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In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

Short Answer

Expert verified
Answer: The percentage reduction in heat loss is 40%.

Step by step solution

01

Find the heat loss ratios

To find the heat loss ratio, we'll need to divide the initial R-value by the final R-value. Heat loss ratio = \(\frac{Initial R-value}{Final R-value}\).
02

Calculate the heat loss ratio with given R-values

Given the initial R-value of 15 and the final R-value of 25, we can calculate the heat loss ratio as: Heat loss ratio = \(\frac{15}{25}\).
03

Simplify the heat loss ratio

Now, we simplify the heat loss ratio: Heat loss ratio = \(\frac{3}{5}\).
04

Calculate the percentage reduction in heat loss

To find the percentage reduction in heat loss, multiply the heat loss ratio by 100 and subtract it from 100%: Percentage reduction in heat loss = \(100\% - (Heat loss ratio \times 100\%)\).
05

Plug in the heat loss ratio and find the answer

We have the heat loss ratio \(\frac{3}{5}\), so the percentage reduction in heat loss is: Percentage reduction in heat loss = \(100\% - (\frac{3}{5}\times 100\%) = 100\% - 60\% = 40\%\). The answer is (b) \(40 \%\).

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Key Concepts

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

Thermal Resistance
Thermal resistance, symbolized by an R-value, is a measurement of a material's ability to resist heat flow. The higher the R-value, the better the insulation's ability to prevent heat transfer.

Imagine a thermal barrier between two environments at different temperatures; the insulation slows down the heat trying to equalize the temperatures across this barrier. In the context of home insulation, materials with higher R-values will slow down the transfer of heat from inside the home to the outside during the winter, and vice versa during the summer.

Real-world Importance

The practical significance of thermal resistance comes into play when choosing materials for insulating homes. The chosen materials should have R-values that align with the climate and the specific heating and cooling needs of the house. In cold climates, a higher R-value is essential to keep heat indoors, while in milder climates, a lower R-value might be sufficient.

When selecting insulation, it's not just the R-value that matters, but also the insulation's thickness, density, and installation quality, as these factors will influence the overall effectiveness of the insulation in resisting heat flow.
Heat Loss Calculation
Calculating heat loss is critical for determining how well a house retains heat, and for making decisions about heating systems and insulation. Heat loss through a structure is influenced by several factors, including the thermal resistance of the materials, the temperature difference between inside and outside, and the surface area through which heat can escape.

Understanding the Calculation

To calculate the heat loss, one must understand the relationship between the R-value and the rate of heat loss. In essence, a low R-value means more heat loss, while a higher R-value suggests less heat escape. The heat loss calculation makes it possible to estimate the efficiency of insulation and the potential savings on energy costs by enhancing insulation.

For homeowners considering upgrades, a heat loss calculation can reveal how much improvement in insulation will reduce the overall heat loss. It solidifies the decision on whether an investment in insulation will yield substantial savings on energy expenses in the long run.
Home Insulation Efficiency
Home insulation efficiency refers to how effectively a building's insulation reduces energy consumption by minimizing heat transfer. The performance of insulation is a key factor in achieving energy efficiency in homes, which not only helps lower utility bills but also lessens environmental impact.

Maximizing Efficiency

To maximize insulation efficiency, one must consider the entire 'envelope' of the home, which includes the attic, walls, floors, windows, and doors. High-efficiency insulation maintains the desired temperature inside the home more consistently, requiring less energy for heating and cooling.

  • Upgrade Insulation: Increasing the R-value of insulation in key areas, like the attic and walls, can significantly reduce energy costs.
  • Seal Gaps: Ensuring that there are no gaps or leaks where air can pass through is crucial for maintaining the insulation's effectiveness.
  • Insulation Maintenance: Regular checks and proper maintenance of insulation can prevent degradation over time.
By focusing on these aspects, homeowners can enhance their home's insulation efficiency, leading to more sustainable living and cost savings.

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Most popular questions from this chapter

Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A \(152 \mathrm{MB}\) memory chip dissipates \(5 \mathrm{~W}\) of heat to air at \(25^{\circ} \mathrm{C}\). If the temperature of this chip is to not exceed \(50^{\circ} \mathrm{C}\), the overall heat transfer coefficient- area product of the finned metal mount must be at least (a) \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (b) \(0.3 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (c) \(0.4 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (d) \(0.5 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (e) \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

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