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-

$E_t=\alpha T^4$

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

${\lambda }_{max}T=2.90\times {10}^3$

Where,

T = temperature in Kelvin

${\lambda }_{max}$ = maximum wavelength

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 4500K is $2.33\times {10}^{7}W{m}^{-2}$.

Explanation of Solution

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

$E_t=\alpha T^4=\left(5.69\times {10}^{-8}\right)T^{4}W{m}^{-2}$ …………….. (1)

Given that-

$\alpha =5.69\times {10}^{-8}W{m}^{-2}$

T = 4500K

The value of total energy E_t is:

$E_t=$$\left(5.69\times {10}^{-8}\right)T^{4}W{m}^{-2}$

$=\left(5.69\times {10}^{-8}\right){\left(4500K\right)}^{4}W{m}^{-2}=2.33\times {10}^{7}W{m}^{-2}$

Thus, the total amount of energy emitted E_t at 4500 K is $2.33\times {10}^{7}W{m}^{-2}$.

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-

$E_t=\alpha T^4$

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

${\lambda }_{max}T=2.90\times {10}^3$

Where,

T = temperature in Kelvin

${\lambda }_{max}$ = maximum wavelength

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 2500K is $2.22\times {10}^{6}W{m}^{-2}$.

Explanation of Solution

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

$E_t=\alpha T^4=\left(5.69\times {10}^{-8}\right)T^{4}W{m}^{-2}$ …………….. (1)

Given that-

$\alpha =5.69\times {10}^{-8}W{m}^{-2}$

T = 2500K

The value of total energy E_t is:

$E_t=$$\left(5.69\times {10}^{-8}\right)T^{4}W{m}^{-2}$

$=\left(5.69\times {10}^{-8}\right){\left(2500K\right)}^{4}W{m}^{-2}=2.22\times {10}^{6}W{m}^{-2}$

Thus, the total amount of energy emitted E_t at 2500 K is $2.22\times {10}^{6}W{m}^{-2}$.

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-

$E_t=\alpha T^4$

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

${\lambda }_{max}T=2.90\times {10}^3$

Where,

T = temperature in Kelvin

${\lambda }_{max}$ = maximum wavelength

Answer to Problem 7.4QAP

The amount of energy emitted by E_t at temperature 1250K is $1.39\times {10}^5{\text{Wm}}^{-2}$.

Explanation of Solution

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

$E_t=\alpha T^4=\left(5.69\times {10}^{-8}\right)T^4{\text{Wm}}^{-2}$ …………….. (1)

Given that-

$\alpha =5.69\times {10}^{-8}W{m}^{-2}$

T = 1250K

The value of total energy E_t is:

$E_t=$$\left(5.69\times {10}^{-8}\right)T^{4}W{m}^{-2}$

$=\left(5.69\times {10}^{-8}\right){\left(1250K\right)}^{4}W{m}^{-2}=1.39\times {10}^{5}W{m}^{-2}$

Thus, the total amount of energy emitted E_t at 1250 K is $1.39\times {10}^{5}W{m}^{-2}$.