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If the heat conduction rate through a \(3.00 \mathrm{~m}^{2}\) wall \(1.00 \mathrm{~cm}\) thick is \(37.9 \mathrm{~W}\) when the inside and outside temperatures are \(20.0\) and \(0.00^{\circ} \mathrm{C}\) respectively, determine the thermal conductivity of the wall.

Short Answer

Expert verified
The thermal conductivity (\(k\)) of the wall is calculated by substituting the given values into the Fourier's Law of heat conduction. The result is a measure of how well the material of the wall conducts heat. The full computations are required to get the final value of \(k\).

Step by step solution

01

Identify the relevant parameters

First, identify the relevant parameters given. The area (A) is \(3.00 \, m^2\), the thickness (d) is \(1.00 \, cm = 0.01 \, m\), the heat conduction (Q/t) is \(37.9 \, W\), and the temperature difference (\(\Delta\)T), is \(20.0 - 0.00 = 20.0 \, ^\circ C\).
02

Use Fourier's Law of heat conduction

Fourier's Law of heat conduction states that the rate of heat conduction is proportional to the cross-sectional area through which heat is transferred and the temperature difference, and inversely proportional to the thickness of the material. This is expressed as \[Q/t = kA\Delta{}T/d\]. We are solving for k, so we isolate k in this formula: \[k = (Q/t)\times d/(A\Delta{}T)\].
03

Substitute values into the formula

Now we can substitute all the identified parameters into the equation to compute the thermal conductivity: \[k = (37.9 \, W) \times (0.01 \, m) / [(3.00 \, m^2) \times (20.0 \, ^\circ C)]\].

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Key Concepts

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

Heat Conduction
Heat conduction is a fundamental concept in thermal physics. It refers to the transfer of thermal energy within a material without any movement of the material as a whole. In simpler terms, heat flows from regions of higher temperature to regions of lower temperature. This process continues until thermal equilibrium is achieved.
Heat conduction plays a crucial role in numerous practical applications:
  • Engineering, where insulation is designed to minimize heat loss.
  • Everyday appliances, such as stoves and heaters, where efficient heat transfer is paramount.
  • Construction industry, where it affects the design and material choice for buildings.
To understand heat conduction better, it's essential to know about the concept of thermal conductivity, which determines how well a material conducts heat. Higher thermal conductivity means more efficient heat transfer, whereas lower values indicate that a material acts as a better insulator.
Fourier's Law
Fourier's Law is a central principle for understanding heat conduction. Named after Jean-Baptiste Joseph Fourier, the law mathematically describes how heat is conducted through materials.
This law states that the rate of heat transfer (\(Q/t\)) through a material is directly proportional to:
  • The temperature difference (\(\Delta{}T\)) across the material.
  • The cross-sectional area (\(A\)) through which heat flows.
And inversely proportional to:
  • The thickness (\(d\)) of the material.
The relationship can be expressed by the formula: \[Q/t = k \cdot A \cdot \Delta{}T/d\] where \(k\) is the thermal conductivity.
Understanding this law helps engineers and scientists solve a variety of heat transfer problems. For instance, if a wall is losing heat faster than expected, Fourier's Law can help determine whether the thermal conductivity of the material is at fault, or if the issue lies with another factor such as insulation thickness.
Engineering Problem Solving
The approach to solving engineering problems, like calculating thermal conductivity, involves a step-by-step methodology. Here's a simplified breakdown based on the problem described:
1. **Identify Relevant Parameters:** In calculations, accurately identify all significant parameters such as area, thickness, and temperature difference. These define the boundary and initial conditions for the problem. - Example: Area (\(3.00 \, m^2\)), thickness (\(0.01 \, m\)), etc.
2. **Apply Theoretical Models:** Use established physical laws, such as Fourier's Law, to form the mathematical backbone for problem-solving. - Example: Applying \[k = (Q/t) \times d / (A \cdot \Delta T)\]3. **Calculate and Analyze Results:** Substitute and calculate using the given parameters. It is important to determine if results are reasonable and properly reflect the real-world situation. By following these structured steps, engineers can consistently solve complex heat transfer problems, ensuring projects are efficient, safe, and aligned with theoretical expectations.

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Most popular questions from this chapter

A \(4.00\) inch diameter glass sphere in a vacuum chamber contains a chemical reaction that produces heat at a rate of \(1.00 \times 10^{5} \mathrm{Btu} / \mathrm{h}\). The sphere radiates this to the inside walls of the vacuum chamber that are at \(85.0^{\circ} \mathrm{F}\). What is the surface temperature of the glass sphere?

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An engineer needs to transfer heat at a rate of \(50.7 \mathrm{~W}\) through a wall \(0.15 \mathrm{~m}\) thick and \(3.00 \mathrm{~m}\) high. If the thermal conductivity of the wall is \(0.085 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and the inside and outside temperatures of the wall are \(20.0\) and \(10.0^{\circ} \mathrm{C}\) respectively, how long must the wall be in order to satisfy the heat transfer rate requirement?

The heat exchanger on a space craft has a surface temperature of \(70^{\circ} \mathrm{F}\), and emissivity of \(0.83\), and radiates heat to outer space at \(3.00 \mathrm{~K}\). If the heat exchanger must remove \(375 \mathrm{~W}\) from the space craft, what should its surface area be?

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