The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

Short Answer

Expert verified
Answer: (c) \(0.25 \mathrm{~m}\)

Step by step solution

01

Use the Temperature function T(x)

We are given the temperature function \(T(x)=52x+25\). We need to find the value of \(x\) for the case when \(T(x) = 38^{\circ} \mathrm{C}\). To do this, simply replace \(T(x)\) with \(38\) in the equation and solve for x.
02

Solve for x

Now, we have the equation \(38=52x+25\). To solve for x, follow these steps: 1. Subtract 25 from both sides of the equation: \(38-25=52x+25-25 \Rightarrow 13=52x\) 2. Divide by 52 on both sides to get x: \(x=\frac{13}{52}=0.25\)
03

Find the correct answer

Now that we have found that x is 0.25 when the temperature is \(38^{\circ} \mathrm{C}\), we can look through the given options and find that the correct answer is: (c) \(0.25 \mathrm{~m}\) This is the thickness of the wall.

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Key Concepts

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

Temperature Distribution
Understanding temperature distribution within solids like a plane wall is crucial for various engineering applications, such as insulation design or heating systems. Temperature distribution refers to how heat is spread throughout a material and is depicted by a temperature function, typically dependent on position and possibly time. In the context of a steady-state scenario, where no changes occur over time, temperature distribution is solely a function of position.

In the exercise, we are presented with a linear temperature function in the form of \(T(x)=52x+25\), where \(T\) denotes the temperature at a given distance \(x\) from a reference point in degrees Celsius. When such linear behavior is observed, it implies a constant rate of temperature change with respect to the position within the solid, which is a characteristic of a uniform material with no internal heat sources.

The formula to find the temperature at a specific position involves substituting the value of \(x\) into the temperature function, which will yield the temperature at that location. In the exercise, we substitute \(T(x)\) with \(38^{\textdegree}\mathrm{C}\) to find the corresponding position \(x\), which denotes the thickness of the wall.
Plane Wall Temperature Profile
A plane wall temperature profile is a graphical representation that shows how temperature varies across the thickness of a wall. The wall is assumed to have a large surface area compared to its thickness, which makes the heat transfer essentially one dimensional. By plotting temperature on the y-axis and the position within the wall (thickness) on the x-axis, one can visualize the temperature gradient.

The linear temperature distribution, as given in the exercise by the function \(T(x)=52x+25\), reflects a straight line on a temperature profile, indicating that the wall has a constant thermal gradient. It simplifies the analysis of heat transfer as one can quickly determine the temperature at any point in the wall by knowing its distance from a reference point.

When assessing the temperature profile of a wall, it's essential to keep in mind that various factors, such as the material properties, boundary conditions, and surrounding temperatures, affect the overall distribution.
Thermal Conductivity
Thermal conductivity is a property of materials that measures their ability to conduct heat. It is typically represented by the symbol \(k\) and measured in units of \(\text{W/(m}\cdot\text{K)}\). In the study of heat conduction in solids, thermal conductivity plays a pivotal role as it dictates the rate at which heat energy transfers through a material due to the temperature gradient.

High thermal conductivity materials, such as metals, facilitate rapid heat transfer, making them ideal for cooling purposes or heat exchangers. Conversely, materials with low thermal conductivity, like insulating foam, are used to prevent heat from easily escaping or entering a space.

The temperature profile within a plane wall is deeply influenced by the thermal conductivity of the wall's material. In a typical heat transfer scenario, \(k\) is used in conjunction with the area through which heat is transferred, the temperature difference, and the wall's thickness to calculate the heat flux. However, in the context of the given exercise, the thermal conductivity is not directly involved as we are looking at temperature distribution rather than evaluating the heat transfer rate.

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Most popular questions from this chapter

Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

The thermal conductivity of stainless steel has been characterized experimentally to vary with temperature as \(k(T)=9.14+0.021 T\) for \(273

Consider uniform heat generation in a cylinder and a sphere of equal radius made of the same material in the same environment. Which geometry will have a higher temperature at its center? Why?

A spherical container, with an inner radius of \(1 \mathrm{~m}\) and a wall thickness of \(5 \mathrm{~mm}\), has its inner surface subjected to a uniform heat flux of \(7 \mathrm{~kW} / \mathrm{m}^{2}\). The outer surface of the container is maintained at \(20^{\circ} \mathrm{C}\). The container wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0023 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the temperature drop across the container wall thickness.

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

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