Question

The temperature of a piece of iron is ${{27}^{0}}C$ and it radiates energy at the rate of Q $kW/{{m}^{2}}$. If its temperature is raised to ${{151}^{0}}C$, the rate of radiation of energy will be approximately –

& \text{A) 2Q }kW/{{m}^{2}} \\\ & \text{B) 4Q }kW/{{m}^{2}} \\\ & \text{C) 6Q }kW/{{m}^{2}} \\\ & \text{D) 8Q }kW/{{m}^{2}} \\\ \end{aligned}$$

Answer 1

The rate of radiation is a temperature-dependent quantity. We need to find the relation between the radiation rate and the temperature from the Stefan-Boltzmann law that governs the rate of radiation of energy from black bodies for the given problem.

Complete step-by-step solution
We are given a situation in which an iron piece radiates the heat energy at the two temperatures ${{27}^{0}}C\text{ and }{{151}^{0}}C$. We need to find the difference in the rate of the radiation of thermal energy due to the temperature difference.
We know that the temperature and the rate of radiation of thermal energy are related. According to Stefan-Boltzmann's law, the rate of radiation of thermal energy is proportional to the emissivity of the object, the Stefan’s constant, the area of the radiating surface, and to the fourth power of the temperature of the system.
We can write the Stefan-Boltzmann’s law mathematically as –
$P=e\sigma A{{T}^{4}}$
Where. P is the rate of radiation of thermal energy by the object,
e is the emissivity of the object (for an ideal blackbody, e=1),
$\sigma$ is the constant of proportionality constant given by $\sigma =5.6703\times {{10}^{-8}}W/{{m}^{2}}{{K}^{4}}$,
A is the area of the radiating body and
T is the temperature of the object.
We understand that, for an object all the values except the temperature are a constant.

& P=e\sigma A{{T}^{4}} \\\ & \Rightarrow P\propto {{T}^{4}} \\\ & \\\ \end{aligned}$$ We are given the initial rate of radiation as Q. Now we can compare the two rates with the temperature in Kelvin scale as – $$\begin{aligned} & P\propto {{T}^{4}} \\\ & \dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{{{T}_{1}}^{4}}{{{T}_{2}}^{4}} \\\ & \Rightarrow \dfrac{Q}{{{P}_{2}}}={{(\dfrac{27+273}{151+273})}^{4}} \\\ & \Rightarrow {{P}_{2}}=Q{{(\dfrac{424}{300})}^{4}} \\\ & \Rightarrow {{P}_{2}}=Q{{(1.413)}^{4}} \\\ & \therefore {{P}_{2}}=4Q\text{ kW/}{{\text{m}}^{2}} \\\ \end{aligned}$$ **The rate of radiation increases by a factor of 4 when the temperature change occurs. The correct answer is option B.** **Note:** The Stefan-Boltzmann’s law gives the direct relationship between the rate of radiation emission of the body and the temperature of the body. The emissivity of the object is a characteristics property of the object. A perfect radiator will have an emissivity equal to 1.