Problem 50 - College Physics

Question

A slab of a thermal insulator with a cross-sectional area of \(100 \mathrm{~cm}^{2}\) is \(3.00 \mathrm{~cm}\) thick. Its thermal conductivity is \(0.075 \mathrm{~W} /(\mathrm{m} \cdot \mathrm{K})\). If the temperature difference between opposite faces is \(80 \mathrm{C}^{\circ},\) how much heat flows through the slab in 1 day?

Step-by-Step Solution

Answer

The heat flow through the slab in 1 day is 17280 Joules.

Step 1: Understand the Problem

We need to determine the amount of heat that flows through a slab over the course of one day. Given the dimensions, thermal conductivity, and temperature difference, we'll use the concept of thermal conduction to calculate the heat flow.

Step 2: Recall the Heat Flow Formula

The formula for heat transfer through a material by conduction is given by \[Q = \frac{k imes A imes ext{Δ}T imes t}{d}\]where \(Q\) is the heat flow in joules, \(k\) is the thermal conductivity, \(A\) is the cross-sectional area, \(\text{Δ}T\) is the temperature difference, \(t\) is the time in seconds, and \(d\) is the thickness of the material.

Step 3: Convert Units

First, we need to ensure that all units are compatible. The cross-sectional area \(A = 100\, \text{cm}^2\) should be converted to \(\text{m}^2\). Similarly, the thickness \(d = 3.00\, \text{cm}\) should be converted to meters:\[A = 100\, \text{cm}^2 = 0.01\, \text{m}^2\]\[d = 3.00\, \text{cm} = 0.03\, \text{m}\]Time \(t\) needs to be in seconds. Since there are 86400 seconds in a day (24 hours x 60 minutes x 60 seconds), we have \(t = 86400\, ext{s}\).

Step 4: Calculate the Heat Flow

Now, substitute all the known values into the formula:\[Q = \frac{0.075 \times 0.01 \times 80 \times 86400}{0.03}\]Simplify and calculate each portion step by step:1. Calculate the numerator part: \[0.075 \times 0.01 \times 80 \times 86400 = 518.4\]2. Divide by the thickness: \[Q = \frac{518.4}{0.03} = 17280\, ext{J}\]The total heat flow through the slab in one day is \(17280\, ext{Joules}\).

Key Concepts

Thermal Insulator

A thermal insulator is a material that reduces the rate of heat transfer. This is achieved by limiting the movement of heat between its molecules. Materials with high thermal resistivity are typically used as insulators.
Examples include foam, fiberglass, and specific types of fabrics. Insulators are used to maintain temperature consistency, either keeping heat in an area or preventing it from entering.

• Commonly used in buildings to improve energy efficiency.
• Help protect sensitive electronic components from overheating.

In the provided exercise, the slab acts as an insulator, impeding the flow of heat from one side to the other. The high resistance to heat transfer means it takes longer for heat to pass through, essential for minimizing energy loss in various applications.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one object or material to another. This can occur through several mechanisms.
In the context of the problem, heat is transferred through conduction. Conduction occurs when heat moves through a material without the material itself moving. It is driven by temperature differences across the material.
This type of heat transfer is direct and happens on the molecular level. Molecules in the warmer part of the object gain kinetic energy and transfer it to neighboring molecules, spreading the energy along the object.

• Conduction is common in solids where molecules are tightly packed.
• Metals are excellent conductors, while materials like wood or glass are poor conductors.

The heat transfer in the exercise reflects this principle, as the slab's material influences how quickly and efficiently heat is conducted through it.

Thermal Conductivity

Thermal conductivity is a crucial property that describes a material's ability to conduct heat. It is denoted by the symbol \(k\) and is measured in watts per meter per Kelvin (W/m·K).
Materials with high thermal conductivity, such as metals, are very effective at conducting heat. Conversely, materials with low thermal conductivity, like our insulator slab, resist the flow of heat.

• Understanding a material's thermal conductivity is important for selecting appropriate materials for thermal management.
• It varies based on factors such as temperature and material structure.

In the problem, the thermal conductivity of the slab, \(0.075 \, \text{W/m} \cdot \text{K}\), is relatively low, indicating it’s a good insulator, which influences how much heat flows through the slab over time.

Temperature Difference

Temperature difference, represented as \(\Delta T\), is the driving force in thermal conduction. It represents the difference in temperature between two points.
In the exercise, there is an 80°C difference between the two faces of the slab. This differential is vital because it determines how vigorously heat will flow through the slab.

• The greater the temperature difference, the higher the rate of heat flow.
• In practical terms, minimizing the temperature difference can reduce energy transfer.

Understanding temperature difference in this context allows us to predict how changes in temperature across an insulator will affect the efficiency of heat transfer.