
The rate of emission of radiation of a black body at ${273^ \circ }C$ is $E$, then what will be the rate of emission of radiation of the body at ${0^ \circ }C$?
A. $\dfrac{E}{{16}}$
B. $\dfrac{E}{4}$
C. $\dfrac{E}{8}$
D. $0$
Answer
483k+ views
Hint- Stefan's law gives us the expression for total power radiated per unit surface area of a black body.
According to this law, the total power radiated per unit surface area of a black body is directly proportional to the fourth power of temperature. In mathematical form this law can be written as,
$E \propto {T^4}$
Here, $T$ is the temperature of the black body.
Step by step solution:
Stefan's law gives us the expression for total power radiated per unit surface area of a black body.
According to this law, the total power radiated per unit surface area of a black body is directly proportional to the fourth power of temperature. In mathematical form this law can be written as,
$E \propto {T^4}$
Or
$E = \sigma {T^4}$
Where, $\sigma $ is the Stefan’s constant and $T$ is the temperature.
Given,
Temperature,
$
{T_1} = {273^ \circ }C \\
= 273 + 273\,K \\
= 556\,K \\
$
Let the rate of emission of radiation of black body at ${273^ \circ }C$ be denoted as ${E_1}$.
Therefore, using Stefan’s law, we can write,
${E_1} = \sigma {T_1}^4$ ……. (1)
Temperature,
$
{T_2} = {0^ \circ }C \\
= 0 + 273\,K \\
= 273\,K \\
$
Let the rate of emission of radiation of black body at ${0^ \circ }C$ be denoted as ${E_2}$.
Therefore, using Stefan’s law, we can write,
${E_2} = \sigma {T_2}^4$ …… (2)
Divide equation (1) by (2). Then, we get
\[
\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{T_1}^4}}{{{T_2}^4}} \\
\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^4} \\
\]
Now substitute the given values. Then, we get
\[
\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{556}}{{273}}} \right)^4} \\
= {\left( 2 \right)^4} \\
= 16 \\
\]
That is,
${E_2} = \dfrac{{{E_1}}}{{16}}$
Hence the correct answer is option A.
Note: Remember to convert the temperatures given in $^ \circ C$ into the corresponding temperature in kelvin.
Formula to remember
Stefan’s law, $E = \sigma {T^4}$
Where, $E$ is the total power radiated per unit surface area of a black body and $T$ is the temperature of black body.
According to this law, the total power radiated per unit surface area of a black body is directly proportional to the fourth power of temperature. In mathematical form this law can be written as,
$E \propto {T^4}$
Here, $T$ is the temperature of the black body.
Step by step solution:
Stefan's law gives us the expression for total power radiated per unit surface area of a black body.
According to this law, the total power radiated per unit surface area of a black body is directly proportional to the fourth power of temperature. In mathematical form this law can be written as,
$E \propto {T^4}$
Or
$E = \sigma {T^4}$
Where, $\sigma $ is the Stefan’s constant and $T$ is the temperature.
Given,
Temperature,
$
{T_1} = {273^ \circ }C \\
= 273 + 273\,K \\
= 556\,K \\
$
Let the rate of emission of radiation of black body at ${273^ \circ }C$ be denoted as ${E_1}$.
Therefore, using Stefan’s law, we can write,
${E_1} = \sigma {T_1}^4$ ……. (1)
Temperature,
$
{T_2} = {0^ \circ }C \\
= 0 + 273\,K \\
= 273\,K \\
$
Let the rate of emission of radiation of black body at ${0^ \circ }C$ be denoted as ${E_2}$.
Therefore, using Stefan’s law, we can write,
${E_2} = \sigma {T_2}^4$ …… (2)
Divide equation (1) by (2). Then, we get
\[
\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{T_1}^4}}{{{T_2}^4}} \\
\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^4} \\
\]
Now substitute the given values. Then, we get
\[
\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{556}}{{273}}} \right)^4} \\
= {\left( 2 \right)^4} \\
= 16 \\
\]
That is,
${E_2} = \dfrac{{{E_1}}}{{16}}$
Hence the correct answer is option A.
Note: Remember to convert the temperatures given in $^ \circ C$ into the corresponding temperature in kelvin.
Formula to remember
Stefan’s law, $E = \sigma {T^4}$
Where, $E$ is the total power radiated per unit surface area of a black body and $T$ is the temperature of black body.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
