(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- E t = α T 4 Wein’s displacement Law which states that the maximum wavelength in micrometers for the radiations of the blackbody is given by: λ m a x T = 2.90 × 10 3 Where, T = temperature in Kelvin λ m a x = maximum wavelength

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Answer 1

Question

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

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Answer 2

Question

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

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Answer 3

Question

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

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Answer 4

Question

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

Answer 6

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

Answer 7

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

Answer 8

Expert Solution

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is 2.33×107Wm−2.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 4500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(4500K)4Wm−2=2.33×107Wm−2

Thus, the total amount of energy emitted E_t at 4500 K is 2.33×107Wm−2.

Answer 13

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

Answer 14

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

Answer 15

Expert Solution

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is 2.22×106Wm−2.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 2500K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(2500K)4Wm−2=2.22×106Wm−2

Thus, the total amount of energy emitted E_t at 2500 K is 2.22×106Wm−2.

Answer 20

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

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

Answer 21

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

Answer 22

Expert Solution

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is 1.39×105Wm−2.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

Et=αT4=(5.69×10−8)T4Wm−2 …………….. (1)

Given that-

α=5.69×10−8Wm−2

T = 1250K

The value of total energy E_t is:

Et=(5.69×10−8)T4Wm−2

=(5.69×10−8)(1250K)4Wm−2=1.39×105Wm−2

Thus, the total amount of energy emitted E_t at 1250 K is 1.39×105Wm−2.

Answer 27

Ch. 7 - Prob. 7.1QAP Ch. 7 - Prob. 7.2QAP Ch. 7 - The Wien displacement law states that the...Ch. 7 - Prob. 7.4QAP Ch. 7 - Prob. 7.5QAP Ch. 7 - Describe the differences and similarities between...Ch. 7 - Prob. 7.7QAP Ch. 7 - Prob. 7.8QAP Ch. 7 - Prob. 7.9QAP Ch. 7 - Prob. 7.10QAP

Answer to Problem 7.4QAP

Explanation of Solution

Answer to Problem 7.4QAP

Explanation of Solution

Answer to Problem 7.4QAP

Explanation of Solution

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