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The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a window that is \(1 \mathrm{~m}\) by \(3 \mathrm{~m}\) on a side? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The heat loss through the window is 8400 W.

Step by step solution

01

Understanding the Problem

We are given a glass window with specific dimensions and temperature differences between its surfaces. We need to calculate the heat loss through this window using the formula for heat conduction.
02

Setting Up the Formula for Heat Transfer

The formula for heat transfer through conduction is \( Q = \frac{k imes A imes \Delta T}{d} \), where \( Q \) is the heat loss, \( k \) is the thermal conductivity, \( A \) is the area of the window, \( \Delta T \) is the temperature difference, and \( d \) is the thickness of the window.
03

Calculating the Area of the Window

The area \( A \) of the window is given by multiplying its dimensions: \( A = 1 \text{ m} \times 3 \text{ m} = 3 \text{ m}^2 \).
04

Calculating the Temperature Difference

The temperature difference \( \Delta T \) between the inner and outer surface is \( 15^{\circ}C - 5^{\circ}C = 10 \text{ K} \).
05

Calculating the Heat Loss

Substitute the values into the formula: \( Q = \frac{1.4 \mathrm{~W/m} \cdot \mathrm{K} \times 3 \mathrm{~m}^2 \times 10 \mathrm{~K}}{0.005 \mathrm{~m}} \). Calculate this to find \( Q \).
06

Performing the Calculation

By performing the calculation, we find \( Q = \frac{1.4 \times 3 \times 10}{0.005} = \frac{42}{0.005} = 8400 \mathrm{~W} \). The heat loss through the window is 8400 W.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. Imagine it like a "road" for heat energy to pass through. Glass, the focus of our exercise, has a thermal conductivity of 1.4 W/(m·K). This means for every square meter of glass, with a 1 Kelvin difference in temperature between its surfaces, 1.4 watts of heat will pass through each meter of thickness.
The concept itself is essential when studying heat transfer because it helps us understand how different materials are more or less efficient in conducting heat. Materials like metals have high thermal conductivity, making them good conductors of heat, while materials like wood have low thermal conductivity, acting as insulators. Knowing the thermal conductivity of a material helps in making informed decisions for applications like building insulation or making cookware.
In our exercise, the thermal conductivity of glass simplifies our calculation by allowing us to find the rate at which heat leaves the warm side of the window and travels to the colder side.
Conduction Formula
The conduction formula is crucial for calculating heat transfer through materials. To find the heat loss, we use the formula: \[ Q = \frac{k \times A \times \Delta T}{d} \]Here's what each term represents:
  • \(Q\) is the heat transfer, in watts (W)
  • \(k\) is the thermal conductivity of the material, in W/(m·K)
  • \(A\) is the area through which the heat is conducted, in square meters (m²)
  • \(\Delta T\) is the temperature difference across the material, in Kelvin (K)
  • \(d\) is the thickness of the material, in meters (m)
Understanding this formula enables us to apply it in real-world scenarios, such as evaluating how much energy is needed to maintain temperatures inside buildings or in engineering applications, where minimizing heat loss is necessary. It's like a recipe—each ingredient (or variable) must be measured correctly to get the expected result.
Temperature Difference
Temperature difference, denoted as \(\Delta T\), is a measure of how much hotter one surface of the material is compared to the other. It is a driving force for heat transfer; without a temperature difference, heat would not flow.

In our example, the inner and outer surfaces of the glass window show temperatures of 15°C and 5°C. Subtracting the two gives us \(\Delta T = 10 \text{ K}\). This 10 K difference indicates potential energy movement—from the hotter inside to the cooler outside of the window.

Conceptually, temperature difference not only affects the rate of heat transfer but also indicates the efficiency of insulation. A smaller \(\Delta T\) would imply less heat loss, which is crucial for energy conservation methods in construction and HVAC design.

Understanding temperature differences helps us design systems that either insulate spaces or maximize heat flow, depending on the desired outcome of the heat exchange process.

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Most popular questions from this chapter

A freezer compartment is covered with a \(2-\mathrm{mm}\)-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2\) \(\mathrm{W} / \mathrm{m}^{2}\) - \(\mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{k} / / \mathrm{kg}\).

A thin electrical heating element provides a uniform air flows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{-} \mathrm{C}\) and the convection hest transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). What heat flux \(q_{0}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of purt (a), what is the temperature \(\left(T_{a}\right)\) of the duct surface next to the heater? (c) With \(T_{1}=85^{\circ} \mathrm{C}\), compute and plot \(q_{n}^{\prime \prime}\) and \(T_{e}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.

The caise of a power transistor, which is of length \(L=10\) num and diameter \(D=12 \mathrm{~mm}\), is cooled by an air stream of temperature \(T_{\mathrm{m}}=25^{\circ} \mathrm{C}\). Lnder conditions for which the air maintains an average convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surfuce of the case, what is the maximum allowable power dissipation if the surface temperature is not to exceed \(85^{\circ} \mathrm{C}\) ?

In the thermal processing of semiconductor materials. annealing is accomplished by heating a silicon wafer according to a temiperature-titue recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follow5, the wafer is in an evacuated chamber whose walls are maintained at \(27 \mathrm{C}\) and wiahin which heating lamps maintain a radiant flux \(q\), at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermul conductivity of 30 W/m \(\cdot \mathrm{K}\), and an emiskivity that cquals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{i}=0.65\right)\). For \(q_{t}^{n}=3.0 \times 10^{5}\) \(\mathrm{W} / \mathrm{m}^{2}\), the temperalure on its lower surface is measured by a ridiation themometer and found to have a value of \(T_{\alpha l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this cundition being met?

A square silicon chip \((k=150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is of width \(w=5 \mathrm{~mm}\) on a side and of thickness \(t=1 \mathrm{~mm}\). The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 4 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

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