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Effect of a Window in a Door. A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{m} \times 0.95 \mathrm{m} \times 5.0 \mathrm{cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8-\mathrm{cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C}\) , and the outside air temperature is \(-8.0^{\circ} \mathrm{C} .\) (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 \(\mathrm{m}\) on a side is inserted in the door? The glass is 0.450 \(\mathrm{cm}\) thick, and the glass has a thermal conductivity of 0.80 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 \(\mathrm{cm}\) of glass.

Step by step solution

01

Calculate Door Area and Resistance

First, calculate the area of the door: \[ A = 2.00 \, \mathrm{m} \times 0.95 \, \mathrm{m} = 1.90 \, \mathrm{m}^2 \]Next, calculate the thickness of the door including the air film equivalent thickness for resistance: \[ d_{total} = 5.0 \, \mathrm{cm} + 1.8 \, \mathrm{cm} = 6.8 \, \mathrm{cm} = 0.068 \, \mathrm{m} \]With a thermal conductivity \(k = 0.120 \, \mathrm{W/m \cdot K}\), calculate the resistance of the door using: \[ R_{door} = \frac{d_{total}}{k \times A} = \frac{0.068}{0.120 \times 1.90} \approx 0.297 \mathrm{K/W} \]

02

Calculate Heat Flow Through Door

Use the temperature difference \(\Delta T = 20.0 - (-8.0) = 28.0 \, \mathrm{K}\) and the resistance to calculate the heat flow: \[ Q_{door} = \frac{\Delta T}{R_{door}} = \frac{28.0}{0.297} \approx 94.28 \, \mathrm{W} \]

03

Calculate Area and Resistance of Glass Window

Calculate the side length area of the glass window: \[ A_{glass} = (0.500 \, \mathrm{m})^2 = 0.250 \, \mathrm{m}^2 \]Total thickness of the window plus air films equivalent: \[ d_{total\,glass} = 0.450 \, \mathrm{cm} + 12.0 \, \mathrm{cm} = 12.45 \, \mathrm{cm} = 0.1245 \, \mathrm{m} \]With thermal conductivity \(k_{glass} = 0.80 \, \mathrm{W/m \cdot K}\), the resistance is: \[ R_{glass} = \frac{d_{total\,glass}}{k_{glass} \times A_{glass}} = \frac{0.1245}{0.80 \times 0.250} \approx 0.6225 \, \mathrm{K/W} \]

04

Calculate Heat Flow Through Glass Window

Again use \(\Delta T = 28.0 \, \mathrm{K}\) to calculate heat flow through the glass: \[ Q_{glass} = \frac{\Delta T}{R_{glass}} = \frac{28.0}{0.6225} \approx 44.97 \, \mathrm{W} \]

05

Calculate New Heat Flow Through Modified Door

Subtract the window area from the door to find the remaining area: \[ A_{remaining} = 1.90 - 0.250 = 1.65 \, \mathrm{m}^2 \]Find the resistance of the modified door area:\[ R_{remaining} = \frac{0.068}{0.120 \times 1.65} \approx 0.343 \mathrm{K/W} \]

06

Calculate Total Heat Flow Through Modified Door

Calculate heat flow through the remaining door area:\[ Q_{remaining} = \frac{\Delta T}{R_{remaining}} = \frac{28.0}{0.343} \approx 81.63 \, \mathrm{W} \]The total heat flow through the door with a window is:\[ Q_{total} = Q_{remaining} + Q_{glass} = 81.63 + 44.97 = 126.60 \, \mathrm{W} \]

07

Find Heat Flow Increase Factor

The initial heat flow through the solid wood door was \(94.28 \, \mathrm{W}\) and the heat flow with window is \(126.60 \, \mathrm{W}\). The increase factor is:\[ \text{Increase Factor} = \frac{Q_{total}}{Q_{door}} = \frac{126.60}{94.28} \approx 1.34 \]

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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 physics that describes the movement of thermal energy from one object to another. This typically occurs when two objects at different temperatures come into contact, and thermal energy naturally flows from the warmer object to the cooler one. In our exercise, the door provides an excellent example, where heat flows from the warm interior of a house to the cooler exterior.

Key points about heat transfer include:

• It is driven by temperature differences, also known as thermal gradients.
• Heat can be transferred via conduction, convection, or radiation. In this exercise, we primarily deal with conduction through the solid door and glass.
• The rate of heat transfer is influenced by the material's thermal properties, such as thermal conductivity (\(k\)).

Equation for heat transfer through conduction: \[ Q = \frac{\Delta T}{R} \] Where \(Q\) is heat transfer rate, \(\Delta T\) is the temperature difference, and \(R\) is thermal resistance.

This equation exemplifies how materials with lower thermal resistance or higher thermal conductivity allow more heat flow.

Thermal Resistance

Thermal resistance is a measure of a material's ability to resist the flow of heat. It plays a crucial role in understanding and calculating the rate of heat transfer, as seen in the original exercise. The door's thermal resistance directly affects how much heat flows from the inside to the outside.

Key components about thermal resistance:

• It is inversely proportional to thermal conductivity. High thermal resistance indicates low thermal conductivity.
• The thicker the material, the higher its thermal resistance. That's why the thickness of the door and the equivalent thickness of air films are factored in the calculations.
• Thermal resistance acts as an insulating barrier, reducing the amount of heat transferring through a material.

The formula for calculating thermal resistance (\(R\)) is:\[ R = \frac{d}{k \cdot A} \] Where \(d\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the surface area.

In our exercise, both the door and the air films possess thermal resistances, which were calculated to understand their ability to insulate and prevent heat loss.

Energy Efficiency

Energy efficiency relates to the ability of a system to utilize energy without unnecessary waste. It is critical in contexts like home insulation, where reducing heat flow enhances energy efficiency, thereby saving on energy costs and reducing environmental impact.

Important considerations in energy efficiency include:

• Materials with high thermal resistance improve energy efficiency by limiting heat flow.
• Adding features such as windows may affect the energy efficiency of a system, as illustrated by the increased heat flow with the window in the door exercise.
• Efficiency can be assessed by comparing the initial and modified states, as seen with the calculation of the increase factor in heat flow.

The formula for efficiency assessment in this context can consider the increase factor calculated by:\[ \text{Increase Factor} = \frac{Q_{total}}{Q_{door}} \] This formula highlights the factor by which the heat flow increases due to changes like the addition of a window.

Optimizing the balance between insulating materials (high thermal resistance) and necessary design elements like windows is key to achieving energy efficiency.