
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
569.4k+ 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.
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