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Explain how to measure each of the following quantities, and list the tools needed to make the measurement. (a) Electrical conductivity (b) Thermal conductivity (c) Peltier coefficient

Short Answer

Expert verified
Electrical conductivity requires a conductivity meter; thermal conductivity needs a thermal conductivity analyzer; Peltier coefficient is measured using a thermoelectric setup. Each quantity has specific tools for precise measurement.

Step by step solution

01

Understanding Electrical Conductivity

Electrical conductivity measures a material's ability to conduct electric current. This property is calculated by the formula \( \sigma = \frac{1}{\rho} \), where \( \sigma \) is the electrical conductivity and \( \rho \) is the resistivity of the material.
02

Tools for Measuring Electrical Conductivity

To measure electrical conductivity, you need a conductivity meter or a multimeter that can measure resistance. A consistent power supply, conductive probes, and the sample material are also necessary to ensure precise measurements.
03

Understanding Thermal Conductivity

Thermal conductivity is the property of a material to conduct heat, denoted by \( k \). It is commonly calculated using the equation \( Q = k \cdot A \cdot \frac{\Delta T}{d} \) where \( Q \) is the heat transfer rate, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference and \( d \) is the thickness of the material.
04

Tools for Measuring Thermal Conductivity

A thermal conductivity analyzer or a setup involving a heat source, a thermometer or thermocouples, and an insulating material to ensure accurate measurement is required. Additionally, a computing or data logging device would be helpful for precise calculations.
05

Understanding Peltier Coefficient

The Peltier coefficient \( \Pi \) describes the amount of heat carried per unit charge when current flows through a junction of two different materials. The formula is \( \Pi = \frac{Q}{I} \), where \( Q \) is the heat absorbed or emitted at the junction, and \( I \) is the current.
06

Tools for Measuring Peltier Coefficient

To measure the Peltier coefficient, you need a thermoelectric setup with a junction, a power supply, an ammeter for measuring current, and a calorimeter or temperature sensors to measure the heat transfer at the junction.

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Key Concepts

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

Electrical Conductivity
Electrical conductivity is a fundamental material property that measures how well a material allows the flow of electrical current. It is commonly used in various fields ranging from electronics to material science. The ability to conduct electricity is assessed by the formula \( \sigma = \frac{1}{\rho} \), where \( \sigma \) represents electrical conductivity and \( \rho \) denotes the material's resistivity.

To properly measure electrical conductivity, specific tools and methods are necessary:
  • A **conductivity meter** or a **multimeter** capable of measuring resistance is essential. These instruments help determine the resistivity of the material, which is then used to calculate conductivity.
  • A **consistent power supply** provides the required energy for measurement.
  • **Conductive probes** are used to make physical contact with the material sample for accurate readings.
  • Finally, the **sample material** itself is crucial for measurement, ensuring the sample is clean and properly prepared can impact the accuracy of your results.

In practice, achieving precise measurements requires calibration and ensuring all equipment is functioning correctly. This ensures that the resistivity, and thus the conductivity, are accurately calculated.
Thermal Conductivity
Thermal conductivity is the ability of a material to transmit heat. This property is vital in understanding and designing systems that involve heat transfer, such as building insulation, electronic devices, and machinery. The formula governing this property is \( Q = k \cdot A \cdot \frac{\Delta T}{d} \), where \( Q \) is the heat transfer rate, \( k \) is the thermal conductivity, \( A \) the cross-sectional area, \( \Delta T \) the temperature difference, and \( d \) the material's thickness.

To measure thermal conductivity, the following tools are typically utilized:
  • A **thermal conductivity analyzer** can directly determine the thermal conductivity of a sample.
  • Alternatively, a **heat source** and **thermometer** or **thermocouples** are used in a controlled setup. These tools help create and measure the necessary temperature gradients across the sample.
  • An **insulating material** is often required to minimize heat loss to the environment, providing more accurate results.
  • A **data logging** device is helpful to capture precise temperature readings over time, which are necessary for calculating the heat transfer rate.

Accuracy in these measurements requires careful consideration of environmental factors and proper calibration of the measuring instruments.
Peltier Coefficient
The Peltier coefficient is a unique property that describes the heat transfer across a junction of two different materials when electrical current flows through it. This characteristic is significant in thermoelectric applications, like cooling and heating elements. The Peltier coefficient is defined through the formula \( \Pi = \frac{Q}{I} \), where \( Q \) is the heat exchanged at the junction and \( I \) is the current flowing through it.

To measure the Peltier coefficient, you will need the following setup:
  • A **thermoelectric setup** involves creating a junction between two distinct materials.
  • A stable **power supply** is required to ensure a consistent current flow.
  • An **ammeter** is necessary to measure the current flowing through the circuit accurately.
  • To assess the heat transfer, a **calorimeter** or **temperature sensors** can be used at the junction to determine how much heat is absorbed or emitted.

It is important to control the experimental conditions carefully to improve the accuracy and reliability of the Peltier coefficient measurements.

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Most popular questions from this chapter

The Carnot efficiency describes the limit of the efficiency for some devices. Does it apply to the following types of devices? (Answer yes or no.) \- Hall effect device \- Semiconductor laser \- Photovoltaic device \- Piezoelectric device \- Pyroelectric device \- Thermoelectric device used as a temperature sensor \- Thermoelectric device used as a refrigerator \- Thermoelectric device used to generate electricity for a sensor system

Qualitatively, explain the difference between each pair of related quantities. (a) Seebeck effect and Peltier effect (b) Thermal conductivity and electrical conductivity

Argon gas is enclosed in a container of a fixed volume. At \(T=300 \mathrm{~K}\), the pressure of the gas is \(50,000 \mathrm{~Pa}\). At \(T=350 \mathrm{~K},\) calculate the pressure of the gas.

A thermoelectric device is used to build a small refrigerator that can hold two pop cans. When the device is operating, the cold side of the device is at \(T=10{ }^{\circ} \mathrm{C}\) while the hot side of the device, outside the refrigerator, is at \(T=42^{\circ} \mathrm{C}\). What is the maximum possible efficiency of this device?

A thermoelectric device is made from a material with resistivity 5 . \(10^{-8} \Omega \cdot \mathrm{m}\) and Seebeck coefficient \(8.5 \cdot 10^{-5} \frac{\mathrm{V}}{\mathrm{K}}\). A cube, \(1 \mathrm{~cm}\) on each side, was used to determine the thermal conductivity. One side of the cube was heated. At a steady state, the rate of energy transfer by conduction through the cube is \(1.8 \mathrm{~W}\). The temperature distribution through the material is linear, and a temperature difference across is measured to be \(20 \mathrm{~K}\) across the cube. Find the thermal conductivity \(\kappa,\) and find the figure of merit \(Z\) for the material.

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