Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
Question
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Chapter 7, Problem 7.4QAP
Interpretation Introduction

(a)

Interpretation:

The total energy output must be calculated for the blackbody heated at 4500 K.

Concept introduction:

Stefan’s law essentially states that the total quantity of heat energy released by a perfect blackbody per unit area per second is directly proportional to the absolute temperature’s fourth power of its surface given by the equation-

Et= αT4

Wein’s displacement Law which states that the maximum wavelength in micrometers for the radiations of the blackbody is given by:

λmaxT = 2.90×103

Where,

T = temperature in Kelvin

λmax = maximum wavelength

Expert Solution
Check Mark

Answer to Problem 7.4QAP

The amount of energy emitted by Et at temperature 4500K is 2.33×107Wm2.

Explanation of Solution

The total energy Et emitted per unit time per unit area is calculated by:

Et=αT4=(5.69×108)T4Wm2 …………….. (1)

Given that-

α=5.69×108Wm2

T = 4500K

The value of total energy Et is:

Et=(5.69×108)T4Wm2

=(5.69×108)(4500K)4Wm2=2.33×107Wm2

Thus, the total amount of energy emitted Et at 4500 K is 2.33×107Wm2.

Interpretation Introduction

(b)

Interpretation:

The total energy output must be calculated for the blackbody heated at 2500 K.

Concept introduction:

Stefan’s law essentially states that the total quantity of heat energy released by a perfect blackbody per unit area per second is directly proportional to the absolute temperature’s fourth power of its surface given by the equation-

Et= αT4

Wein’s displacement Law which states that the maximum wavelength in micrometers for the radiations of the blackbody is given by:

λmaxT = 2.90×103

Where,

T = temperature in Kelvin

λmax = maximum wavelength

Expert Solution
Check Mark

Answer to Problem 7.4QAP

The amount of energy emitted by Et at temperature 2500K is 2.22×106Wm2.

Explanation of Solution

The total energy Et emitted per unit time per unit area is calculated by:

Et=αT4=(5.69×108)T4Wm2 …………….. (1)

Given that-

α=5.69×108Wm2

T = 2500K

The value of total energy Et is:

Et=(5.69×108)T4Wm2

=(5.69×108)(2500K)4Wm2=2.22×106Wm2

Thus, the total amount of energy emitted Et at 2500 K is 2.22×106Wm2.

Interpretation Introduction

(c)

Interpretation:

The total energy output must be calculated for the blackbody heated at 1250K.

Concept introduction:

Stefan’s law essentially states that the total quantity of heat energy released by a perfect blackbody per unit area per second is directly proportional to the absolute temperature’s fourth power of its surface given by the equation-

Et= αT4

Wein’s displacement Law which states that the maximum wavelength in micrometers for the radiations of the blackbody is given by:

λmaxT = 2.90×103

Where,

T = temperature in Kelvin

λmax = maximum wavelength

Expert Solution
Check Mark

Answer to Problem 7.4QAP

The amount of energy emitted by Et at temperature 1250K is 1.39×105Wm2.

Explanation of Solution

The total energy Et emitted per unit time per unit area is calculated by:

Et=αT4=(5.69×108)T4Wm2 …………….. (1)

Given that-

α=5.69×108Wm2

T = 1250K

The value of total energy Et is:

Et=(5.69×108)T4Wm2

=(5.69×108)(1250K)4Wm2=1.39×105Wm2

Thus, the total amount of energy emitted Et at 1250 K is 1.39×105Wm2.

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