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Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of \(10.0 \mathrm{m}^{2}\) and a thermal conductivity twice that of glass wool with the rate of heat conduction through a 0.750 -cm-thick window that has an area of \(2.00 \mathrm{m}^{2},\) assuming the same temperature difference across each.

Short Answer

Expert verified
The rate of heat conduction through the wall is approximately 1.92 times that of the window, assuming the same temperature difference across each of them.

Step by step solution

01

Recall the formula and identify the given parameters

We are given the following information for the wall: - Thickness: 13.0 cm; - Area: 10.0 m²; - Thermal conductivity: Twice that of glass wool. And for the window: - Thickness: 0.750 cm; - Area: 2.00 m²; - Thermal conductivity: same as that of glass wool. We will use these parameters and the formula for the rate of heat conduction to find the rate of heat conduction through the wall and the window.
02

Calculate the rate of heat conduction through the wall

We know that the thermal conductivity of the wall is twice that of glass wool. The thermal conductivity of glass wool is approximately \(0.04 \frac{W}{m*K}\). Let's calculate the rate of heat conduction through the wall using the given formula: \[Q_{wall} = kA\frac{dT}{dx}\] Given, - \(k_{wall} = 2 * 0.04 \frac{W}{m*K} = 0.08 \frac{W}{m*K}\) (Since it's twice that of glass wool) - \(A_{wall} = 10.0 \; m^2\) - \(dx_{wall} = 13.0 \; cm = 0.13 \; m\) (Converting to meters) Assuming the same temperature difference across the wall and window, we have: \[Q_{wall} = 0.08 * 10.0 * \frac{dT}{0.13}\]
03

Calculate the rate of heat conduction through the window

Now let's calculate the rate of heat conduction through the window: \[Q_{window} = kA\frac{dT}{dx}\] Given, - \(k_{window} = 0.04 \frac{W}{m*K}\) (Thermal conductivity of glass wool) - \(A_{window} = 2.00 \; m^2\) - \(dx_{window} = 0.750 \; cm = 0.00750 \; m\) (Converting to meters) Assuming the same temperature difference across the wall and window, we have: \[Q_{window} = 0.04 * 2.00 * \frac{dT}{0.00750}\]
04

Compare the rates of heat conduction for the wall and the window

Now that we have the two expressions for the rates of heat conduction through the wall and window, we can compare them directly: \[\frac{Q_{wall}}{Q_{window}} = \frac{0.08 * 10.0 * \frac{dT}{0.13}}{0.04 * 2.00 * \frac{dT}{0.00750}}\] Simplifying the above expression: \[\frac{Q_{wall}}{Q_{window}} = \frac{0.08 * 10.0 * 0.00750}{0.04 * 2.00 * 0.13}\] Now, calculate the ratio: \[\frac{Q_{wall}}{Q_{window}} = \approx 1.92\] This means that the rate of heat conduction through the wall is approximately 1.92 times that of the window, assuming the same temperature difference across each of them.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It quantifies the ease with which thermal energy, or heat, passes through a material. When considering different substances, those with high thermal conductivity, such as metals like copper or aluminum, transfer heat rapidly, while materials with low thermal conductivity, such as glass wool or styrofoam, are much slower at conducting heat and are therefore used as thermal insulators.

Understanding thermal conductivity is crucial in tasks ranging from designing energy-efficient buildings to creating heat-resistant components in electronics. The thermal conductivity of a material is typically denoted by the symbol \(k\) and expressed in the units watts per meter per Kelvin \(W/m\cdot K\). A higher \(k\) value indicates that a material is more efficient at conducting heat across a temperature gradient. For example, in the exercise given, the wall has a thermal conductivity double that of glass wool, which implies that it will conduct heat at twice the rate given the same conditions.
Fourier's Law of Heat Conduction
Fourier's law of heat conduction is a fundamental principle that describes how heat flows through materials. This law states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area through which the heat is flowing. Mathematically, the law is expressed as \[Q = -kA\frac{dT}{dx}\], where \(Q\) is the rate of heat conduction, \(k\) is the thermal conductivity, \(A\) is the area through which heat is being transferred, \(dT\) is the temperature difference across the material, and \(dx\) is the thickness of the material.

Following Fourier's law, a larger temperature difference or area will increase the heat conduction rate, while a thicker material (greater \(dx\)) will reduce it. This relationship is crucial when comparing rates of heat conduction, as seen in the exercise. By manipulating variables such as thickness, area, or thermal conductivity, the exercise illustrates how different materials and dimensions influence the rate of heat transfer.
Thermal Insulation Materials
Thermal insulation materials are designed to reduce the rate of heat transfer between two areas. Their effectiveness is largely determined by their thermal conductivity; materials with low thermal conductivity are better insulators. Common examples of thermal insulation materials include glass wool, polystyrene, and fiberglass.

Insulation is a pivotal component in constructing energy-efficient buildings, allowing for temperature control with minimal energy loss. In the context of the exercise, glass wool serves as a point of comparison for understanding the efficiency of insulation. With a lower rate of heat conduction, a material like glass wool is preferable for insulation purposes, as it helps maintain a consistent temperature by slowing the loss or gain of heat. This choice of material and understanding its properties directly affects the thermal efficiency of a building and, consequently, its energy consumption and costs.

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Most popular questions from this chapter

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0 -L capacity when at \(10.0^{\circ} \mathrm{C}\). What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of \(95.0^{\circ} \mathrm{C},\) given that the fluid's volume coefficient of expansion is \(\beta=400 \times 10^{-6} /^{\circ} \mathrm{C} ?\) (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than \(95.0^{\circ} \mathrm{C}\) ).

The brakes in a car increase in temperature by \(\Delta T\) when bringing the car to rest from a speed \(v\). How much greater would \(\Delta T\) be if the car initially had twice the speed? You may assume the car stops fast enough that no heat transfers out of the brakes.

The goal in this problem is to find the growth of an ice layer as a function of time. Call the thickness of the ice layer \(L\). (a) Derive an equation for \(d L / d t\) in terms of \(L\), the temperature \(T\) above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at \(t=0,\) you have \(L=0 .\) If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get \(d L / d t\) multiplied by a (very simple) function of \(L\) on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on \(t\). (c) Will the water eventually freeze to the bottom of the flask?

A 1.28-kg sample of water at \(10.0^{\circ} \mathrm{C}\) is in a calorimeter. You drop a piece of steel with a mass of 0.385 \(\mathrm{kg}\) at \(215^{\circ} \mathrm{C}\) into it. After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don't affect the outcome, as we'll see in the next section.)

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of \(150 \mathrm{MW}\) by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails (1 watt \(=1\) joule/second or \(1 \mathrm{W}=1 \mathrm{J} / \mathrm{s}\) and \(1 \mathrm{MW}=1 \text { megawatt }) . \quad\) (a) Calculate the rate of temperature increase in degrees Celsius per second ( \(^{\circ} \mathrm{C} / \mathrm{s}\) ) if the mass of the reactor core is \(1.60 \times 10^{5} \mathrm{kg}\) and it has an average specific heat of \(0.3349 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\). (b) How long would it take to obtain a temperature increase of \(2000^{\circ} \mathrm{C},\) which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the \(500,000-\mathrm{kg}\) steel containment vessel would also begin to heat up.)

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