Question

A small hole in a furnace acts like a black body. Its area is $1\,{\text{c}}{{\text{m}}^2}$, and its temperature is the same as that of the interior of the furnace, $1727^\circ {\text{C}}$. The energy radiated out of the hole per second is nearly;
(Stefan’s constant=$5.67 \times {10^{ - 8}}\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}} \cdot {{\text{K}}^{{\text{ - 4}}}}$)
A.$79\,{\text{J}}$
B.$60\,{\text{J}}$
C.$91\,{\text{J}}$
D.$104\,{\text{J}}$

Answer 1

Use the formula for the energy radiated by a body per unit area. This formula gives the relation between the energy radiated, area of the body, Stefan’s constant, emissivity of the body and the surface temperature of the body.

Formula used:
The energy $E$ radiated per unit area of a body is given by
$\dfrac{E}{A} = \sigma \varepsilon {T^4}$ …… (1)
Here, $A$ is the area of the body, $T$ is the temperature of the body in Kelvin, $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity of the body.

Complete step by step answer:
The temperature of the hole is the same as that of the interior furnace which is $1727^\circ {\text{C}}$.
$T = 1727^\circ {\text{C}}$
The area of the hole is $1\,{\text{c}}{{\text{m}}^2}$.
$A = 1\,{\text{c}}{{\text{m}}^2}$
The small hole in the furnace acts like a black body. The emissivity $\varepsilon$ of the black body is 1.
$\varepsilon = 1$
Convert the unit of the temperature of the hole from degree Celsius to degree kelvin.
$T = \left( {1727^\circ {\text{C}}} \right) + 273$
$\Rightarrow T = 2000^\circ {\text{K}}$
Hence, the temperature of the hole is $2000^\circ {\text{K}}$.
Convert the unit of area of the hole to the SI system of units.
$A = \left( {1\,{\text{c}}{{\text{m}}^2}} \right)\left( {\dfrac{{{{10}^{ - 4}}\,{{\text{m}}^2}}}{{1\,{\text{c}}{{\text{m}}^2}}}} \right)$
$\Rightarrow A = {10^{ - 4}}\,{{\text{m}}^2}$
Hence, the area of the hole is ${10^{ - 4}}\,{{\text{m}}^2}$.
Determine the energy radiated $E$ per second by the hole.
Rearrange equation (1) for the energy radiated by the hole.
$E = \sigma \varepsilon A{T^4}$
Substitute $5.67 \times {10^{ - 8}}\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}} \cdot {{\text{K}}^{{\text{ - 4}}}}$ for $\sigma$, $1$ for $\varepsilon$, ${10^{ - 4}}\,{{\text{m}}^2}$ for $A$ and $2000^\circ {\text{K}}$ for $T$ in the above equation.
$E = \left( {5.67 \times {{10}^{ - 8}}\,{\text{W}} \cdot {{\text{m}}^{{\text{ - 2}}}} \cdot {{\text{K}}^{{\text{ - 4}}}}} \right)\left( 1 \right)\left( {{{10}^{ - 4}}\,{{\text{m}}^2}} \right){\left( {2000^\circ {\text{K}}} \right)^4}$
$\Rightarrow E = 90.72\,{\text{J}}$
$\therefore E \approx 91\,{\text{J}}$

Therefore, the energy radiated out of the hole per second is nearly $91\,{\text{J}}$.

So, the correct answer is “Option C”.

Note:
Convert the unit of temperature to Kelvin as the formula for energy radiated per unit area states that temperature must be in degree Kelvin. Also don’t forget to convert the unit of area of the hole in the SI system of units.