
If the temperature of a hot body is increased by 50% then the increase in the quantity of emitted heat radiation will be
A. 125%
B. 200%
C. 300%
D. 400%
Answer
545.4k+ views
Hint: Use the radiative energy equation to find the percentage of change in heat radiation. As we know temperature is a crucial parameter in heat radiation. Its about fourth power of the temperature.
Complete step by step answer:
Consider the initial temperature as \[{{T}_{1}}\] and final temperature as \[{{T}_{2}}\]. From the question we can understand that the final temperature is increased by 50% of initial temperature. So that,
\[{{T}_{2}}={{T}_{1}}(1+\dfrac{50}{100})\]
According to the heat radiation formula, emissivity is proportional to the fourth power of temperature. If suppose the initial radiation is \[{{e}_{1}}\] and final radiation is \[{{e}_{2}}\].
Then the percentage of increase in radiation will be equal to,
\[\Rightarrow \dfrac{{{e}_{2}}-{{e}_{1}}}{{{e}_{1}}}\times 100\]
According to the question, temperature is the quantity that only varies. So, we don’t have to take other terms in the equation. Therefore, the Percentage of increase in radiation is equal to,
\[\dfrac{{{e}_{2}}-{{e}_{1}}}{{{e}_{1}}}\times 100=\left[ \dfrac{T_{2}^{4}-T_{1}^{4}}{T_{1}^{4}} \right]\times 100\]
\[\Rightarrow \left[ \dfrac{{{\left[ \dfrac{15}{10}{{T}_{1}} \right]}^{4}}-{{T}_{1}}^{4}}{{{T}_{1}}^{4}} \right]\times 100\]
We can convert this equation into simpler form by considering \[{{T}_{1}}\] as 10. Substituting \[{{T}_{1}}=10\] into the equation. So, the answer will be,
\[\Rightarrow \left[ \dfrac{{{15}^{4}}-{{10}^{4}}}{{{10}^{4}}} \right]\times 100\]
\[=406.25%\], which is approximately equal to the 400%.
Hence the option D is correct.
Additional information:
Radiation is a method in which heat gets transferred from a hot body to a cold body. But for this transfer we don’t need any medium. The energy associated with radiation is known as radiated energy. As we know black surfaces are better absorbers of radiations than any other surfaces. Polished surfaces are not that much radiative. That’s why outside of cooking utensils are made with that design. In summer it is not advised to wear black coloured dresses since they absorb radiations and provide heat to the person.
Note: You don’t have to take all the units of heat radiation for this comparison. Since temperature is the only parameter, which is changing in this question and it will be easy for the calculation of percent of increase of heat radiation.
Complete step by step answer:
Consider the initial temperature as \[{{T}_{1}}\] and final temperature as \[{{T}_{2}}\]. From the question we can understand that the final temperature is increased by 50% of initial temperature. So that,
\[{{T}_{2}}={{T}_{1}}(1+\dfrac{50}{100})\]
According to the heat radiation formula, emissivity is proportional to the fourth power of temperature. If suppose the initial radiation is \[{{e}_{1}}\] and final radiation is \[{{e}_{2}}\].
Then the percentage of increase in radiation will be equal to,
\[\Rightarrow \dfrac{{{e}_{2}}-{{e}_{1}}}{{{e}_{1}}}\times 100\]
According to the question, temperature is the quantity that only varies. So, we don’t have to take other terms in the equation. Therefore, the Percentage of increase in radiation is equal to,
\[\dfrac{{{e}_{2}}-{{e}_{1}}}{{{e}_{1}}}\times 100=\left[ \dfrac{T_{2}^{4}-T_{1}^{4}}{T_{1}^{4}} \right]\times 100\]
\[\Rightarrow \left[ \dfrac{{{\left[ \dfrac{15}{10}{{T}_{1}} \right]}^{4}}-{{T}_{1}}^{4}}{{{T}_{1}}^{4}} \right]\times 100\]
We can convert this equation into simpler form by considering \[{{T}_{1}}\] as 10. Substituting \[{{T}_{1}}=10\] into the equation. So, the answer will be,
\[\Rightarrow \left[ \dfrac{{{15}^{4}}-{{10}^{4}}}{{{10}^{4}}} \right]\times 100\]
\[=406.25%\], which is approximately equal to the 400%.
Hence the option D is correct.
Additional information:
Radiation is a method in which heat gets transferred from a hot body to a cold body. But for this transfer we don’t need any medium. The energy associated with radiation is known as radiated energy. As we know black surfaces are better absorbers of radiations than any other surfaces. Polished surfaces are not that much radiative. That’s why outside of cooking utensils are made with that design. In summer it is not advised to wear black coloured dresses since they absorb radiations and provide heat to the person.
Note: You don’t have to take all the units of heat radiation for this comparison. Since temperature is the only parameter, which is changing in this question and it will be easy for the calculation of percent of increase of heat radiation.
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