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Two surfaces of a 2 -cm-thick plate are maintained at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) respectively. If it is determined that heat is transferred through the plate at a rate of \(500 \mathrm{W} / \mathrm{m}^{2}\), determine its thermal conductivity.

Short Answer

Expert verified
Answer: The thermal conductivity of the plate is 0.1 W/m・K.

Step by step solution

01

Recall Fourier's law of heat conduction

Fourier's law of heat conduction is given by the formula: $$q = -kA\frac{dT}{dx}$$ Where: - \(q\) is the heat transfer rate per unit area (\(\mathrm{W/m^2}\)), - \(k\) is the thermal conductivity of the material (\(\mathrm{W/m\cdot K}\)), - \(A\) is the surface area (\(\mathrm{m^2}\)), - \(dT\) is the temperature difference between the two surfaces (\(\mathrm{K}\) or \(\mathrm{C}\); both are equivalent due to the difference relation), - \(dx\) is the thickness of the plate (\(\mathrm{m}\)). Since, we are given the heat transfer rate per unit area \(q\), the temperatures of the two surfaces, and the thickness of the plate, we can rearrange the above formula to solve for thermal conductivity \(k\).
02

Rearranging the formula to solve for \(k\)

Rearrange the equation to isolate \(k\) on one side: $$k = -\frac{q dx}{A dT}$$
03

Plug in the known values

Now, we plug in the known values to the equation: - \(q = 500 \mathrm{W/m^2}\) - \(dx = 2 \mathrm{cm} = 0.02 \mathrm{m}\) - \(dT = 100 ^{\circ}\mathrm{C} - 0^{\circ}\mathrm{C} = 100 ^{\circ}\mathrm{C}\) Since we are finding thermal conductivity \(k\) which is independent of the area, we do not need the value of \(A\). So, the equation becomes: $$k = -\frac{500 \cdot 0.02}{100}$$
04

Calculate the thermal conductivity \(k\)

Finally, calculate the value of \(k\): $$k = -\frac{500 \cdot 0.02}{100} = -\frac{10}{100} = -0.1 \mathrm{W/m\cdot K}$$ However, a negative value for thermal conductivity does not make physical sense. Since the negative sign in Fourier's law signifies the direction of heat transfer, but we are interested in the magnitude of thermal conductivity, we can simply take the absolute value of our result: $$k = 0.1 \mathrm{W/m\cdot K}$$ So the thermal conductivity of the plate is \(0.1 \mathrm{W/m\cdot K}\).

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Key Concepts

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

Fourier's Law of Heat Conduction
Understanding heat transfer in materials is crucial for various engineering applications, and this is where Fourier's law of heat conduction comes into play. It describes how heat energy moves through a material based on the temperature difference and the material's properties. Specifically, it states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area through which heat is flowing. In simpler terms, it means heat flows from regions of high temperature to regions of low temperature, and the rate at which it does this depends on the material's thermal conductivity.

Fourier's law is mathematically described by the equation: \[q = -kA\frac{dT}{dx}\] where \(k\) is the thermal conductivity, a measure of a material's ability to conduct heat. Applying this law within the context of practical problems, such as determining the necessary thickness of insulation for a pipe or the rate at which a freezer can cool down goods, is a standard application in thermodynamics.
Heat Transfer Rate
  • The rate at which heat is transferred through a material is a fundamental concept in thermodynamics and directly affects how we design and manage thermal systems. It is represented by the variable \(q\) and is usually measured in watts per square meter (\(W/m^2\)).
  • According to Fourier’s law of heat conduction, the heat transfer rate depends on the thermal conductivity of the material, the area perpendicular to the heat transfer direction, and the temperature difference across the material.
Understanding the heat transfer rate is crucial for engineers to predict how quickly a material will reach thermal equilibrium and to ensure the safety and efficiency of thermal systems. Devices and structures must often be designed with a certain heat transfer rate in mind to prevent overheating or to ensure rapid heat removal.
Temperature Difference
  • Temperature difference, denoted as \(dT\), is the driving force behind heat transfer. It's the disparity in temperature between two points, which prompts the movement of thermal energy from the hotter region to the cooler one.
  • Fourier's law of heat conduction shows us that the larger the temperature difference, the higher the heat transfer rate, assuming the medium through which heat is being transferred remains consistent. This principle is leveraged in numerous applications, including heating systems, refrigeration, and even electronic devices' thermal management.

When solving heat transfer problems, accurately determining the temperature difference is essential. In the given textbook problem, the temperature difference creates a heat flow through the plate, and by knowing the rate at which heat is transferred and the plate’s thickness, one can determine the material’s thermal conductivity.

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Most popular questions from this chapter

A grist mill of the 1800 s employed a water wheel that was \(14 \mathrm{m}\) high; 320 liters per minute of water flowed on to the wheel near the top. How much power, in \(\mathrm{kW},\) could this water wheel have produced?

A 7 -cm-external-diameter, 18 -m-long hot-water pipe at \(80^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(5^{\circ} \mathrm{C}\) by natural convection with a heat transfer coefficient of \(25 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) Determine the rate of heat loss from the pipe by natural convection, in kW.

A 75 -hp compressor in a facility that operates at full load for \(2500 \mathrm{h}\) a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is \(\$ 0.06 / \mathrm{kWh}\), the annual electricity cost of this compressor is \((a) \$ 7802\) \((b) \$ 9021\) \((c) \$ 12,100\) \((d) \$ 8389\) \((e) \$ 10,460\)

An oil pump is drawing \(44 \mathrm{kW}\) of electric power while pumping oil with \(\rho=860 \mathrm{kg} / \mathrm{m}^{3}\) at a rate of \(0.1 \mathrm{m}^{3} / \mathrm{s}\) The inlet and outlet diameters of the pipe are \(8 \mathrm{cm}\) and \(12 \mathrm{cm},\) respectively. If the pressure rise of oil in the pump is measured to be \(500 \mathrm{kPa}\) and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump.

How is the combined pump-motor efficiency of a pump and motor system defined? Can the combined pumpmotor efficiency be greater than either the pump or the motor efficiency?

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