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# The radiant power of a furnace of surface area of $0.6\;\;{{\text{m}}^{\text{2}}}$ is $34\;{\text{kW}}$. The temperature of the furnace is nearly: ($\sigma = 5.7 \times {10^{ - 8}}\;{\text{W}}{{\text{m}}^{{\text{ - 2}}}}{{\text{K}}^{{\text{ - 4}}}}$)A) $3400\;{\text{K}}$B) $1512\;{\text{K}}$ C) $1000\;{\text{K}}$ D) $5700\;{\text{K}}$

Last updated date: 13th Jun 2024
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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.