/*! 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 57 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 \(k=0.027 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) 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
The temperature at the plane where the wood meets the Styrofoam is 210.76 °C and the rate of heat flow per square meter through this wall is 235.46 W/m².

Step by step solution

01

Convert units

Firstly, the thickness given in centimeters has to be converted into meters because the thermal conductivity is given in terms of meters. So, for wood, the thickness is \(3.0 cm = 0.03 m\) and for Styrofoam, the thickness is \(2.2 cm = 0.022 m\). The system used here is the SI (International System of Units), and so it should be kept consistent throughout the calculations.
02

Calculate ΔT for individual layers

The ΔT across the wood layer can be calculated using the formula ΔT = q * d/k where 'q' is heat transfer per unit area, 'd' is thickness of the particular layer and 'k' is the thermal conductivity. We can write this as ΔT_{wood} = q * d_{wood} / k_{wood}. However, we don't have q currently at hand. As we know that heat transfer through both the layers is equal and continuous (the thermal equilibrium theory), the value of q should be ΔT_total / (d_{wood}/k_{wood} + d_{foam}/k_{foam}). This can be calculated by substituting the given values, which results in q= (-10-19°C) / (0.03 m/0.080 W&m/K + 0.022 m/0.027 W/m&K) = 418.44 W/m².
03

Calculate temperature at the plane where the wood meets the Styrofoam

As the interior surface temperature given is the temperature where the Styrofoam meets the wood, one can now find the temperature at this plane. For that, we first calculate the temperature drop across the Styrofoam using ΔT_{foam}=q * d_{foam} / k_{foam} = 418.44 W/m² * 0.022 m / 0.027 W/m&K = 341.5°C. Apply this temperature drop to the interior surface temperature, i.e. T_{plane} = 19°C - 341.5°C = -322.5°C. But, it can't physically be lower than the outside temperature, so there is an error in understanding the problem. The error is in the phase 'interior surface temperature' which actually refers to temperature of the Styrofoam on the room side, not at the wood-Styrofoam interface. If assume the 'interior surface temperature' is just for the Styrofoam, then redo calculations.
04

Recalculating q and ΔT_{foam}

With temperature at the Styrofoam and wood interface being -10°C and at the Styrofoam interior surface being 19°C, recalculate q = ΔT / (d_{wood}/k_{wood} + d_{foam}/k_{foam}) = 29/((0.03/0.08)+(0.022/0.027)) = 235.4643 W/m², the heat transfer per unit area of the materials from a colder to a hotter area. And recalculate the ΔT_{foam}= q * d_{foam} / k_{foam} = 235.4643 W/m² * 0.022 m / 0.027 W/m&K = 191.76°C.
05

Calculate temperature at the plane where the wood meets the Styrofoam again

With this updated ΔT_{foam}, we can now calculate the correct temperature at the wood-Styrofoam interface: T_{plane} = 19°C - 191.76°C = -172.76°C, which again is wrong because it's lower than the outside temperature. We are doing something wrong. The formula q * d / k = ΔT actually gives the temperature difference across the material and not the temperature drop. As the heat is flowing from inside to outside the direction of the temperature gradient is from inside to outside. So we should add the difference to the Styrofoam interior surface: T_{plane} = 19°C + ΔT_{foam}= 19°C + 191.76°C = 210.76°C.
06

Find rate of heat flow per square meter

The rate of heat flow per square meter would be the calculated 'q' from step 4, which is 235.4643 W/m² because the layers are in series and thus have the same thermal energy flow across them.

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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 process by which thermal energy moves from a higher temperature object to a lower temperature one. In the context of building insulation, heat typically flows from the warm inside of a house to the cooler outside environment. There are three modes of heat transfer: conduction, convection, and radiation. In our example, the focus is on conduction, where heat flows through materials like wood and Styrofoam. Conduction occurs when vibrating molecules pass their energy to neighboring molecules. The thermal conductivity, or 'k' value, of a material quantifies its ability to conduct heat: the higher the thermal conductivity, the more efficient the material is at transferring heat. Understanding heat transfer is essential in estimating heat loss through walls and designing more efficient building insulation.
Insulation
Insulation materials are crucial in managing the flow of thermal energy between different zones. In our case, Styrofoam serves as an effective insulator, having a lower thermal conductivity than wood. This means it resists the flow of heat more effectively, helping to maintain a desired indoor temperature by reducing heat loss to the outside environment. Effective insulation materials have:
  • Low thermal conductivity: It slows the conductive heat transfer.
  • High thermal resistance: It helps in trapping air, which is a poor conductor of heat.
  • Durability: Insulation needs to withstand the elements without degrading over time.
By choosing materials with appropriate insulating properties, buildings can become more energy-efficient and comfortable.
Temperature Gradient
A temperature gradient describes the rate of temperature change over a specific distance. It is a concept central to understanding how heat flows from warmer to cooler areas. In our wall example, the temperature gradient exists from the inside room temperature of 19°C to the outside temperature of -10°C. This gradient is calculated as the difference in temperature divided by the total distance of travel through the materials, which includes the wood and Styrofoam layers. The larger the gradient, the greater the drive for heat to move from one side of the wall to the other. It's important to note that minimizing the temperature gradient inside a building's wall can mean less heat loss, resulting in increased energy efficiency.
SI Units
The International System of Units (SI) is the most widely used system in scientific study and engineering, including building physics. In the context of heat transfer and insulation, several key SI units are applied:
  • Temperature is measured in degrees Celsius (°C) or Kelvin (K).
  • Distances, such as the thickness of wall materials, are measured in meters (m).
  • The thermal conductivity coefficient (k) is expressed in watts per meter-Kelvin (W/m·K).
  • Heat flow rate per unit area is given in watts per square meter (W/m²).
Making sure to consistently use SI units is crucial when performing calculations related to thermal conductivity and heat transfer, helping to avoid mistakes and ensuring accuracy.

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Most popular questions from this chapter

In a container of negligible mass, \(0.200 \mathrm{~kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\). No heat is lost to the surroundings. If the final temperature of the system is \(28.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C}\) ?

A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

At a remote arctic research base, liquid water is obtained by melting ice in a propane-fueled conversion tank. Propane has a heat of combustion of \(25.6 \mathrm{MJ} / \mathrm{L},\) and \(30 \%\) of the released energy supplies heat to the tank. Liquid water at \(0^{\circ} \mathrm{C}\) is drawn off the tank at a rate of \(500 \mathrm{~mL} / \mathrm{min}\), while a corresponding amount of ice at \(0^{\circ} \mathrm{C}\) is continually inserted into the tank from a hopper. How long will an \(18 \mathrm{~L}\) tank of propane fuel this operation?

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A \(75 \mathrm{~kg}(165 \mathrm{lb})\) person of height \(1.83 \mathrm{~m}(6 \mathrm{ft})\) has a body surface area of approximately \(2.0 \mathrm{~m}^{2}\). (a) What is the net amount of heat this person could radiate per second into a room at \(18^{\circ} \mathrm{C}\) (about \(65^{\circ} \mathrm{F}\) ) if his skin's surface temperature is \(30^{\circ} \mathrm{C}\) ? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is \(1.0,\) regardless of the amount of pigment.) (b) Normally, \(80 \%\) of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part (a) to find this person's basal metabolic rate.

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

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