Question

Consider a 3-m-high, 6-m-wide, and \(0.3\)-m-thick brick wall whose thermal conductivity is \(k=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be \(14^{\circ} \mathrm{C}\) and \(2^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss through the wall on that day.

Short answer

Answer: The rate of heat loss through the brick wall on the given day is 576 W.

Step-by-step solution

Understand the details of the exercise

In this exercise, our goal is to determine the rate of heat loss or heat transfer through the brick wall. We are given the wall's dimensions, thermal conductivity value, and the temperatures of the inner and the outer surfaces of the wall. We know that the heat transfer rate can be calculated using the formula: $$\dot{Q} = k \cdot A \cdot \frac{\Delta T}{d}$$, where \(\dot{Q}\) represents the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the area of the wall, \(\Delta T\) is the temperature difference between the inner and outer surfaces, and \(d\) is the thickness of the wall.

Calculate the wall's area

Since we know the dimensions of the wall (width and height), we can easily find the area of the surface through which the heat is transferred. The formula to calculate the area is: $$A = w \cdot h$$, where \(w\) represents the width and \(h\) represents the height of the wall. We are given \(w=6\,\text{m}\) and \(h=3\,\text{m}\). Now we can calculate the area: $$A = 6\,\text{m} \cdot 3\,\text{m} = 18\,\text{m}^2$$.

Calculate the temperature difference between the inner and outer surfaces

We are given the inner surface temperature as \(T_{i} = 14^{\circ} \mathrm{C}\), and the outer surface temperature as \(T_{o} = 2^{\circ} \mathrm{C}\). To calculate the temperature difference \(\Delta T\), we simply subtract the two temperatures: $$\Delta T = T_{i} - T_{o} = 14^{\circ} \mathrm{C} - 2^{\circ} \mathrm{C} = 12^{\circ} \mathrm{C}$$.

Calculate the rate of heat loss through the wall

Now that we have all the required values, we can plug them into the heat transfer rate formula: $$\dot{Q} = k \cdot A \cdot \frac{\Delta T}{d}$$ We are given the thermal conductivity value, \(k = 0.8\,\mathrm{W}/\mathrm{m}\cdot\mathrm{K}\), and the wall thickness, \(d = 0.3\,\text{m}\). The calculation is now as follows: $$\dot{Q} = 0.8\,\mathrm{W}/\mathrm{m}\cdot\mathrm{K} \cdot 18\,\text{m}^2 \cdot \frac{12^{\circ} \mathrm{C}}{0.3\,\text{m}}$$ $$\dot{Q} = 576\,\mathrm{W}$$ The rate of heat loss through the wall on the given day is \(576\,\mathrm{W}\).

Key concepts

Thermal Conductivity

When we talk about thermal conductivity, we refer to the property of a material to conduct heat. In simple terms, it is a measure of how well heat can move through a material. Materials with high thermal conductivity, like metals, allow heat to pass through quickly, whereas those with low thermal conductivity, such as wood or brick, are much slower at transferring heat.

Thermal conductivity is represented by the symbol k and its units are Watts per meter-Kelvin (W/m·K). When solving problems related to heat transfer, it's essential to know the thermal conductivity of the material involved to calculate how much heat is lost or gained over time. In our exercise, we looked at a brick wall with a thermal conductivity of 0.8 W/m·K, indicating how the thermal energy will conduct through the wall material.

Temperature Difference

The temperature difference, often denoted as ΔT, is a driving force in heat transfer. It is simply the difference in temperature between two areas, such as the inner and outer surfaces of a wall. The greater the temperature difference, the greater the heat transfer rate.

In the context of our problem, we examined the temperature on either side of a brick wall. One side was at 14°C and the other at 2°C, resulting in a temperature difference of 12°C. This difference is crucial for calculating the rate of heat transfer because it represents the potential for heat to flow from the warmer to the cooler side.

Conduction Heat Transfer

The concept of conduction heat transfer relates to the way heat moves within and through materials. It occurs at a molecular level when faster-moving (hotter) molecules transfer energy to slower-moving (cooler) molecules. The rate at which this energy is transferred depends on the material's ability to conduct heat, which, as we know, is quantified by its thermal conductivity.

Conduction is significant in our wall scenario because it explains how heat from the warmer interior of a structure is eventually lost to the colder exterior.

Fourier's Law of Conduction

Understanding Fourier's law of conduction is key to solving heat transfer problems like the one presented. This law mathematically describes how heat conduction is related to the thermal conductivity k, the area A through which heat is passing, the temperature difference ΔT, and the material's thickness d. It's written as:
\[\dot{Q} = k \cdot A \cdot \frac{\Delta T}{d}\]
By using Fourier's law, we found out the rate of heat loss through our hypothetical brick wall. This law underpins the calculations for any scenario where heat is transferred by conduction and is a fundamental principle in understanding how materials and structures respond to temperature differences.