/*! 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 12 A cubical box of volume \(216 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A cubical box of volume \(216 \mathrm{~cm}^{3}\) is made up of \(0.1 \mathrm{~cm}\) thick wood. The inside is heated electrically by a \(100 \mathrm{~W}\) heater. It is found that the temperature difference between the inside and the outside surface is \(5^{\circ} \mathrm{C}\) in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.

Short Answer

Expert verified
The thermal conductivity of the material is approximately 0.93 W/(m·K).

Step by step solution

01

Calculate the Side Length of the Cube

The formula for the volume of a cube is \( V = a^3 \), where \( a \) is the side length of the cube. Given the volume \( V = 216 \ \mathrm{cm}^3 \), we solve for \( a \):\[ a = \sqrt[3]{216} \ \mathrm{cm} = 6 \ \mathrm{cm}. \]
02

Determine the Surface Area

The surface area of a cube with side length \( a \) is given by the formula \( A = 6a^2 \). With \( a = 6 \ \mathrm{cm} \), the surface area is:\[ A = 6 \times (6)^2 \ \mathrm{cm}^2 = 216 \ \mathrm{cm}^2. \]
03

Calculate the Heat Transfer

The rate of heat transfer \( Q \) through a material is given by the equation:\[ Q = \frac{k \cdot A \cdot \Delta T}{d}, \] where \( k \) is the thermal conductivity, \( A \) is the surface area, \( \Delta T \) is the temperature difference, and \( d \) is the thickness of the material. Given that \( Q = 100 \ \mathrm{W} \), \( A = 216 \ \mathrm{cm}^2 = 0.0216 \ \mathrm{m}^2 \), \( \Delta T = 5 \ \mathrm{K} \), and \( d = 0.1 \ \mathrm{cm} = 0.001 \ \mathrm{m} \), we solve for \( k \):\[ 100 = \frac{k \cdot 0.0216 \cdot 5}{0.001}. \]
04

Solve for Thermal Conductivity

Rearrange the heat transfer equation to solve for \( k \):\[ k = \frac{100 \times 0.001}{0.0216 \times 5}. \]Calculate \( k \):\[ k = \frac{0.1}{0.108} \approx 0.9259 \ \mathrm{W/(m \cdot K)}. \]

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that describes the movement of heat energy from one place to another. It can occur through conduction, convection, or radiation. Conduction happens when heat moves through a solid substance from a high-temperature area to a lower-temperature area. In this exercise, heat is transferred through the wooden box by conduction. The heat generated by the heater inside the box moves through the wood to the outside surface.

To quantify heat transfer, we use a key formula:
  • The equation is: \( Q = \frac{k \, \cdot \, A \, \cdot \, \Delta T}{d} \), where:
    • \( Q \) is the rate of heat transfer.
    • \( k \) is the thermal conductivity of the material.
    • \( A \) is the surface area through which heat is transferred.
    • \( \Delta T \) is the temperature difference across the material.
    • \( d \) is the thickness of the material the heat transfers through.
By understanding this formula, you can determine how effectively a material conducts heat, which is essential in designing objects that either conserve heat or allow for efficient heat dissipation.
Temperature Difference
The temperature difference, denoted as \( \Delta T \), refers to the disparity in temperature between two areas. In our exercise, this is the difference between the inside temperature of the heated box and the ambient room temperature outside. This difference is crucial because it drives the heat transfer process.

The greater the temperature difference, the more rapidly heat will travel from the warmer area to the cooler area. This relationship is why temperature difference is included as a multiplicative factor in the heat transfer equation. It highlights how slight changes in the internal or external temperature can significantly affect the rate of heat movement.

In the context of everyday life, understanding temperature difference helps in energy conservation. For instance, in heating systems or refrigeration, optimizing the temperature gradient can lead to more energy-efficient designs.
Surface Area Calculation
Surface area plays a vital role in the heat transfer equation because it determines the size of the interface through which heat is conducted. For a cube, the surface area is calculated using the formula: \( A = 6a^2 \), where \( a \) is the length of one side of the cube.

In this exercise, since the side of the cube is \( 6 \ \mathrm{cm} \), the surface area is computed as \( 6 \times 6^2 \ \mathrm{cm}^2 = 216 \ \mathrm{cm}^2 \). Converting this into meters squared gives us the area for efficient calculation in the SI unit system, i.e., \( 0.0216 \ \mathrm{m}^2 \).

The larger the surface area, the more space is available for heat to be transferred through the substance. By maximizing the surface area through design, engineers can either enhance heat gain or loss, tailoring to specific needs, like in radiators or insulation materials.
Volume of a Cube
The volume of a cube is a key starting point in calculating other dimensions, such as its side length and surface area. Volume helps us understand how much space an object occupies in three dimensions.

For a cube, the volume is determined by the formula: \( V = a^3 \), where \( a \) is the side length. In the given problem, the cube's volume is \( 216 \ \mathrm{cm}^3 \). Solving for \( a \), we find \( a = \sqrt[3]{216} \ \mathrm{cm} = 6 \ \mathrm{cm} \).

Once we have the side length, we can find additional properties of the cube that relate to its thermal behavior, such as its surface area. Understanding these dimensions is critical in applications such as manufacturing and packaging, where both the size and the thermal properties of materials are crucial considerations.

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Most popular questions from this chapter

One end of a rod of length \(20 \mathrm{~cm}\) is inserted in a furnace at \(800 \mathrm{~K}\). The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is \(750 \mathrm{~K}\) in the steady state. The temperature of the surrounding air is \(300 \mathrm{~K}\). Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conduetivity of the rod. Stefan constant \(\sigma=6 \cdot 0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\).

A metal ball of mass \(1 \mathrm{~kg}\) is heated by means of a \(20 \mathrm{~W}\) heater in a room at \(20^{\circ} \mathrm{C}\). The temperature of the ball becomes steady at \(50^{\circ} \mathrm{C}\). (a) Find the rate of loss of heat to the surrounding when the ball is at \(50^{\circ} \mathrm{C}\). (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at \(30^{\circ} \mathrm{C}\). (c) Assume that the temperature of the ball rises uniformly from \(20^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal.

A hollow tube has a length \(l\), inner radius \(R_{1}\) and outer radius \(R_{2}\). The material has a thermal conductivity \(K\). Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperatures \(T_{1}\) and \(T_{2}\left(T_{2}>T_{1}\right)\) (b) the inside of the tube is maintained at temperature \(T_{1}\) and the outside is maintained at \(T_{2}\)

Steam at \(120^{\circ} \mathrm{C}\) is continuously passed through a \(50-\mathrm{cm}\) long rubber tube of inner and outer radii \(1 \cdot 0 \mathrm{~cm}\) and \(1.2 \mathrm{~cm}\). The room temperature is \(30^{\circ} \mathrm{C}\). Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber \(=0 \cdot 15 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\)

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of (a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and \((\mathrm{b})\) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium \(=900 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) and that of copper \(=390 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{-1} \mathrm{C}^{-1}\). The density of copper \(=3-4\) times the density of aluminium.

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