Question

An object is at a temperature of $400^\circ C$ . The temperature at which it can radiate energy twice as faster (Neglect the temperature of surroundings) is $\left( {{2^{\dfrac{1}{4}}} = {\text{ }}1.189} \right)$:
A. $800K$
B. $673K$
C. $500K$
D. $400K$

Answer 1

The given question can be solved using the Stefan-Boltzmann law which describes the radiated power from the black body in terms of temperature. It states that The thermal energy radiated by a black body radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by the formula - $P = \sigma {T^4}$.

Complete step by step answer:
Using the Stefan-Boltzmann law, and its mathematical formula, i.e. $P = \sigma {T^4}$.Let the initial energy be ${P_1}$ and temperature be ${T_1}$. The value of temperature in kelvin will be ${T_1} = {400^ \circ }C + 273 = 673K$. Now, calculating ${P_1}$,
${P_1} = \sigma {(673)^4}$
Let this be the value of ${P_1}$. Now, let's assume the final energy be ${P_2}$ and temperature be ${T_2}$,
The ${P_2} = 2{P_1}$ ,
So, ${P_2} = 2{P_1} = \sigma {({T_2})^4}$
Solving the equation, we have,
$2{P_2} = \left( {\dfrac{{{P_1}}}{{{{(673)}^4}}}} \right){({T_2})^4}$
Now solving for ${T_2}$, we have,
${({T_2})^4} = 2{(673)^4} \\\ \therefore {T_2} = 800.33K \approx 800K$

Hence, the correct answer is A.

Note: A hypothetical article that can be completely productive at engrossing and emanating radiation is known as a blackbody, and it along these lines transmits what's known as blackbody radiation. A blackbody is dark at room temperature, thus the name.Nonetheless, at higher temperatures, it can really sparkle at apparent frequencies. Along these lines, you ought to know that space experts and physicists utilize the term 'blackbody' to likewise allude to objects that sparkle.