Question

There is a small hole in a hollow container. At what temperature should it be maintained in order that it emits one calorie of energy per second per $mete{r^2}$ ?
A.$10K$
B.$100K$
C.$200K$
D.$500K$

Answer 1

The small hole in the hollow container acts like a black body.Therefore we will be using Stefan’s law for a perfect black body. It states that Radiant emittance or the energy radiated per second per unit area by a perfectly black body is proportional to the fourth power of its absolute temperature.

Complete step by step answer:
The theoretical proof of Stefan’s was given by Boltzmann. So, the law is also known as Stefan-Boltzmann law. The above law is only valid for a perfect black body.
According to the question we consider the small hole in the given hollow container as a perfect black body.
According to Stefan’s law, $E = \sigma {T^4}$
Where $E$ is the energy radiated per second per unit area by a perfect black body
$T$ is the absolute temperature
$\sigma$ is Stefan's constant.
It is given that $E = 1cal{\operatorname{s} ^{ - 1}}{m^{ - 2}}$
As $1cal = 4.2joule$
$E = 4.2J{\operatorname{s} ^{ - 1}}{m^{ - 2}}$
We know Stefan’s constant $\sigma = 5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$
Putting these values in the above equation we get
$4.2 = 5.67 \times {10^{ - 8}} \times {T^4}$
$T \approx 100K$
We can see the value of temperature is nearly equal to $100K$ in order that it emits one calorie of energy per second per $mete{r^2}$

Hence option B) $100K$ is the correct option.

Note:
A perfect black body is a material that absorbs all the incident electromagnetic radiations regardless of frequency. It neither reflects or transmits any incident radiation. It is an ideal body that does not exist in reality. The radiation emitted by a perfectly black body is known as black body radiation.