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 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) \(10.2^{\circ} \mathrm{C}\)

Short answer

Answer: The temperature of the water 1 meter below the surface after 400 hours is approximately 6.3°C.

Step-by-step solution

Write down the heat equation and required constants

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

Calculate the thermal diffusivity

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}

Calculate the amplitude A

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}

Calculate the temperature at the given depth and time

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) \(6.3^{\circ} \mathrm{C}\).

Key concepts

Thermal Diffusivity

Thermal diffusivity is a measure of how quickly heat moves through a material. In simpler terms, it describes the rate at which temperature changes occur within a substance as heat is applied or removed. This property is crucial in understanding heat conduction, especially in fluids such as water.

To calculate thermal diffusivity, we use the formula \( \alpha = \frac{k}{c\rho} \), where \( k \) is the thermal conductivity, \( c \) is the specific heat capacity, and \( \rho \) is the density of the fluid.

This relationship tells us that thermal diffusivity depends on how easily heat can travel through the material (thermal conductivity), how much energy the material can store (specific heat capacity), and the material's density. This is why thermal diffusivity is a vital concept when analyzing how temperature changes in a lake or any body of water over time.

For water in a lake, a low thermal diffusivity value means the heat takes a longer time to penetrate deep into the water, explaining why temperature changes slowly at greater depths.

Conduction in Fluids

Heat conduction in fluids, like water, is a process where thermal energy is transferred from hotter regions to cooler regions. This transfer occurs via the movement of molecules in the fluid.

In the context of a lake, when the surface temperature suddenly increases due to warm air, the heat starts to conduct downwards. However, as water has relatively low thermal conductivity, this process is not very rapid. The heat gradually diffuses deeper into the water, creating layers of differing temperatures with depth.

Layers form because warmer water remains on top of the colder water, given that it is less dense. This stratification is common in large bodies of water and affects how quickly or slowly temperature changes at different depths.

Understanding this process in fluids helps predict temperature changes and manage natural water bodies effectively, keeping in mind that factors like turbulence and wind can alter these patterns significantly.

Temperature Distribution

Temperature distribution in a medium like water is the variation of temperature across its depth. When heat is applied to the surface of a lake, analyzing how the temperature changes as you go deeper is essential.

This concept is often analyzed using the heat equation, which describes how heat diffuses through a homogeneous medium over time. In the example of a lake, the heat equation helps calculate the temperature at different depths by considering factors such as the thermal diffusivity of water, the initial temperature, and the change in surface temperature. For a significant time, after a sudden change in surface temperature, the temperature at a given depth can be determined by factors like time elapsed, depth, and initial conditions. Understanding temperature distribution helps predict environmental changes and make necessary adjustments to sustain aquatic life and ecosystem balance.