A 10-cm-inner diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at \(3^{\circ} \mathrm{C}\). The heat transfer coefficient on the surface of the can is \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

We have been given the following data: - Inner diameter of the can: \(d = 10\) cm - Length of the can: \(L = 30\) cm - Initial water temperature: \(T_{i} = 25^{\circ}\)C - Final water temperature: \(T_{f} = 5^{\circ}\)C - Refrigerator temperature: \(T_{r} = 3^{\circ}\)C - Heat transfer coefficient: \(h = 14\) W/m²K.

We need to convert the given dimensions to meters: - Inner diameter: \(d = \frac{1}{100}\) m - Length: \(L = 0.3\) m

We can consider the can as a cylinder and calculate its surface area. The formula for the surface area of a cylinder \(A_{cylinder}\) is given by: $$A_{cylinder} = 2\pi r_{cylinder}(r_{cylinder} + h_{cylinder})$$ where \(r_{cylinder} = \frac{d}{2}= 0.05\) m \(h_{cylinder} = L = 0.3\) m Substitute the values: $$A_{cylinder} = 2\pi(0.05)(0.05 + 0.3) = 0.035 \mathrm{m}^{2}$$

To find the mass of the water, we can first find the volume of the cylinder and then use the density of water (ρ) to calculate its mass: $$V_{cylinder} = \pi r_{cylinder}^{2}h_{cylinder} = \pi (0.05)^{2}(0.3)$$ $$m_{water} = \rho V_{cylinder} = 1000 \times \pi (0.05)^{2}(0.3)$$ The specific heat of water \(c_{\mathrm p_{water}}\) is 4.18 kJ/(kg·K) or 4180 J/(kg·K).

Newton's law of cooling states that the rate of heat loss is proportional to the difference in temperature between the object and its surrounding environment. We can write it as: $$Q = H \Delta T t$$ Where \(Q\) is the heat lost by the water, \(H\) is the heat transfer coefficient multiplied by the surface area of the can, \(\Delta T\) is the difference between the initial and final temperature of the water, and \(t\) is the time required for the cooling process. The heat lost by the water can also be calculated as: $$Q = m c_\mathrm{p} \Delta T_{water}$$ Equating the two expressions for \(Q\) and solving for \(t\): $$H \Delta T t = m c_\mathrm{p} \Delta T_{water}$$ $$t = \frac{m c_\mathrm{p} \Delta T_{water}}{H \Delta T}$$ Substitute the values: $$t = \frac{1000 \times \pi (0.05)^{2}(0.3) \times 4180 (25-5)}{14 \times 0.035 (25-3)} = 1.17 \ \mathrm{h}$$

The time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is approximately 1.17 hours, so the correct answer would be option (b) \(1.17 \mathrm{~h}\).