Question

A black body at a temperature of ${227}^{\circ }C$ radiates heat energy at the rate $5calc{m}^{-2}{s}^{-1}$. At a temperature of ${727}^{\circ }C$, the rate of heat radiated per unit area in $calc{m}^{-2}{s}^{-1}$ will be:
A. 80
B. 160
C. 250
D. 500

Answer 1

Hint:The radiation energy emitted by a black body per unit surface area of the body in one unit of time is directly proportional to the fourth power of the temperature of the body. Use this and form equations for the two rates of energy and find the unknown rate.
Formula used:
$P=k{T}^4$

Complete answer:
A black body is a body that absorbs and radiates heat radiations in the form of electromagnetic waves of all the possible wavelengths.
The rate of emission of the radiation of energy by a black body is directly proportional to the fourth power of the temperature of the body. That is the radiation energy emitted by a black body per unit of the surface area of the body in one unit of time is directly proportional to the fourth power of the temperature of the body.
i.e. $P\propto T^4$.
$\Rightarrow P=k{T}^4$, where k is the proportionality constant.
It is given that the rate of the heat energy radiated by the black body at temperature of ${227}^{\circ }C$ is $5calc{m}^{-2}{s}^{-1}$.
Let $P_1=5calc{m}^{-2}{s}^{-1}$ and $T={227}^{\circ }C=(227+273)K=500K$.
$\Rightarrow P_1=k{T}_1^4$
$\Rightarrow 5=k{\left(500\right)}^4$ …… (i).
Let the rate of the heat energy radiated by the black body at temperature of $T_2={727}^{\circ }C$ be $P_2$.
Convert the temperature in kelvin.
$T_2={727}^{\circ }C=(727+273)K=1000K$.
$\Rightarrow P_2=k{T}_2^4$
$\Rightarrow P_2=k{(1000)}^2$ ….. (ii).
Divide equation (ii) by equation (i).
 $\Rightarrow P_2=5{\left({\displaystyle \frac{1000}{500}}\right)}^4$
$\Rightarrow P_2=5{\left(2\right)}^4=5\times 16=80calc{m}^{-2}{s}^{-1}$.

So, the correct answer is “Option A”.

Note:
Always remember to substitute the value of temperature in kelvin in the formulae of thermodynamics or problems related to temperature. This is because all the formulae are derived considering the kelvin scale for temperature.