When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake \(\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface 400 hours after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

The heat equation is given by: \begin{equation} u(z,t)=Ae^{-(\frac{z}{\sqrt{4\alpha t}})^2}(T_{s}-T_{0})+T_{0} \end{equation} where \(u(z,t)\) is the temperature at point \(z\) and time \(t\), \(A\) represents the amplitude, \(\alpha =\frac{k}{c\rho}\) is the thermal diffusivity, \(T_{s}\) is the surface temperature, \(T_{0}\) is the initial temperature, and \(z\) is the depth below the surface. We are given the following constants: - Thermal conductivity, \(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Heat capacity, \(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} = 4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) - The density of water, \(\rho \approx 1000 \mathrm{~kg} / \mathrm{m}^{3}\) - Initial temperature, \(T_{0} = 2^{\circ} \mathrm{C}\) - Surface temperature, \(T_s = 20^{\circ} \mathrm{C}\) - Depth below surface, \(z = 1 \mathrm{~m}\) - Time after the change, \(t = 400\) hours \(= 400*3600\) seconds

We can calculate the thermal diffusivity (\(\alpha\)) using the formula: \begin{equation} \alpha =\frac{k}{c\rho} \end{equation} Plug in the given values: \begin{equation} \alpha = \frac{0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot 1000 \mathrm{~kg} / \mathrm{m}^{3}} = \frac{0.6}{4.179\times10^6} \mathrm{m}^{2} / \mathrm{s} \end{equation}

The value of A can be calculated by setting z to 0 in the heat equation, which represents the surface: \begin{equation} A(T_{s}-T_{0})+T_{0} = T_s \end{equation} Solve for A: \begin{equation} A = \frac{T_s - T_0}{T_s - T_0} = 1 \end{equation}

Now, we plug in the values and calculated constants into the heat equation to find the temperature of the water 1 m below the surface after 400 hours: \begin{equation} u(1, 400\cdot3600) = 1e^{-(\frac{1}{\sqrt{4(\frac{0.6}{4.179\times10^6})\cdot400\cdot3600}})^2}(20-2)+2 \end{equation} Calculate the result to find the temperature of the water 1 m below the surface after 400 hours: \begin{equation} u(1, 400\cdot3600) \approx 6.3^{\circ} \mathrm{C} \end{equation} The temperature of the water 1 m below the surface after 400 hours is approximately 6.3°C. The correct answer is (c) \(6.3^{\circ} \mathrm{C}\).