/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 175 When water, as in a pond or lake... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

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

Step by step solution

01

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
02

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}
03

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}
04

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

What are the factors that affect the quality of frozen fish?

A thermocouple, with a spherical junction diameter of \(0.5 \mathrm{~mm}\), is used for measuring the temperature of hot air flow in a circular duct. The convection heat transfer coefficient of the air flow can be related with the diameter \((D)\) of the duct and the average air flow velocity \((V)\) as \(h=2.2(V / D)^{0.5}\), where \(D, h\), and \(V\) are in \(\mathrm{m}, \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{m} / \mathrm{s}\), respectively. The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Determine the minimum air flow velocity that the thermocouple can be used, if the maximum response time of the thermocouple to register 99 percent of the initial temperature difference is \(5 \mathrm{~s}\).

Consider a curing kiln whose walls are made of \(30-\mathrm{cm}-\) thick concrete with a thermal diffusivity of \(\alpha=0.23 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Initially, the kiln and its walls are in equilibrium with the surroundings at \(6^{\circ} \mathrm{C}\). Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to \(42^{\circ} \mathrm{C}\) and the temperature is maintained at that level for \(2.5 \mathrm{~h}\). The curing kiln is then opened and exposed to the atmospheric air after the steam flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for \(2.5 \mathrm{~h}\) only, or would it make no difference? Base your answer on calculations.

A body at an initial temperature of \(T_{i}\) is brought into a medium at a constant temperature of \(T_{\infty}\). How can you determine the maximum possible amount of heat transfer between the body and the surrounding medium?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.