Question

The unit of Stefan’s constant $\sigma $ is:
$(A)W{m^{ - 2}}{K^{ - 1}}$
$(B)W{m^2}{K^{ - 4}}$
$(C)W{m^{ - 2}}{K^{ - 4}}$
$(D)W{m^{ - 2}}{K^4}$

Answer 1

Hint:In order to get the unit of Stefan’s constant $\sigma $, we need to find an expression in terms of Stefan’s constant. So, firstly we will make use of the equation for a perfectly black body and then with the help of the concept of power, we will get the required expression.

Complete step by step solution:
Stefan’s constant is basically a proportionality constant in the Stefan-Boltzmann law. This law states that the total heat power radiated from any surface is proportional to the fourth power of the absolute temperature.
For any perfectly black body in this universe, the energy which is radiated per unit area is,
$E = \sigma {T^4}$
In the above expression,
$\sigma $ is the Stefan’s constant
And $T$ is the temperature which is measured in the Kelvin scale.
We are also aware of the fact that the power is defined as the ratio of energy and the time.
So, the power dissipated per unit area $ = \sigma {T^4}$
On taking \[\sigma \] on one side and all the other terms on the other side, we get,
$\sigma = \dfrac{P}{{A{T^4}}}......(1)$
We know that,
the unit of power is $W$
the unit of area is ${m^2}$
the unit of temperature is $K$
On putting these values in the equation (1), we get,
$\sigma = \dfrac{W}{{{m^2}{K^4}}}$
So, the unit of Stefan's constant $\sigma $ is $W{m^{ - 2}}{K^{ - 4}}$.
So, the correct answer is $(C)W{m^{ - 2}}{K^{ - 4}}$.

Note:
A perfect blackbody is the body that absorbs all incoming light which falls on it and does not reflect any light. At room temperature, these types of objects appear to be perfectly black and this is the reason that it is called a black body. But, if the black body is heated to a higher temperature, then a black body will begin to glow with the help of thermal radiation.