Question

The temperature of a perfectly black body is ${{727}^{\circ }}$C and its area is $0.1{{m}^{2}}$. If Stefan’s constant is $5.67\times {{10}^{-8}}$ $w/{{m}^{2}}$ then heat radiated by it ( in $J$ ) in $0.3$ minutes is :
A. 1701
B. 17010
C. 102060
D. 102.60

Answer 1

n the question we have given the value of Stefan’s constant, temperature and area. So use Stefan’s law, from which you can get formulas. Use a reliable formula which will give a relationship between temperature, area and energy radiation. Time is given in minutes so convert units of time in seconds then only you will get the desired answer.

Formula used:
According to Stephan’s law rate of flow of heat through body is given as
$\dfrac{dt}{dq}=\sigma A {{\tau }^{4}}={{\epsilon }_{b}}$
Where,
${{\epsilon }_{b}}$ = emissive power of black body
$\sigma$ = Stefan’s constant
$\tau$ = Absolute temperature of perfectly black body
$A$ = area of perfectly black body

Complete answer:
The amount of radiation of energy emitted in a given per unit time and per unit surface area (A) of a perfectly black body is directly proportional to the fourth power of its absolute temperature. This is known as Stefan’s law.

According to Stefan’s law,

The rate of emitted heat flow through a perfectly black body is given as .
$\dfrac{dq}{dt}=\sigma A {{\tau }^{4}}$ ----- (1)
Since we know that,
Mathematically Stefan’s law can be written as

$\dfrac{q}{A t}\propto {{\tau }^{4}}$ or ${{\epsilon }_{b}}\propto {{\tau }^{4}}$
$\Rightarrow {{\epsilon }_{b}}=\dfrac{q}{A t}$
$\Rightarrow \dfrac{q}{A t}\propto {{\tau }^{4}}$
$\Rightarrow \dfrac{q}{A t}=\sigma {{\tau }^{4}}$
$\Rightarrow \dfrac{q}{t}=A \sigma {{\tau }^{4}}$

Now, we have given value of Stefan’s constant $\left( \sigma \right)=5.67\times {{10}^{-8}}w/{{m}^{2}}{{k}^{4}}$ and temperature of black body as ${{727}^{\circ }}C$ and surface area $\left(A \right)=0.1{{m}^{2}}$
Convert temperature of perfectly black body from degree to Kelvin , we get
$\tau ={{727}^{\circ }}C=727+273=1000K$
Now put all value in equation (1) , we get

$\dfrac{dq}{dt}=\sigma A{{\tau }^{4}}=5.67\times {{10}^{-8}}\times 0.1\times {{\left( 1000 \right)}^{4}}=5670w$ ----(2)
Finally we need to calculate heat radiated by perfectly black body in joule in $0.3$ minutes and it is given as
Heat radiated $=\dfrac{dq}{dt}\times 0.3\min =\dfrac{dq}{dt}\times 0.3\times 60$
Put value from equation (2),
Heat radiated $=5670\times 0.3\times 60=102060joule\left( j \right)$
Hence heat radiated by a perfectly black body in $0.3\operatorname{mins}$ is $102060j$ .

So, the correct answer is “Option C”.

Note:
Given formula, in this question, is applicable for a perfectly black body only. It is not applicable for ordinary bodies. For ordinary body, ths Stefan’s law is modified as,
$\epsilon =e\sigma {{\tau }^{4}}$
Where, $e=$ coefficient of emission.
If $T$ is absolute body temperature and ${{T}_{0}}$ is the lower absolute temperature and if perfectly black body is placed in surrounding then energy radiated per unit time per unit area is given as equal to $\sigma {{\tau }^{4}}$.