Problem 99 - University Physics with Modern Physics

Question

A carpenter builds a solid wood door with dimensions 2.00 m \(\times\) 0.95 m \(\times\) 5.0 cm. Its thermal conductivity is k = 0.120 W/m \(\cdot\) 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\(^\circ\)C, and the outside air temperature is -8.0\(^\circ\)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 m on a side is inserted in the door? The glass is 0.450 cm thick, and the glass has a thermal conductivity of 0.80 W/m \(\cdot\) 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 cm of glass.

Step-by-Step Solution

Answer

Heat flow is 93.91 W; increased by a factor of 1.333 with window.

Step 1: Calculate the thermal resistance of the wooden door

The thermal resistance of an object is given by the formula \( R = \frac{L}{kA} \), where \( L \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the area. First, calculate the area of the door:\[ A = 2.00 \times 0.95 = 1.90 \text{ m}^2. \]Next, convert the thickness of the door to meters:\[ L = 5.0 \text{ cm} = 0.05 \text{ m}. \]Then, calculate the thermal resistance of the wooden part of the door using the given thermal conductivity, \( k = 0.120 \text{ W/m} \cdot \text{K} \):\[ R_{wood} = \frac{0.05}{0.120 \times 1.90} = 0.2193 \text{ K/W}. \]

Step 2: Calculate the thermal resistance of the air film

The air films' thermal resistance is equivalent to an additional 1.8 cm of wood:\[ L_{air} = 1.8 \text{ cm} = 0.018 \text{ m}. \]Calculate the thermal resistance of the equivalent wood:\[ R_{air} = \frac{0.018}{0.120 \times 1.90} = 0.0789 \text{ K/W}. \]

Step 3: Calculate the total thermal resistance of the door

Add the thermal resistances of the wooden door and the air films:\[ R_{total} = R_{wood} + R_{air} = 0.2193 + 0.0789 = 0.2982 \text{ K/W}. \]

Step 4: Calculate the rate of heat flow through the door

The rate of heat flow can be calculated using the formula\[ Q = \frac{\Delta T}{R_{total}}, \]where \( \Delta T \) is the temperature difference across the door. Here, \( \Delta T = 20.0 - (-8.0) = 28.0 \text{ K}. \) Substituting the values:\[ Q = \frac{28.0}{0.2982} = 93.91 \text{ W}. \]Thus, the rate of heat flow through the door is approximately 93.91 W.

Step 5: Calculate the thermal resistance of the glass and air

First, convert the thickness of the glass to meters:\[ L_{glass} = 0.450 \text{ cm} = 0.0045 \text{ m}. \]Calculate the area of the glass window:\[ A_{glass} = 0.500 \times 0.500 = 0.25 \text{ m}^2. \]Calculate the thermal resistance of the glass:\[ R_{glass} = \frac{0.0045}{0.80 \times 0.25} = 0.0225 \text{ K/W}. \]The air film around the glass has the equivalent thermal resistance of 12.0 cm of glass:\[ L_{air ext{_glass}} = 12.0 \text{ cm} = 0.12 \text{ m}. \]Calculate the thermal resistance of this equivalent glass thickness:\[ R_{air ext{_glass}} = \frac{0.12}{0.80 \times 0.25} = 0.60 \text{ K/W}. \]The total resistance is:\[ R_{total ext{_glass}} = R_{glass} + R_{air ext{_glass}} = 0.0225 + 0.60 = 0.6225 \text{ K/W}. \]

Step 6: Calculate the total heat flow with the glass window inserted

First, calculate the new total thermal resistance of the door with the window:- Remove the area of the glass from the wood: \[ A_{wood ext{_new}} = 1.90 - 0.25 = 1.65 \text{ m}^2. \]- Calculate the resistance of the new wooden part: \[ R_{wood ext{_new}} = \frac{0.05}{0.120 \times 1.65} = 0.2525 \text{ K/W}. \]- Calculate air film resistance for new wood area (also changes due to area change): \[ R_{air ext{_new}} = \frac{0.018}{0.120 \times 1.65} = 0.0913 \text{ K/W}. \]- Total new wood and air resistance: \[ R_{total ext{_wood ext{_new}}} = R_{wood ext{_new}} + R_{air ext{_new}} = 0.2525 + 0.0913 = 0.3438 \text{ K/W}. \]- Combine with glass: \[ R_{total ext{_combined}} = \frac{1}{\left(\frac{1}{0.3438} + \frac{1}{0.6225}\right)} = 0.2240 \text{ K/W}. \]Finally, calculate the new rate of heat flow:\[ Q ext{_new} = \frac{28.0}{0.2240} = 125.0 \text{ W}. \]

Step 7: Calculate the increase factor for the heat flow

To find the factor by which the heat flow is increased, take the ratio of the new heat flow to the original heat flow:\[ \text{Increase Factor} = \frac{Q ext{_new}}{Q} = \frac{125.0}{93.91} \approx 1.333. \]

Step 8: Conclusion: Summary of Results

The rate of heat flow through the original door is approximately 93.91 W, and with the window inserted, the heat flow increases by a factor of approximately 1.333.

Key Concepts

Thermal Conductivity

Thermal conductivity is a measure of how well a material can conduct heat. It is an intrinsic property that quantifies the ease with which thermal energy passes through a material. In mathematical terms, thermal conductivity, denoted as \( k \), is expressed in watts per meter per kelvin (W/m·K). A higher value of thermal conductivity means a material can transfer heat more efficiently.

For instance, in the exercise, the wooden door has a thermal conductivity of 0.120 W/m·K, which is relatively low compared to glass, which has a higher thermal conductivity of 0.80 W/m·K.- **Wood** has lower thermal conductivity, making it a better insulator, as it does not allow heat to pass through easily.- **Glass**, on the other hand, conducts heat more readily, making it less effective as an insulator in this context.

Understanding thermal conductivity is crucial for calculating how much heat flows through materials like doors and windows.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or material to another. This movement occurs due to a temperature difference, and there are three primary modes: conduction, convection, and radiation. The exercise focuses on conduction, which is the transfer of heat through a solid material.In this exercise, heat is transferred through the wooden door by conduction. The rate of heat flow \( Q \) through the door can be calculated using the equation:\[Q = \frac{\Delta T}{R_{total}},\]where \( \Delta T \) is the temperature gradient across the door and \( R_{total} \) is the total thermal resistance.

- **Conductive Heat Transfer**: The thicker the material or the lower its thermal conductivity, the more resistance it offers to heat flow.

Proper understanding of heat transfer is essential for designing energy-efficient buildings.

Temperature Gradient

The temperature gradient is the difference in temperature across a material. It is represented as \( \Delta T \) in the context of heat transfer. A larger temperature gradient means a greater potential for heat to flow from one side to the other.

In the wooden door scenario, the inside temperature is 20.0°C, while the outside is -8.0°C, creating a temperature difference of 28.0 K. This difference drives the heat flow through the door.

• **Greater Difference**: A higher temperature difference results in an increased rate of heat transfer.
• **Direct Proportionality**: The rate of heat flow is directly proportional to the temperature gradient, influencing how quickly thermal equilibrium is reached.

Understanding temperature gradients helps in designing materials and structures that either maximize or minimize heat transfer, based on the need for insulation or conduction.