The 40-cm-thick roof of a large room made of concrete \(\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.88 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). After a heavy snow storm, the outer surface of the roof remains covered with snow at \(-5^{\circ} \mathrm{C}\). The roof temperature at \(18.2 \mathrm{~cm}\) distance from the outer surface after a period of 2 hours is (a) \(14^{\circ} \mathrm{C}\) (b) \(12.5^{\circ} \mathrm{C}\) (c) \(7.8^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-5^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Use the one-dimensional semi-infinite solid formula and the given values of thermal conductivity (2.12 W/mK) and thermal diffusivity (5.88 × 10^{-7} m²/s). Answer: (c) \(7.8^{\circ} \mathrm{C}\). Solution: Following the given steps, we found that the penetration depth is approximately 0.0203 m, which is much smaller than the 0.4 m thickness of the roof. The temperature change at a distance of 0.182 m from the outer surface was calculated using the initial and outer temperatures and the error function. The final temperature at the given distance was found to be approximately 7.8°C by adding the temperature change to the outer temperature.

Step by step solution

01

Calculate Penetration Depth

Calculate the penetration depth (δ) using the formula: δ = √(αt), where α is the thermal diffusivity, and t is the time elapsed (2 hours). Convert the time to seconds: 2 × 60 × 60 = 7200 s. δ = √(5.88 × 10^{-7} m²/s × 7200 s) ≈ 0.0203 m. 2. Calculate the temperature change at the given distance from the outer surface.
02

Calculate Temperature Change

Now we can calculate the temperature change (ΔT) using the formula: ΔT = (T_i - T_o) × erf(x / (2 × √(αt))), where T_i is the initial temperature, T_o is the outer temperature, x is the given distance (18.2 cm), and erf is the error function. Let's convert the distance to meters: 18.2 cm = 0.182 m ΔT = (15 - (-5)) × erf(0.182 / (2 × √(5.88 × 10^{-7} m²/s × 7200 s))) 3. Calculate the final temperature at the given distance.
03

Calculate Final Temperature

Finally, calculate the roof temperature (T) at the given distance by adding the temperature change to the outer temperature: T = T_o + ΔT Using the previous step result: T = -5 + ΔT ≈ 7.8°C So, the correct answer is (c) \(7.8^{\circ} \mathrm{C}\).

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 Vaia!

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 can spread through a material. It plays a critical role in understanding how temperature changes within an object over time. Represented by the symbol \( \alpha \), thermal diffusivity is a material-specific property and is mathematically defined as the ratio of the thermal conductivity \( k \) to the product of density \( \rho \) and specific heat capacity \( c_p \), or simply \( \alpha = \frac{k}{\rho c_p} \).

When dealing with heat transfer problems, such as the temperature change within a concrete roof after a snowstorm, knowing the thermal diffusivity of concrete allows us to calculate how deeply heat will penetrate the material over a given period. This concept is crucial for solving problems related to transient heat conduction, where the temperatures are constantly changing with time. In our exercise, the thermal diffusivity of concrete helps us to determine the penetration depth and ultimately the temperature at a certain depth and time, given the initial and boundary conditions.
Temperature Distribution
Temperature distribution refers to the spatial variation of temperature within a material or across different regions of a space. It's significant for engineers and scientists because it can affect the structural integrity of materials, the comfort levels in buildings, or the efficiency of thermal systems.

The temperature distribution is governed by the heat transfer mechanisms within a given material—namely conduction, convection, and radiation. However, in cases like the one described in our exercise, where a concrete roof's temperature profile changes over time due to an external weather event, conduction is the predominant mode of heat transfer. Mathematics aids in mapping out this temperature distribution using various equations and considering the material's thermal properties, with the heat equation being central in such calculations. In the textbook problem, the temperature profile of the concrete roof following the snowstorm allows us to calculate the temperature at a specific depth using the methods of transient heat conduction analysis.
Error Function in Heat Transfer
The error function, often denoted as \( \text{erf} \), plays a pivotal role in heat transfer, particularly in solutions to transient heat conduction problems. In our context, when temperature changes with time, the error function comes into play to describe the temperature distribution resulting from diffusion processes.

In our exercise, the error function is used as part of a formula to calculate the temperature change at a certain depth within the roof. The use of \( \text{erf} \) stems from the integral of the Gaussian distribution, which is a foundational element in the probability and statistics fields. It helps us quantify the variation in temperature from the surface to a given point inside a material. The calculation in the exercise utilizes the error function to provide us with the temperature change (\( \Delta T \)) so that we can find the exact temperature at a specific location and time after the heat has been applied or, as in our case, after the cooling effect of snow. This usage of the error function is integral to understanding how heat propagates through objects and is essential in predicting the thermal response of materials under varying temperature conditions.

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

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

A long roll of 2-m-wide and \(0.5\)-cm-thick 1-Mn manganese steel plate coming off a furnace at \(820^{\circ} \mathrm{C}\) is to be quenched in an oil bath \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(45^{\circ} \mathrm{C}\). The metal sheet is moving at a steady velocity of \(15 \mathrm{~m} / \mathrm{min}\), and the oil bath is \(9 \mathrm{~m}\) long. Taking the convection heat transfer coefficient on both sides of the plate to be \(860 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at \(45^{\circ} \mathrm{C}\).

Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\), and \(\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and \(h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The alumina powder is made of particles that are spherical in shape with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of \(20^{\circ} \mathrm{C}\), to reach their melting point from the moment they are injected into the plasma jet.

Large steel plates \(1.0\)-cm in thickness are quenched from \(600^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by submerging them in an oil reservoir held at \(30^{\circ} \mathrm{C}\). The average heat transfer coefficient for both faces of steel plates is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Average steel properties are \(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Calculate the quench time for steel plates.

4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.

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.

Sign-up for free