Problem 83 - College Physics

Question

\(\bullet\) One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 \(\mathrm{W}\) is required to keep the interior surface of the box 65.0 \(\mathrm{C}^{\circ}\) (about 120 \(\mathrm{F}^{\circ} )\) above the temperature of the outer surface. The total area of the box is \(2.18 \mathrm{m}^{2},\) and the wall thickness is 3.90 \(\mathrm{cm} .\) Find the thermal conductivity of the material in SI units.

Step-by-Step Solution

Answer

Thermal conductivity is approximately 0.0495 W/m°C.

Step 1: Understand the Given Data

We are given the power input required, which is \(180\, \text{W}\), the temperature difference between the inside and outside of the box, \( \Delta T = 65.0\, \text{°C}\), the surface area of the box, \( A = 2.18\, \text{m}^2\), and the wall thickness, \( d = 3.90\, \text{cm}\) which needs to be converted into meters (\(0.039\, \text{m}\)). We need to find the thermal conductivity \( k \).

Step 2: Identify the Formula

The formula for thermal conductivity \( k \) can be derived from the formula for heat transfer across a material, \( P = \frac{kA \Delta T}{d} \). Rearranging this formula will allow us to solve for \( k \).

Step 3: Rearrange the Formula

Rearrange the formula to find \( k \):\[ k = \frac{P \times d}{A \times \Delta T} \] where \( P \) is the power in watts, \( d \) is the thickness in meters, \( A \) is the area in square meters, and \( \Delta T \) is the temperature difference in Celsius.

Step 4: Substitute the Values

Substitute the given values into the rearranged formula:\[ k = \frac{180 \times 0.039}{2.18 \times 65}\]

Step 5: Calculate the Thermal Conductivity

Perform the calculation:\[ k = \frac{7.02}{141.7} \approx 0.0495\, \text{W/m°C} \]. Therefore, the thermal conductivity of the material is approximately \(0.0495\, \text{W/m°C} \).

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept that describes the movement of thermal energy from one object to another. It occurs through conduction, convection, or radiation. In this particular exercise, we focus on conduction, which happens when heat moves through a material. When you have a temperature difference across the material, heat will flow from the hotter side to the cooler one.

Heat transfer by conduction can be quantified using the formula:

• \[ P = \frac{kA \Delta T}{d} \]

Here,

• \( P \) is the power in watts,
• \( k \) is the thermal conductivity,
• \( A \) is the area through which heat is being transferred,
• \( \Delta T \) is the temperature difference,
• \( d \) is the thickness of the material.

This formula helps us understand how effectively a material can transfer heat.

Insulating Materials

Insulating materials are crucial for controlling heat flow. They work by reducing the rate of heat transfer due to their low thermal conductivity. This property allows them to maintain stable temperatures in different environments.

When designing buildings or devices, choosing the right insulating material can significantly affect energy efficiency. Common examples include fiberglass, foam, and various types of plastic.

In the context of the exercise, the insulating box’s walls keep the inside hot while minimizing heat loss to the outside. Understanding the material's thermal conductivity allows us to predict how well the insulation will perform under certain conditions.

SI Units

The International System of Units (SI) provides a standard way to express measurements, ensuring consistency and clarity in scientific communication. In heat transfer calculations, we use several SI units:

• Power is measured in watts (W), representing energy flow per unit time.
• Area is measured in square meters (m²), indicating the extent of the surface through which heat is transferred.
• Temperature differences are expressed in degrees Celsius (°C).
• Thickness is given in meters (m). This requires converting measurements if they are provided in other units, such as centimeters.

Understanding these units helps us perform accurate calculations and interpret the results correctly.

Temperature Difference

In heat transfer, the temperature difference between two sides of an insulating material drives the flow of heat. The greater the temperature difference (\( \Delta T \)), the more intense the heat transfer.

In the exercise, a 65°C difference is maintained between the inside and outside of the box. This significant temperature gradient indicates that a lot of heat needs to be managed to maintain the interior temperature.

The temperature difference is a vital factor in calculating thermal conductivity because it directly affects how much heat will flow through the material. Keeping track of this difference helps us understand and optimize thermal performance in various applications.