Problem 55 - College Physics

Question

A picture window has dimensions of \(1.40 \mathrm{~m} \times 2.50 \mathrm{~m}\) and is made of glass \(5.20 \mathrm{~mm}\) thick. On a winter day, the outside temperature is \(-20.0^{\circ} \mathrm{C},\) while the inside temperature is a comfortable \(19.56^{\circ} \mathrm{C}\). (a) At what rate is heat being lost through the window by conduction? (b) At what rate would heat be lost through the window if you covered it with a \(0.750-\mathrm{mm}\) -thick layer of paper (thermal conductivity \(0.0500 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})) ?\)

Step-by-Step Solution

Answer

(a) 21,330 Watts; (b) 6,440 Watts.

Step 1: Understand the Formula for Heat Conduction

The rate of heat loss (in Watts) through a material by conduction is determined using the formula: \[ Q = \frac{kA(T_{hot} - T_{cold})}{d} \]Where:- \( Q \) is the rate of heat loss through the material.- \( k \) is the thermal conductivity of the material.- \( A \) is the area of the surface through which heat is being conducted.- \( T_{hot} \) and \( T_{cold} \) are the temperatures on each side of the material.- \( d \) is the thickness of the material.

Step 2: Calculate Parameters for the Glass Window

First, find the surface area \( A \) of the window: \[ A = 1.40 \times 2.50 = 3.50 \text{ m}^2 \]Given thermal conductivity \( k_{glass} = 0.80 \text{ W/m·K} \) and thickness of the glass \( d = 5.20 \times 10^{-3} \text{ m} \).Temperatures are \( T_{hot} = 19.56 \text{ °C} \) and \( T_{cold} = -20.0 \text{ °C} \).

Step 3: Calculate Heat Loss Through the Glass

Substitute the values into the heat conduction formula:\[Q_{glass} = \frac{0.80 \times 3.50 \times (19.56 - (-20.0))}{5.20 \times 10^{-3}} \]Evaluating this gives:\[Q_{glass} = \frac{0.80 \times 3.50 \times 39.56}{5.20 \times 10^{-3}} = 21,330 \text{ Watts}\]

Step 4: Calculate Parameters for the Paper Layer

When a paper layer is added, the thickness \( d_{paper} = 0.750 \times 10^{-3} \text{ m} \) and the thermal conductivity \( k_{paper} = 0.0500 \text{ W/m·K} \).

Step 5: Calculate Heat Loss Through the Composite System

The total thermal resistance \( R \) of the composite system (glass + paper) is:\[R = \frac{d_{glass}}{k_{glass}} + \frac{d_{paper}}{k_{paper}} = \frac{5.20 \times 10^{-3}}{0.80} + \frac{0.750 \times 10^{-3}}{0.0500}\]Calculate each term:\[R = 0.0065 + 0.015 = 0.0215 \text{ W}^{-1} \text{m}^2 \text{K}\]So, the heat loss is:\[Q_{total} = \frac{A(T_{hot} - T_{cold})}{R} = \frac{3.50 \times 39.56}{0.0215} = 6,440 \text{ Watts}\]

Step 6: Conclusion

Based on our calculations: - The rate of heat loss through just the glass is approximately 21,330 Watts. - When a paper layer is added, the rate of heat loss decreases to approximately 6,440 Watts.

Key Concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. Imagine it as the highway for heat transfer. Different materials possess different thermal conductivities. For example, metals tend to have high thermal conductivity, making them great conductors of heat, while materials like wood or paper are poor conductors and have low thermal conductivity. This property is crucial because it tells us how fast or slow heat can travel through a given material when there is a temperature difference. In our exercise, the glass window and the paper covering have different thermal conductivities.

• The glass has a thermal conductivity of 0.80 W/m·K.
• The paper has a thermal conductivity of 0.0500 W/m·K.

Low thermal conductivity is beneficial when you want to prevent heat flow, such as in insulating materials. Knowing the thermal conductivity helps in calculating how much heat will be lost or gained through different materials, which is essential in designing energy-efficient buildings.

Temperature Difference

The temperature difference between two sides of a material is a driving force for heat transfer. In simple terms, heat always flows from a warmer area to a cooler area, trying to reach equilibrium. The greater the temperature difference, the faster the heat will flow.

• In the exercise, we see the inside temperature is 19.56°C, and the outside is -20.0°C.
• This creates a temperature difference of 39.56°C.

This temperature difference is used in the heat conduction formula to determine the rate at which heat is lost through the window. Maintaining a larger temperature difference generally results in more energy loss, reinforcing the need for effective insulation in homes and buildings, especially during extreme weather conditions.

Composite Systems

Composite systems consist of multiple layers of materials through which heat must pass. These systems take advantage of the different thermal properties of materials to control heat flow. When you combine two different materials, like glass and paper in our window problem, each layer adds resistance to heat flow, which is called thermal resistance. The combined effect reduces the overall heat transfer through the system.

• For glass, the thermal resistance is given by its thickness divided by its thermal conductivity.
• For the paper layer, the process is similar but with its respective properties.

In these composite systems, the total thermal resistance is the sum of the resistances of each layer. By reducing the rate at which heat is lost, composite systems help conserve energy, making buildings more comfortable and efficient.

Heat Loss Calculation

Calculating heat loss through materials is critical for understanding and improving energy efficiency in structures. By applying the heat conduction formula, you can determine how much heat is being lost and how to mitigate it. The formula takes into account the area through which heat is lost, the temperature difference, and the thermal resistance of the materials.

• Initially, the window loses 21,330 Watts through the glass alone.
• Adding a paper layer reduces the heat loss to 6,440 Watts.

This significant reduction shows the effectiveness of additional insulation. By knowing the rate of heat loss, you can make informed decisions about the materials and design strategies you use in building and retrofitting efforts to improve energy efficiency, comfort, and savings on energy bills.