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(a) Calculate the rate of heat conduction through house walls that are \(13.0 \mathrm{cm}\) thick and have an average themal conductivity twice that of glass wool. Assume there are no windows or doors. The walls' surface area is \(120 \mathrm{m}^{2}\) and their inside surface is at \(18.0^{\circ} \mathrm{C}\), while their outside surface is at \(5.00^{\circ} \mathrm{C} .\) (b) How many \(1-\mathrm{k} \mathrm{W}\) room heaters would be needed to balance the heat transfer due to conduction?

Short Answer

Expert verified
After completing these steps, you'll have calculated the rate of heat conduction through the walls in J/s (or kW), and the number of 1 kW room heaters required to balance this heat transfer.

Step by step solution

01

Calculate the thermal conductivity of the wall

Firstly, we need to find the thermal conductivity (γ) of the house walls. This is given by multiplying the thermal conductivity of glass wool by 2, or \(\gamma = 2 \cdot 0.038 \, W/m \cdot K\). That gives us \(\gamma = 0.076 \, W/m \cdot K\). This value represents how efficiently the wall material conducts heat.
02

Calculate the rate of heat conduction

Now we find the thermal conduction (Q) using the formula mentioned earlier: \(Q = \frac{kA\Delta T}{d}\). Substituting the values gives us: \[Q = \frac{0.076 \, W/m \cdot K \cdot 120 \, m^2 \cdot (18.0 \, C - 5.00\, C)}{0.13 \, m}\] Calculate this expression to find Q.
03

Calculate the number of heaters

We know the heat produced by each heater is 1 kWh. To balance the heat transferred due to conduction, we divide the heat conduction Q by 1 kW. Remember that 1 kW corresponds to \(1000 \, J/s\), so convert Q to kW if needed before performing the division.

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Key Concepts

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

Thermal Conductivity
Understanding thermal conductivity is essential when we talk about heat conduction. It's a material property that describes the ability to conduct heat. Every substance has a unique thermal conductivity, usually denoted as \( k \) or sometimes \( \text{gamma} \), and it's measured in watts per meter-kelvin (\( W/m \cdot K \)). A high value means the material can rapidly transfer heat, which is typical for metals, while a low value indicates an insulating property, like that of rubber or glass wool.

When we refer to thermal conductivity in the context of building materials, it directly influences how energy efficient a building is. For instance, the higher the thermal conductivity of the wall material, the more heat it will lose to the outside during the winter or gain from the outside during the summer. This is why insulating materials with low thermal conductivity are used in constructing energy-efficient homes.
Thermal Conduction Calculation
To calculate the rate of heat conduction, also known as heat flux (\( Q \)), we use the Fourier's law of heat conduction. The general formula is simple:

\[ Q = \frac{kA\Delta T}{d} \]
Here, \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area through which heat is being transferred, \( \Delta T \) is the temperature difference across the material, and \( d \) is the thickness of the material.

In an educational context, understanding this formula is crucial. For better concept visualization, imagine the heat flow through a wall as water flow through a pipe; the bigger the pipe (area) and the greater the pressure difference (temperature difference), the more water (heat) flows through. Similarly, the thicker the pipe (wall), the less water (heat) flows, akin to resistance. Thus, to ensure students fully grasp the concept, real-world analogies alongside the formula significantly aid in their understanding.
Heat Transfer in Buildings
Heat transfer in buildings is a real-world application of the principles of thermal conductivity and heat conduction calculation. It occurs by conduction through walls, roofs, and floors, as well as by convection and radiation through spaces and openings like windows and doors.

For buildings, maintaining a comfortable indoor temperature regardless of external conditions is the goal. To achieve this, materials with appropriate thermal conductivity values are selected to reduce unwanted heat transfer. Insulating materials with a low thermal conductivity are installed in walls, attics, and around windows to limit heat loss or gain.

Understanding this concept helps not only in calculating the rate at which heat is transferred through building components but also in designing efficient heating and cooling systems. For example, in the original exercise, assuming no windows or doors simplifies the calculation, but in reality, these factors must be considered as they significantly impact overall heat transfer. Still, the simplified calculation is an excellent start for students to learn the underlying physics before tackling more complex scenarios.

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Most popular questions from this chapter

How many grams of coffee must evaporate from 350 g of coffee in a \(100-\mathrm{g}\) glass cup to cool the coffee and the cup from \(95.0^{\circ} \mathrm{C}\) to \(45.0^{\circ} \mathrm{C}\) ? Assume the coffee has the same thermal properties as water and that the average heat of vaporization is \(2340 \mathrm{kJ} / \mathrm{kg}(560 \mathrm{kcal} / \mathrm{g}) .\) Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answer?

Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while the surface only a few centimeters away is safe to touch. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?

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.)

Satellites designed to observe the radiation from cold (3 K) dark space have sensors that are shaded from the Sun, Earth, and the Moon and are cooled to very low temperatures. Why must the sensors be at low temperature?

Loose-fitting white clothing covering most of the body, shown below, is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night.

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