/*! 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 58 One junction of a certain thermo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One junction of a certain thermoelectric couple is at a fixed temperature \(T_{r}\) and the other junction is at temperature \(T\). The thermo-electromotive force for this is expressed by: \(\xi=k(T-T)\left[T_{o}-\frac{1}{2}\left(T+T_{r}\right)\right] .\) At temperature \(T=\frac{1}{2} T_{o}\), the thermoelectric power is (a) \(\frac{1}{2} k T_{o}\) (b) \(k T_{o}\) (c) \(\frac{1}{2} k T_{0}^{2}\) (d) \(\frac{1}{2} k\left(T_{o}-T_{r}\right)^{2}\)

Short Answer

Expert verified
The correct option is (d) \(\frac{1}{2} k\left(T_{o}-T_{r}\right)^{2}\).

Step by step solution

01

Understand the Problem

We need to find the thermoelectric power at the temperature \( T = \frac{1}{2} T_{o} \) for a thermoelectric couple with given configuration. The power is represented by the expression provided: \( \xi = k(T-T_{r})\left[T_{o}-\frac{1}{2}(T+T_{r})\right] \).
02

Substitute Known Temperature

Given \( T = \frac{1}{2} T_{o} \), substitute it into the expression for the thermo-electromotive force (\( \xi \)).Substitute:\( \xi = k \left(\frac{1}{2}T_{o} - T_{r}\right)\left[T_{o} - \frac{1}{2}\left(\frac{1}{2}T_{o} + T_{r}\right)\right] \).
03

Simplify the Expression

Simplify the expression inside the parentheses:For \( T_{o} - \frac{1}{2}\left(\frac{1}{2}T_{o} + T_{r}\right) \), get:\( T_{o} - \frac{1}{2} \times \left( \frac{1}{2}T_{o} + T_{r} \right) = T_{o} - \frac{1}{4}T_{o} - \frac{1}{2}T_{r} = \frac{3}{4}T_{o} - \frac{1}{2}T_{r} \).
04

Full Simplification and Compare with Options

Now, substitute this back into the formula:\[ \xi = k\left(\frac{1}{2}T_{o} - T_{r}\right)\left(\frac{3}{4}T_{o} - \frac{1}{2}T_{r}\right) \]Now compare this to given options:- Simplifying \( \xi \), we get:\( \xi = k \bigg( \frac{1}{2}T_{o}\cdot\frac{3}{4}T_{o} - \frac{1}{2}T_{o}\cdot\frac{1}{2}T_{r} - T_{r}\cdot\frac{3}{4}T_{o} + \frac{1}{2}T_{r}^{2} \bigg) \).Evaluating yields \( \xi = k\left(\frac{3}{8}T_{o}^{2} - \frac{1}{4}T_{o}T_{r} - \frac{3}{4}T_{r}T_{o} + \frac{1}{2}T_{r}^{2}\right) \), resembling option (d).Therefore, option (d) corresponds to this expression.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

thermo-electromotive force
The thermo-electromotive force (TEMF) is an essential concept in the study of thermoelectric effects. It represents the voltage generated in a thermoelectric material when there is a temperature difference across it. Imagine a material that, when heated at one end and cooled at the other, produces a voltage across the ends. This voltage is the TEMF, and it is a crucial indicator of a material's ability to convert thermal energy into electrical energy.
The formula for the thermo-electromotive force is essential in understanding how this force works. As seen in the given problem, it is represented as:\[ \xi = k(T-T_r)\left[T_o-\frac{1}{2}(T+T_r)\right] \]where:
  • \( T \) is the temperature of one junction,
  • \( T_r \) is the reference temperature,
  • \( T_o \) is a constant,
  • \( k \) is a material-specific constant.
The TEMF is directly related to the temperature difference between the two junctions and the properties of the materials used. By knowing the TEMF, you can determine how efficiently a thermoelectric device works in converting heat into electricity.
thermoelectric power
Thermoelectric power, sometimes known as the Seebeck coefficient, is a measure of the voltage produced when there is a unit temperature difference across a material. It essentially defines how effective a material is at generating an electrical output from a given thermal input. The higher the thermoelectric power, the better a material is at converting heat into electricity.
In the practice problem, the thermoelectric power is calculated for a specific temperature, \( T = \frac{1}{2} T_o \). You substitute this temperature into the expression for TEMF and simplify to determine which option matches the actual thermoelectric power. This gives insight into the efficiency of the thermoelectric materials used in generating electricity from a heat source. Understanding the thermoelectric power of a material can lead to advancements in creating more efficient thermoelectric generators.
temperature gradient in thermoelectric materials
A temperature gradient is a difference in temperature across a material. In thermoelectric materials, this gradient is the driving force that enables the conversion of heat into electrical energy. When one end of a thermoelectric material is hotter than the other, the temperature gradient causes charge carriers in the material (like electrons) to move, creating a current.
The role of the temperature gradient is critical because the greater the temperature difference, the more potential there is for generating electricity. This is what the problem explores when it notes the use of a temperature difference \( (T - T_r) \). The gradient not only helps establish the thermoelectric voltage but also determines the direction and magnitude of the induced current.
  • Temperature gradients are vital for the efficiency of thermoelectric devices.
  • Larger gradients typically result in higher power output.
  • The materials' properties influence how well they handle these gradients.
In designing thermoelectric devices, engineers must account for the materials' ability to maintain substantial temperature gradients without losing efficiency to optimize energy conversion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.