Answer
64.8k+ views
Hint: In this question, first discuss the radiant power formula. Now as the value of sigma (\[\sigma \]) is given and we know the emissivity property of a perfect hot body hence we can directly use the formula of radiant power, i.e. Radiant energy per unit time.
Complete step by step solution:
As we know that the radiation is defined as that mode of transmission of heat in which heat travels from hot body to cold body in straight lines without heating the intervening medium. It is the energy emitted by matter in the form of electromagnetic waves.
We can directly use the formula for radiant power which is-
\[E = A\sigma \varepsilon \left( {{{T_r}^4} - {{T_s}^4}} \right)..........................\left( 1 \right)\]
Where, \[{T_r}\] is the temperature of radiator, \[{T_s}\] is the temperature of surroundings, \[\sigma \] is the Stefan’s constant and \[\varepsilon \] is emissivity of body, and $A$ is the area of the cross-section.
But here in this case the radiator is a hot body hence we can say that the temperature \[{T_r}\] is much greater than \[{T_s}\]. Hence, we can ignore the temperature of the surroundings. So, equation \[\left( 1 \right)\] becomes,
\[E = A\sigma \varepsilon {T^4}\]
Now, we rearrange the above equation as,
\[ \Rightarrow {T^4} = \dfrac{E}{{\sigma A}}\]
Now, we substitute the data in the equation we get
\[{T^4} = \dfrac{{34 \times {{10}^3}}}{{0.6 \times 5.7 \times {{10}^{ - 8}}}}\]
After simplification we get,
\[\therefore T = 998.3\;{\text{K}}\; \approx \;1000\;{\text{K}}\]
So, the temperature of the furnace is $1000\;{\text{K}}$.
Hence, the correct option is C.
Note: If the radiator would not have been a perfect hot body than the equation used would have had both the temperatures, i.e. the temperature of the surroundings and the radiator. We also remember that the idealized surface which emits radiation at maximum rate is called the black body.
Complete step by step solution:
As we know that the radiation is defined as that mode of transmission of heat in which heat travels from hot body to cold body in straight lines without heating the intervening medium. It is the energy emitted by matter in the form of electromagnetic waves.
We can directly use the formula for radiant power which is-
\[E = A\sigma \varepsilon \left( {{{T_r}^4} - {{T_s}^4}} \right)..........................\left( 1 \right)\]
Where, \[{T_r}\] is the temperature of radiator, \[{T_s}\] is the temperature of surroundings, \[\sigma \] is the Stefan’s constant and \[\varepsilon \] is emissivity of body, and $A$ is the area of the cross-section.
But here in this case the radiator is a hot body hence we can say that the temperature \[{T_r}\] is much greater than \[{T_s}\]. Hence, we can ignore the temperature of the surroundings. So, equation \[\left( 1 \right)\] becomes,
\[E = A\sigma \varepsilon {T^4}\]
Now, we rearrange the above equation as,
\[ \Rightarrow {T^4} = \dfrac{E}{{\sigma A}}\]
Now, we substitute the data in the equation we get
\[{T^4} = \dfrac{{34 \times {{10}^3}}}{{0.6 \times 5.7 \times {{10}^{ - 8}}}}\]
After simplification we get,
\[\therefore T = 998.3\;{\text{K}}\; \approx \;1000\;{\text{K}}\]
So, the temperature of the furnace is $1000\;{\text{K}}$.
Hence, the correct option is C.
Note: If the radiator would not have been a perfect hot body than the equation used would have had both the temperatures, i.e. the temperature of the surroundings and the radiator. We also remember that the idealized surface which emits radiation at maximum rate is called the black body.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)