Question

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

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

Step-by-step solution

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: \(q=25\mathrm{~W}\).

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: $$A = (10\mathrm{~cm} \times 10\mathrm{~cm}) \times \left( \frac{1\mathrm{~m}}{100\mathrm{~cm}} \right)^2 = 0.01\mathrm{~m^2}$$

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: $$\Delta T = 82^{\circ}\mathrm{C} - 74^{\circ}\mathrm{C} = 8^{\circ}\mathrm{C}$$

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: $$\Delta x = 0.5\mathrm{~cm} \times \frac{1\mathrm{~m}}{100\mathrm{~cm}} = 0.005\mathrm{~m}$$

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 \(1.5625\mathrm{~W/(m~K)}\).

Key concepts

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.