Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use \(k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, v=0.894 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}\) )

To move forward, we need to find the cross-sectional area (A) of the tube. Using the provided diameter (D) of 5 mm, we can find A: $$ A = \frac{\pi D^2}{4} $$ $$ A = \frac{\pi (0.005)^2}{4} $$ $$ A = 1.963 \times 10^{-5} \mathrm{m}^2 $$

Next, we will calculate the mass flow rate (m_dot) of the water using the provided velocity (v) and water properties such as density (rho): $$ m_{\text{dot}} = \rho v A $$ $$ m_{\text{dot}} = (997 \mathrm{~kg/m}^3)(0.3 \mathrm{~m/s})(1.963 \times 10^{-5} \mathrm{m}^2) $$ $$ m_{\text{dot}} = 5.863 \times 10^{-3} \mathrm{kg/s} $$

We will now calculate the Reynolds number (Re) by using the formula: $$ \text{Re} = \frac{\rho v D}{\mu} $$ $$ \text{Re} = \frac{(997 \mathrm{~kg/m}^3)(0.3 \mathrm{~m/s})(0.005 \mathrm{~m})}{(0.894 \times 10^{-6} \mathrm{m}^2/\mathrm{s})} $$ $$ \text{Re} = 16640 $$ As we have the Prandtl number (Pr) given, we can find the Nusselt number (Nu) using the following equation for laminar flow: $$ \text{Nu} = 0.023 \cdot \text{Re}^{0.8} \cdot \operatorname{Pr}^{0.4} $$ $$ \text{Nu} = 0.023 \cdot (16640)^{0.8} \cdot (6.14)^{0.4} $$ $$ \text{Nu} = 59.87 $$

Using the Nusselt number (Nu), we can now find the heat transfer coefficient (h) using the thermal conductivity (k) of water: $$ h = \frac{k \cdot \text{Nu}}{D} $$ $$ h = \frac{(0.607 \mathrm{W/m \cdot K})(59.87)}{0.005 \mathrm{m}} $$ $$ h = 7280 \mathrm{W/m}^2 \cdot \mathrm{K} $$

Using the energy balance equation, we can find the exit temperature (T_exit): $$ m_{\text{dot}}c_p(T_{\text{exit}} - T_{\text{inlet}}) = h A (T_{\text{tube}} - T_{\text{exit}})L $$ $$ T_{\text{exit}} = \frac{h A (T_{\text{tube}} - T_{\text{exit}})L + m_{\text{dot}}c_pT_{\text{inlet}}}{m_{\text{dot}}c_p} $$ Now, we need to find the \(T_{\text{exit}}\) and replace all given and calculated values in the above equation. $$ T_{\text{exit}} = \frac{(7280 \mathrm{W/m}^2 \cdot \mathrm{K})(1.963 \times 10^{-5} \mathrm{m}^2)(8 - T_{\text{exit}})(13 \mathrm{m}) + (5.863\times 10^{-3}\mathrm{kg/s})(4180 \mathrm{J/kg} \cdot \mathrm{K})(45)}{(5.863\times 10^{-3}\mathrm{kg/s})(4180 \mathrm{J/kg} \cdot \mathrm{K})} $$ Solving for \(T_{\text{exit}}\), we get: $$ T_{\text{exit}} \approx 10.6^{\circ}\mathrm{C} $$ Therefore, the answer is (c) \(10.6^{\circ}\mathrm{C}\).