Problem 107 - University Physics with Modern Physics

Question

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 -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

Answer

(a) 93.8 W (b) Heat flow increases by a factor of 1.35.

Step 1: Calculate Total Thermal Resistance for Solid Wood Door

First, calculate the total thickness including the air films for the wood part of the door. The actual wood thickness is 5.0 cm or 0.050 m. Adding the equivalent thickness for the air films, which is 1.8 cm or 0.018 m, gives total thickness:\[ d_{total, wood} = 0.050 + 0.018 = 0.068 \, \text{m} \]The thermal resistance per unit area for this thickness is given by:\[ R_{wood} = \frac{d_{total, wood}}{k_{wood}} = \frac{0.068}{0.120} = 0.567 \, \text{m}^2 \cdot \text{K} / \text{W} \]

Step 2: Calculate Rate of Heat Flow Through Wood Door

Next, calculate the rate of heat flow, which is defined as:\[ Q = \frac{A \cdot \Delta T}{R} \]where \( A \) is the area of the door, \( \Delta T \) is the temperature difference, and \( R \) is the thermal resistance.The area \( A \) is:\[ A = 2.00 \times 0.95 = 1.90 \, \text{m}^2 \]The temperature difference \( \Delta T \) is:\[ \Delta T = 20 - (-8) = 28 \, \text{K} \]Substituting these into the formula gives the heat transfer rate:\[ Q_{wood} = \frac{1.90 \times 28}{0.567} \approx 93.8 \, \text{W} \]

Step 3: Calculate Thermal Resistance with Added Window

Calculate the thermal resistance for the section with the glass. The thickness of the glass is 0.450 cm or 0.0045 m. With the air films, the total thickness is:\[ d_{total, glass} = 0.0045 + 0.12 = 0.1245 \, \text{m} \]The thermal resistance of the glass section is:\[ R_{glass} = \frac{d_{total, glass}}{k_{glass}} = \frac{0.1245}{0.80} = 0.1556 \, \text{m}^2 \cdot \text{K} / \text{W} \]

Step 4: Calculate Rate of Heat Flow Through Window Section

Determine the heat flow rate through the window section, where the window area \( A \) is:\[ A_{window} = 0.5 \times 0.5 = 0.25 \, \text{m}^2 \]The heat transfer rate for the window section is:\[ Q_{glass} = \frac{0.25 \times 28}{0.1556} \approx 45.02 \, \text{W} \]

Step 5: Combine and Reevaluate Heat Flow for Door with Window

Calculate the heat flow rate for the remaining wood section, which has area \( A_{wood_{new}} = 1.90 - 0.25 = 1.65 \, \text{m}^2 \). The new rate of heat flow through this area is:\[ Q_{wood_{new}} = \frac{1.65 \times 28}{0.567} \approx 81.50 \, \text{W} \]Therefore, the total heat flow with a window is:\[ Q_{total} = Q_{wood_{new}} + Q_{glass} = 81.50 + 45.02 = 126.52 \, \text{W} \]

Step 6: Determine the Factor Increase in Heat Flow

The factor by which the heat flow increases is:\[ \text{Factor Increase} = \frac{Q_{total}}{Q_{wood}} = \frac{126.52}{93.8} \approx 1.35 \]

Key Concepts

Heat Flow

Heat flow is the movement of thermal energy from one object or area to another. It occurs due to a temperature difference between two regions. The flow of heat can happen through conduction, convection, or radiation.
In the exercise, heat flow through the door was calculated using the formula: \[ Q = \frac{A \cdot \Delta T}{R} \]Here, - **Q** is the rate of heat flow, measured in watts (W). - **A** is the area over which the heat is flowing, measured in square meters (m²). - **ΔT** is the temperature difference across the material, measured in Kelvin (K). - **R** is the thermal resistance of the material, measured in m²·K/W.
Good understanding of these variables is crucial when solving such problems. Heat flow is higher when the temperature difference is greater or when thermal resistance decreases.

Thermal Resistance

Thermal resistance refers to a material's ability to resist the flow of heat. It measures how well a material insulates against heat transfer.
This concept is similar to electrical resistance in that a high thermal resistance means less heat flow. The exercise utilized thermal resistance as follows: \[ R = \frac{d_{total}}{k} \]Here, - **R** is the thermal resistance. - **d_total** is the total thickness of the material, including any equivalent thickness for air films, measured in meters (m). - **k** is the thermal conductivity of the material, measured in watts per meter per Kelvin (W/m·K).
The inclusion of air films increases the total thermal resistance, providing more insulation. Understanding thermal resistance helps in designing buildings or objects that require insulation against heat loss or gain.

Temperature Difference

Temperature difference (ΔT) is the driving factor for heat flow from warmer to cooler areas. It is the difference in temperature between the inside and outside environments in the context of the problem.
The formula used was \[ \Delta T = T_{inside} - T_{outside} \] where - **T_inside** is the temperature inside the house, - **T_outside** is the outside temperature.
In this scenario, the temperature difference was \[ \Delta T = 20^{\circ}C - (-8^{\circ}C) = 28^{\circ}C = 28 \, K \]. Greater temperature differences can lead to higher rates of heat flow. This concept is crucial in understanding how insulation materials work to maintain desired indoor temperatures despite varying external conditions.