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

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

Step by step solution

01

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.
02

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: $$A = 6\,\text{m} \cdot 3\,\text{m} = 18\,\text{m}^2$$.
03

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: $$\Delta T = T_{i} - T_{o} = 14^{\circ} \mathrm{C} - 2^{\circ} \mathrm{C} = 12^{\circ} \mathrm{C}$$.
04

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 \(576\,\mathrm{W}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A \(152 \mathrm{MB}\) memory chip dissipates \(5 \mathrm{~W}\) of heat to air at \(25^{\circ} \mathrm{C}\). If the temperature of this chip is to not exceed \(50^{\circ} \mathrm{C}\), the overall heat transfer coefficient- area product of the finned metal mount must be at least (a) \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (b) \(0.3 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (c) \(0.4 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (d) \(0.5 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (e) \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)

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