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(a) What is the rate of thermal energy transfer in watts per square meter through a glass window \(3.0 \mathrm{~mm}\) thick if the outside temperature is \(-20^{\circ} \mathrm{F}\) and the inside temperature is \(+72^{\circ} \mathrm{F} ?\) (b) A storm window having the same thickness of glass is installed parallel to the first window, with an air gap of \(7.5 \mathrm{~cm}\) between the two windows. What now is the rate of energy loss if conduction is the only important energytransfer mechanism?

Short Answer

Expert verified
a) 14280 W/m², b) 16.28 W/m².

Step by step solution

01

Convert temperatures to Celsius

First, convert the outside and inside temperatures from Fahrenheit to Celsius using the formula: \( T_{C} = \frac{5}{9} (T_{F} - 32) \).For outside temperature \( T_{\text{outside}} = \frac{5}{9} (-20 - 32) = \frac{5}{9} (-52) = -29^{\circ}C \).For inside temperature \( T_{\text{inside}} = \frac{5}{9} (72 - 32) = \frac{5}{9} (40) = 22^{\circ}C \).
02

Calculate the temperature difference

Find the temperature difference \( \Delta T \) between inside and outside: \( \Delta T = T_{\text{inside}} - T_{\text{outside}} = 22^{\circ}C - (-29^{\circ}C) = 51^{\circ}C \).
03

Determine thermal conductivity of glass

Thermal conductivity \( k \) for glass is given as \( 0.84 \text{ W/(m·K)} \).
04

Calculate the heat transfer rate through one glass pane

Using Fourier's Law for heat conduction, \( Q = \frac{k \cdot A \cdot \Delta T}{d} \), where:\( Q \) is the rate of thermal energy transfer,\( A \) is the area (we can use a reference area per square meter, A = 1 m²),\( \Delta T \) is the temperature difference,\( d \) is the thickness in meters, so \( d = 3.0 \text{ mm} = 0.003 \text{ m} \).Plug in the values: \( Q = \frac{0.84 \cdot 1 \cdot 51}{0.003} = 14280 \text{ W/m}^2 \).
05

Calculate the heat transfer rate with storm window

For the combined glass and air gap, calculate separately and then combine: 1. Air gap conductivity \( k_{\text{air}} = 0.024 \text{ W/(m·K)} \).Conversion of air gap to meters: \( d_{\text{air}} = 7.5 \text{ cm} = 0.075 \text{ m} \).2. Total resistance \( R = R_{\text{glass}} + R_{\text{air}} \), where: \( R_{\text{glass}} = \frac{d_{\text{glass}}}{k_{\text{glass}}} = \frac{0.003}{0.84} \approx 0.0036 \text{ m}^2\text{K/W} \), and \( R_{\text{air}} = \frac{d_{\text{air}}}{k_{\text{air}}} = \frac{0.075}{0.024} \approx 3.125 \text{ m}^2\text{K/W} \).3. Total thermal resistance: \( R_{\text{total}} = 2 \cdot R_{\text{glass}} + R_{\text{air}} = 2 \cdot 0.0036 + 3.125 = 3.1322 \text{ m}^2\text{K/W} \).4. Use the formula: \( Q_{\text{total}} = \frac{A \cdot \Delta T}{R_{\text{total}}} = \frac{1 \cdot 51}{3.1322} \approx 16.28 \text{ W/m}^2 \).

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Key Concepts

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

heat conduction
Heat conduction is a process where thermal energy transfers from a region of higher temperature to a region of lower temperature within a material or between materials in thermal contact. This process occurs because of the random motion and collisions of microscopic particles such as atoms or molecules. In the context of our problem, heat conduction happens through a glass window from the warm inside to the cold outside.
The ability of a material to conduct heat is quantified by its thermal conductivity. In our problem, we deal with a glass window, having a given thermal conductivity. Let's delve deeper into thermal conductivity next.
thermal conductivity
Thermal conductivity, denoted by the symbol k, is a material-specific property that measures a material's ability to conduct heat. A higher thermal conductivity means a material can transfer heat more efficiently. For example, metals like copper have high thermal conductivity, while materials like glass or air have lower thermal conductivity.
In the given problem, the thermal conductivity of glass is provided as 0.84 W/(m·K). This means that for every meter of glass thickness and every degree Celsius of temperature difference, 0.84 watts of heat energy is transferred per second.
Understanding the thermal conductivity helps in calculating the rate of heat transfer using Fourier's Law of Heat Conduction: \[ Q = \frac{k \times A \times \triangle T}{d} \]
where:
  • Q is the rate of thermal energy transfer (W).
  • A is the area through which heat is flowing (m²).
  • \( \triangle T \) is the temperature difference across the material (K).
  • d is the thickness of the material (m).
Let's proceed to discuss the importance of the temperature difference in heat conduction.
temperature difference
Temperature difference, symbolized as \( \triangle T \), plays a crucial role in the process of heat conduction. It represents the driving force behind the flow of thermal energy. A larger temperature difference results in a higher rate of heat transfer. In the problem, the temperature difference between the inside (22°C) and outside (-29°C) is calculated to be 51°C.
This significant temperature difference means there is a strong thermal gradient driving heat from the warmer interior to the colder exterior. In practical terms, this is why insulation or additional barriers like storm windows are used—they reduce the effective temperature difference and thus the overall heat transfer. In our problem, adding the storm window with an air gap significantly reduces the rate of thermal energy transfer as air is a poor conductor of heat, providing extra resistance.
Using both the glass and air gap, represented by their respective thermal resistances, we can calculate the total resistance and subsequently the reduced heat transfer rate, highlighting the effectiveness of adding insulation. This illustrates how structures can be optimized for thermal efficiency.

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Most popular questions from this chapter

A cylindrical copper rod of length \(1.2 \mathrm{~m}\) and cross-sectional area \(4.8 \mathrm{~cm}^{2}\) is insulated to prevent thermal energy from being transferred through its surface. The ends are maintained at a temperature difference of \(100^{\circ} \mathrm{C}\) by having one end in a water-ice mixture and the other in boiling water and steam. (a) Find the rate at which thermal energy is conducted along the rod. (b) Find the rate at which ice melts at the cold end.

(a) Two \(50 \mathrm{~g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a thermally insulated container. If the water is initially at \(25^{\circ} \mathrm{C}\), and the ice comes directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature of the drink when the drink reaches thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

version: How long does a \(2.0 \times 10^{5} \mathrm{Btu} / \mathrm{h}\) water heater take to raise the temperature of \(40 \mathrm{gal}\) of water from \(70^{\circ} \mathrm{F}\) to \(100^{\circ} \mathrm{F}\) ? Metric version: How long does a \(59 \mathrm{~kW}\) water heater take to raise the temperature of \(150 \mathrm{~L}\) of water from \(21^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\) ?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)

what temperature do the following pairs of scales read the same, if ever: (a) Fahrenheit and Celsius (verify the listing in Table \(19-1\) ), (b) Fahrenheit and Kelvin, and (c) Celsius and Kelvin?

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