25}{12 \times 0.0148\times 25}\ \mathrm{s}\). Now, we have the time it takes for each geometry to rise to 25°C: \(t_\text{sphere}\), \(t_\text{cube}\), and \(t_\text{rectangular\_prism}\).