Question

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 \(400 \mathrm{~h}\) 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) \(10.2^{\circ} \mathrm{C}\)

Short answer

Answer: (b) 4.2°C

Step-by-step solution

Identify the relevant formula for heat conduction

For this problem, we will use the heat conduction formula, which is given by: \begin{equation} T(x,t) = T_i + (T_{\text{surface}} - T_{i}) \times \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right) \end{equation} where \(T(x,t)\) is the temperature at the depth \(x\) and time \(t\), \(T_i\) is the initial temperature, \(T_{\text{surface}}\) is the surface temperature, and \(\text{erf}\) is the error function (a standard mathematical function). The parameter α is the thermal diffusivity, which is defined as \(\frac{k}{\rho c_p}\), where \(k\) is the thermal conductivity, \(\rho\) is the density of water (approximately 1000 kg/m³), and \(c_p\) is the specific heat capacity.

Calculate the thermal diffusivity α

Given the values of \(k\), \(\rho\), and \(c_p\), we can calculate the thermal diffusivity α: \begin{align*} \alpha &= \frac{k}{\rho c_p} \\ &= \frac{0.6 \text{ W/m} \cdot \text{K}}{(1000 \text{ kg/m}^3)(4.179 \times 10^3 \text{ J/kg} \cdot \text{K})} \\ &\approx 1.43 \times 10^{-7} \text{ m}^2 / \text{s} \end{align*}

Calculate x and t in the appropriate units

We need to find the temperature 1m below the surface (x) after 400 hours (t). Convert these values to the SI units: \begin{align*} x &= 1 \text{ m} \\ t &= 400 \text{ h} \times 3600 \text{ s/h} = 1.44 \times 10^6 \text{ s} \end{align*}

Calculate the temperature T(x,t)

Now, let's plug in the values into the heat conduction equation and calculate the temperature \(1 \text{ m}\) below the surface after \(400 \text{ h}\): \begin{align*} T(x,t) &= T_i + (T_{\text{surface}} - T_i) \times \text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right) \\ &= 2 + (20 - 2) \times \text{erf}\left(\frac{1}{2\sqrt{1.43 \times 10^{-7} \text{ m}^2 / \text{s} \times 1.44 \times 10^6 \text{ s}}}\right) \\ &\approx 2 + 18 \times \text{erf}(0.143) \\ &\approx 2 + 18 \times 0.156 \\ &\approx 4.8 \end{align*} Since \(4.8\) is closer to \(4.2\) than any of the other choices, the correct answer is (b) \(4.2^\circ \text{C}\).