/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A carpenter builds an exterior h... [FREE SOLUTION] | 91Ó°ÊÓ

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A carpenter builds an exterior house wall with a layer of wood 3.0 \(\mathrm{cm}\) thick on the outside and a layer of Styrofoam insulation 2.2 \(\mathrm{cm}\) thick on the inside wall surface. The wood has \(k=0.080 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) , and the Styrofoam has \(k=0.010 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The interior surface temperature is \(19.0^{\circ} \mathrm{C}\) , and the exterior surface temperature is \(-10.0^{\circ} \mathrm{C}\) (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

Short Answer

Expert verified
(a) The temperature at the wood-Styrofoam interface is approximately 14.8°C. (b) The heat flow rate is about 11.3 W/m².

Step by step solution

01

Identify Heat Transfer Formula

For materials in series, the rate of heat flow per square meter, \( q \), through the wall can be determined using the formula: \[ q = \frac{(T_1 - T_2)}{R_{\text{total}}} \]where \( T_1 \) is the interior temperature, \( T_2 \) is the exterior temperature, and \( R_{\text{total}} \) is the total thermal resistance of the wall.
02

Calculate Individual Thermal Resistances

Thermal resistance, \( R \), for each layer is calculated using the formula: \[ R = \frac{d}{k} \]where \( d \) is the thickness in meters and \( k \) is the thermal conductivity. - For wood: \[ R_{\text{wood}} = \frac{0.030\, \text{m}}{0.080}\, \text{W/m} \cdot \text{K} = 0.375 \, \text{m}^2\cdot\text{K/W} \] - For Styrofoam: \[ R_{\text{styrofoam}} = \frac{0.022\, \text{m}}{0.010}\, \text{W/m} \cdot \text{K} = 2.2 \, \text{m}^2\cdot\text{K/W} \]
03

Calculate Total Thermal Resistance

The total thermal resistance of the wall is the sum of the resistances of the wood and Styrofoam layers:\[ R_{\text{total}} = R_{\text{wood}} + R_{\text{styrofoam}} = 0.375 + 2.2 = 2.575 \, \text{m}^2\cdot\text{K/W} \]
04

Find Rate of Heat Flow

Using the total thermal resistance in the heat transfer equation, we can find \( q \):\[ q = \frac{(19.0 - (-10.0))}{2.575} = \frac{29.0}{2.575} \approx 11.26 \, \text{W/m}^2 \]This is the rate of heat flow per square meter through the wall.
05

Determine Temperature at Interface of Materials

To find the temperature where the wood meets the Styrofoam, use the heat flow rate and the resistance of only the wood to calculate the temperature difference across the wood:\[ q = \frac{T_i - T_m}{R_{\text{wood}}} \]\[ 11.26 = \frac{19.0 - T_m}{0.375} \]\[ T_m = 19.0 - (11.26 \times 0.375) \approx 14.77\, ^\circ \text{C} \]

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy between objects or systems due to a temperature difference. In the context of our exercise, this concept is crucial as it helps us understand how heat flows through the materials of a wall.
Heat transfer usually occurs in three different methods:
  • Conduction: This is the process through which heat energy is transferred within a body or from one body to another within solid media. In the carpenter's wall, heat conducts through the wood and Styrofoam.
  • Convection: This method involves the movement of fluid (liquid or gas) that carries heat away as it moves. For walls, this is often more relevant on the surface level where air moves along each side.
  • Radiation: This mechanism involves the transfer of heat through electromagnetic waves, which can happen even through a vacuum.
In our example, the focus is on conduction, specifically through materials in series, such as inner wood and Styrofoam layers. The resistance each material presents impacts the overall heat flow through the wall.
Calculating the heat flow allows us to determine how effective these materials are in preventing heat from escaping or entering the building.
Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. Materials with high thermal conductivity transfer heat effectively, while those with low thermal conductivity do not. In the wall built by the carpenter, there are two materials to consider: wood and Styrofoam. The thermal conductivities for these materials are:
  • Wood: 0.080 W/m·K
  • Styrofoam: 0.010 W/m·K
These values tell us that wood allows heat to pass through it more easily than Styrofoam does. Styrofoam, with its low thermal conductivity, acts as a better insulator, slowing down the rate of heat transfer.
This property is vital when designing insulation systems, as its knowledge helps to select materials that will maintain desired temperatures within a building by resisting unwanted heat transfer.
Building Insulation
Building insulation is a crucial topic for maintaining energy efficiency and comfort within homes and buildings. Insulation works by reducing the flow of heat into or out of a building, thereby maintaining a pleasant indoor temperature regardless of the extremes outside. In this scenario, the carpenter added layers of wood and Styrofoam as insulation. Though both materials contribute to insulation, their effectiveness varies because of differing thermal resistances. Each material has specific thickness and thermal conductivity that affect the wall's total thermal resistance:
  • Wood: Offers moderate resistance due to its thermal property and thickness.
  • Styrofoam: Provides higher resistance, being a more effective insulator.
This layered approach in construction helps in achieving better thermal performance, preventing significant heat loss in winter or heat gain in summer. By selecting materials with suitable thermal conductivities and thicknesses, building efficiency can be optimized to save energy and maintain indoor comfort levels more effectively.

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Most popular questions from this chapter

A \(25,000-\mathrm{kg}\) subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) .

A metal rod that is 30.0 \(\mathrm{cm}\) long expands by 0.0650 \(\mathrm{cm}\) when its temperature is raised from \(0.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\) . A rod of a different metal and of the same length expands by 0.0350 \(\mathrm{cm}\) for the same rise in temperature. A third rod, also 30.0 \(\mathrm{cm}\) long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 \(\mathrm{cm}\) between \(0.0^{\circ} \mathrm{C}\) and \(100.0^{\circ} \mathrm{C}\) . Find the length of each portion of the composite rod.

An aluminum tea kettle with mass 1.50 \(\mathrm{kg}\) and containing 1.80 \(\mathrm{kg}\) of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from \(20.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

A worker pours 1.250 kg of molten lead at a temperature of \(365.0^{\circ} \mathrm{C}\) into 0.5000 \(\mathrm{kg}\) of water at a temperature of \(75.00^{\circ} \mathrm{C}\) in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Arod is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) throughout. One end is kept at \(0^{\circ} \mathrm{C}\) , and the other is brought into contact with a steam bath at \(100^{\circ} \mathrm{C}\) . The surface of the rod is insulated so that heat can flow only lengthwise along the rod. The cross-sectional area of the rod is \(2.50 \mathrm{cm}^{2},\) its length is 120 \(\mathrm{cm}\) , its thermal conductivity is \(380 \mathrm{W} / \mathrm{m} \cdot \mathrm{K},\) its density is \(1.00 \times 10^{4} \mathrm{kg} / \mathrm{m}^{3},\) and its specific heat is 520 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . Consider a short cylindrical element of the rod 1.00 \(\mathrm{cm}\) in length. (a) If the temperature gradient at the cooler end of this element is 140 \(\mathrm{C}^{\circ} / \mathrm{m}\) , how many joules of heat energy flow across this end per second? (b) If the average temperature of the element is increasing at the rate of 0.250 \(\mathrm{C} \%\) /s, what is the temperature gradient at the other end of the element?

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