Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 28

The resistance of a conductor is \(19.8 \Omega\) at \(15.0^{\circ} \mathrm{C}\) and \(25.0 \Omega\) at \(85.0^{\circ} \mathrm{C} .\) What is the temperature coefficient of resistance of the material?

Short Answer

Expert verified
Answer: The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

Step by step solution

01

Identify the given values

We are given the resistances of the conductor at two temperatures: - \(R_1 = 19.8 \Omega\) at \(T_1 = 15.0^{\circ} \mathrm{C}\) - \(R_2 = 25.0 \Omega\) at \(T_2 = 85.0^{\circ} \mathrm{C}\)
02

Convert the temperatures to Kelvin scale

To calculate the temperature coefficient of resistance, it is essential to use the Kelvin scale for temperature. The conversion from Celsius to Kelvin is as follows: \(K = ^{\circ}\mathrm{C} + 273.15\). So, the temperatures in Kelvin are: - \(T_1 = 15.0^{\circ} \mathrm{C} + 273.15 = 288.15 \mathrm{K}\) - \(T_2 = 85.0^{\circ} \mathrm{C} + 273.15 = 358.15 \mathrm{K}\)
03

Substitute the given values into the temperature coefficient of resistance formula

Now that we have the resistances and corresponding temperatures in Kelvin, we can calculate the temperature coefficient of resistance (\(\alpha\)) using the following formula: \(\alpha = \dfrac{(R_2 - R_1)}{R_1(T_2 - T_1)}\) Substitute the given values into the formula as follows: \(\alpha = \dfrac{(25.0 \Omega - 19.8 \Omega)}{19.8 \Omega(358.15 \mathrm{K} - 288.15 \mathrm{K})}\)
04

Simplify and solve for the temperature coefficient of resistance

First, simplify the equation: \(\alpha = \dfrac{5.2 \Omega}{19.8 \Omega(70 \mathrm{K})}\) Next, cancel out the units and solve for \(\alpha\): \(\alpha = \dfrac{5.2}{19.8 \times 70}\) \(\alpha \approx 3.72 \times 10^{-3} \ \mathrm{K}^{-1}\)
05

Write down the final answer

The temperature coefficient of resistance of the material is approximately \(3.72 \times 10^{-3} \ \mathrm{K}^{-1}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Suppose a collection of five batteries is connected as shown. (a) What is the equivalent emf of the collection? Treat them as ideal sources of emf. (b) What is the current through the resistor if its value is \(3.2 \Omega ?\)
A voltmeter has a switch that enables voltages to be measured with a maximum of \(25.0 \mathrm{V}\) or \(10.0 \mathrm{V}\). For a range of voltages to $25.0 \mathrm{V},\( the switch connects a resistor of magnitude \)9850 \Omega$ in series with the galvanometer; for a range of voltages to \(10.0 \mathrm{V}\), the switch connects a resistor of magnitude \(3850 \Omega\) in series with the galvanometer. Find the coil resistance of the galvanometer and the galvanometer current that causes a full-scale deflection. [Hint: There are two unknowns, so you will need to solve two equations simultaneously.]
A \(500-\) W electric heater unit is designed to operate with an applied potential difference of \(120 \mathrm{V}\). (a) If the local power company imposes a voltage reduction to lighten its load, dropping the voltage to $110 \mathrm{V}$, by what percentage does the heat output of the heater drop? (Assume the resistance does not change.) (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than calculated in part (a)?
A person can be killed if a current as small as \(50 \mathrm{mA}\) passes near the heart. An electrician is working on a humid day with hands damp from perspiration. Suppose his resistance from one hand to the other is $1 \mathrm{k} \Omega$ and he is touching two wires, one with each hand. (a) What potential difference between the two wires would cause a 50 -mA current from one hand to the other? (b) An electrician working on a "live" circuit keeps one hand behind his or her back. Why?
A current of \(10.0 \mathrm{A}\) is carried by a copper wire of diameter $1.00 \mathrm{mm} .\( If the density of the conduction electrons is \)8.47 \times 10^{28} \mathrm{m}^{-3},$ how long does it take for a conduction electron to move \(1.00 \mathrm{m}\) along the wire?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation

Famous Physicists

Read Explanation

Linear Momentum

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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.

Sign-up for free