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One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

Short Answer

Expert verified
Answer: The thermal conductivity of the material at the average temperature is 1.5625 W/(m K).

Step by step solution

01

Determine the heat transfer rate

We know the electric power consumed by the heater is \(25\mathrm{~W}\). In steady-state conditions, the heat generated by the heater is transferred to the outer surface by conduction, thus the heat transfer rate \(q\) is equal to the electric power consumed: \(q=25\mathrm{~W}\).
02

Calculate the surface area of the material

The dimensions of each sample are given as \(10 \mathrm{~cm} \times 10 \mathrm{~cm}\). To calculate the surface area of the material, convert the dimensions to meters and multiply the length and width: $$A = (10\mathrm{~cm} \times 10\mathrm{~cm}) \times \left( \frac{1\mathrm{~m}}{100\mathrm{~cm}} \right)^2 = 0.01\mathrm{~m^2}$$
03

Calculate the temperature difference across the sample

The temperature of each sample drops from \(82^{\circ}\mathrm{C}\) at the inner surface to \(74^{\circ}\mathrm{C}\) at the outer surface, which gives a temperature difference of: $$\Delta T = 82^{\circ}\mathrm{C} - 74^{\circ}\mathrm{C} = 8^{\circ}\mathrm{C}$$
04

Calculate the thickness of the material

Given that each sample has a thickness of \(0.5\mathrm{-cm}\), we need to express the thickness in meters: $$\Delta x = 0.5\mathrm{~cm} \times \frac{1\mathrm{~m}}{100\mathrm{~cm}} = 0.005\mathrm{~m}$$
05

Determine the thermal conductivity

Now we can use the heat conduction formulation to calculate the thermal conductivity: $$q = kA \frac{\Delta T}{\Delta x} \Rightarrow k = \frac{q\Delta x}{A\Delta T}$$ $$k = \frac{25\mathrm{~W} \times 0.005\mathrm{~m}}{0.01\mathrm{~m^2} \times 8^{\circ}\mathrm{C}} = \frac{0.125\mathrm{~W}}{0.08\mathrm{~Km^2}} = 1.5625\mathrm{~W/(m~K)}$$ Therefore, the thermal conductivity of the material at the average temperature is \(1.5625\mathrm{~W/(m~K)}\).

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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 physical act of thermal energy being exchanged between systems or bodies due to temperature differences. It's fundamental to the functioning of heaters, engines, and even our own bodies. There are three primary modes of heat transfer: conduction, convection, and radiation.

In the context of measuring thermal conductivity, we're interested in conduction, the process where heat is transferred through a solid material from the high temperature side to the lower temperature side. Imagine a metal rod with one end heated - the heat will travel down the rod to the cooler end because of conduction. The rate at which heat is transferred via conduction is influenced by the material's thermal conductivity, the temperature difference, the cross-sectional area through which heat is being transferred, and the distance the heat travels.
Steady-State Conduction
Steady-state conduction refers to the condition where the temperature distribution in the conducting medium does not change over time. In other words, despite continuous heat transfer, the temperatures at any given point in the material remain constant. This is a common assumption for many engineering problems because it greatly simplifies the analysis of thermal systems.

In steady-state conduction scenarios, such as the textbook exercise in question, the rate of heat input into the system equals the rate of heat output, leading to a balance that maintains constant temperatures throughout the material over time. The assumption of steady-state is crucial when applying thermal conductivity formulas since these are derived based on time-invariant conditions.
Thermal Conductivity Formula
The thermal conductivity formula embodies how materials conduct heat and is critical for solving heat transfer problems. The formula used to determine a material's thermal conductivity, often denoted as 'k', incorporates several factors: the rate of heat transfer 'q', the thickness of the material 'Δx', the surface area 'A' through which heat is transferred, and the temperature difference 'ΔT' across the material.

The mathematical expression is given by Fourier’s law of conduction: \[ k = \frac{q\Delta x}{A\Delta T} \]

In essence, a material with high thermal conductivity will transfer heat quickly (think metals like copper), while materials with low thermal conductivity do so much more slowly (such as wood or foam). The unit of thermal conductivity is Watts per meter-Kelvin (\(\frac{W}{mK}\)), signifying the amount of heat that passes through a unit area of a material with a unit temperature gradient per unit distance.
Thermocouples
Thermocouples are devices composed of two different types of metal wires joined at one end, used to measure temperature. When the join point, or 'junction', is exposed to a certain temperature, a voltage is created that can be measured and translated into a temperature reading. This phenomenon is known as the Seebeck effect.

They are important in experiments that measure thermal conductivity because they can provide accurate and continuous temperature readings at various points in the material. By attaching thermocouples to different surfaces of the samples in the textbook exercise, one can track the temperature gradient needed to determine thermal conductivity. Due to their reliability and simplicity, thermocouples are an essential tool in many industrial and scientific applications where temperature monitoring is critical.

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Most popular questions from this chapter

The north wall of an electrically heated home is 20 \(\mathrm{ft}\) long, \(10 \mathrm{ft}\) high, and \(1 \mathrm{ft}\) thick, and is made of brick whose thermal conductivity is \(k=0.42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\). On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about \(62^{\circ} \mathrm{F}\) and \(25^{\circ} \mathrm{F}\), respectively, for a period of \(8 \mathrm{~h}\). Determine \((a)\) the rate of heat loss through the wall that night and \((b)\) the cost of that heat loss to the home owner if the cost of electricity is \(\$ 0.07 / \mathrm{kWh}\).

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour). An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of \(150 \mathrm{~m}^{2}\) and an average height of \(3 \mathrm{~m}\) at \(1000 \mathrm{~m}\) elevation, where the standard atmospheric pressure is \(89.6 \mathrm{kPa}\). The house is maintained at a temperature of \(22^{\circ} \mathrm{C}\), and the infiltration losses are estimated to amount to \(0.7 \mathrm{ACH}\). Assuming the pressure and the temperature in the house remain constant, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is \(5^{\circ} \mathrm{C}\). Also, determine the cost of this energy loss for that day if the unit cost of electricity in that area is $$\$ 0.082 / \mathrm{kWh}$$

How does forced convection differ from natural convection?

The inner and outer surfaces of a \(25-\mathrm{cm}\)-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\), and convection with ambient air also at \(40^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

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