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Consider heat transfer through a windowless wall of a house on a winter day. Discuss the parameters that affect the rate of heat conduction through the wall.

Short Answer

Expert verified
Answer: The main parameters that affect the rate of heat conduction are the thermal conductivity of the wall material, the wall thickness, the temperature difference between the inside and outside, and the surface area of the wall. Thermal conductivity determines how well a material conducts heat, with lower values being desirable for insulation. Wall thickness impacts heat resistance, with thicker walls reducing heat loss. A larger temperature difference causes a higher heat flow through the wall, while a larger surface area allows for more significant heat loss. Analyzing these parameters helps optimize a house wall to minimize heat loss during winter.

Step by step solution

01

Understand the concept of heat conduction.

Heat conduction is the process by which heat energy is transferred through a material due to a temperature difference. In the case of a house wall, heat flow occurs from the warm interior to the cold exterior. The rate of heat transfer through the wall is determined by the thermal conductivity, wall thickness, temperature difference, and the area of the wall.
02

Examine the role of thermal conductivity.

Thermal conductivity (k) is a measure of how well a material conducts heat. It is dependent on the material's composition and structure. A higher thermal conductivity means heat can transfer more quickly through the material. In the case of heat transfer through a wall, materials with low thermal conductivity values are desirable to reduce heat loss to the external environment.
03

Investigate the influence of wall thickness.

The thickness of the wall (d) is another factor that affects the rate of heat conduction. A thicker wall increases the distance that heat must travel to reach the other side, resulting in a reduction of heat loss. Since heat resistance is directly proportional to the thickness of the wall, a thicker wall is more insulating and offers better energy-saving benefits.
04

Consider temperature difference and surface area.

The temperature difference (ΔT) between the inside and outside of the house is a driving force for heat transfer. A larger temperature difference results in a higher heat flow through the wall. Additionally, the area (A) of the wall plays a role in heat transfer, with a larger surface area allowing for more significant heat loss.
05

Apply Fourier's Law of heat conduction.

Fourier's Law describes the relationship between the rate of heat transfer (Q), thermal conductivity (k), temperature difference (ΔT), wall thickness (d), and surface area (A) of the wall: Q = -k * A * (ΔT / d) This formula shows that the rate of heat transfer depends on the thermal conductivity, temperature difference, wall thickness, and surface area. By analyzing this formula and understanding the role of each parameter, it becomes easier to understand how to optimize a house wall to minimize heat loss during winter. By considering the parameters of thermal conductivity, wall thickness, temperature difference, and surface area, the rate of heat conduction through a windowless wall on a winter day can be analyzed and understood. This knowledge is crucial for designing effective insulation strategies and constructing energy-efficient homes.

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Key Concepts

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

Thermal Conductivity
When discussing heat conduction through the walls of a house, thermal conductivity is at the forefront. It’s essentially a property that reflects how well a material can transfer heat. Think of thermal conductivity as a highway for heat. Some materials, like metals, have high thermal conductivity and allow heat to road trip through easily. Others, like bricks or fiberglass, are not so generous, thereby holding onto heat.
  • Materials with high thermal conductivity allow rapid heat transfer.
  • Materials with low thermal conductivity are good insulators.
So, for a house wall, materials with low thermal conductivity help keep the warmth inside on a chilly day, making them essential for reducing energy costs and keeping homes snug.
Fourier's Law
Understanding how heat conduction works wouldn’t be complete without Fourier’s Law. This law looks at the specifics, giving us a formula to calculate the rate at which heat transfers through materials.
Fourier's Law is expressed as:
\[ Q = -k \times A \times (\Delta T / d) \]
  • Here, \(Q\) represents the rate of heat transfer.
  • \(k\) stands for thermal conductivity.
  • \(A\) is the surface area through which heat is being transferred.
  • \(\Delta T\) is the temperature difference.
  • \(d\) is the thickness of the wall.
Using this formula, we can understand how various factors, like the material of the wall or external temperatures, affect how much heat leaves or enters a home. This knowledge allows us to develop better insulation practices, enhancing energy efficiency.
Insulation Strategies
To combat unwanted heat loss, particularly in homes without windows or areas highly exposed to the elements, insulation strategies are key. These are techniques or materials used to minimize heat transfer.
  • Use materials with low thermal conductivity such as foam boards or fiberglass.
  • Increase wall thickness to create more resistance to heat flow.
These strategies help keep homes warmer in winter and cooler in summer, maintaining a balanced indoor climate and potentially reducing heating and cooling costs. Effective insulation is a cornerstone of designing energy-efficient homes.
Temperature Difference
In the realm of heat conduction, temperature difference is the driving force. It’s the disparity between two points that prompts heat to move.
  • A higher temperature difference results in a higher rate of heat transfer.
  • This difference can be managed through insulation and effective building design.
A large temperature gap between the warm interior and cold exterior of a house means heat wants to escape quickly. By minimizing this difference inside the home, energy loss can be controlled.
Energy-efficient Homes
Creating an energy-efficient home is more than just a trendy architectural goal; it’s a necessity in today’s eco-conscious world. To achieve this, understanding heat conduction is vital as it allows for smarter choices in design and materials.
  • Use of low conductivity materials reduces energy consumption.
  • Applying Fourier’s Law to optimize building design effectively controls indoor climate.
This attentiveness in home construction means not only reduced utility bills but also a decreased environmental footprint. When the principles of thermal management are considered in building plans, anyone can enjoy a comfortable indoor temperature with reduced energy use.

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Most popular questions from this chapter

Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.

Heat is lost through a brick wall \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is \(4 \mathrm{~m}\) long, \(3 \mathrm{~m}\) wide, and \(25 \mathrm{~cm}\) thick at a rate of \(500 \mathrm{~W}\). If the inner surface of the wall is at \(22^{\circ} \mathrm{C}\), the temperature at the midplane of the wall is (a) \(0^{\circ} \mathrm{C}\) (b) \(7.5^{\circ} \mathrm{C}\) (c) \(11.0^{\circ} \mathrm{C}\) (d) \(14.8^{\circ} \mathrm{C}\) (e) \(22^{\circ} \mathrm{C}\)

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

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