The temperature of a piece of iron is 27^{cir | vedclass

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(B) According to Stefan-Boltzmann Law,the rate of energy radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \pr

Vedclass · Accepted Answer

$4$

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(B) According to Stefan-Boltzmann Law,the rate of energy radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.Given initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 	ext{ K}$.Final temperature $T_2 = 151^{\circ}C = 151 + 273 = 424 	ext{ K}$.The ratio of the rates of radiation is given by $\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} ight)^4$.Substituting the values: $\frac{E_2}{Q} = \left( \frac{424}{300} ight)^4$.Calculating the ratio: $\frac{424}{300} \approx 1.413$.Then,$1.413^4 \approx 3.99 \approx 4$.Therefore,the new rate of radiation is approximately $4Q 	ext{ kW m}^{-2}$.

The temperature of a piece of iron is $27^{\circ}C$ and it is radiating energy at the rate of $Q \text{ kW m}^{-2}$. If its temperature is raised to $151^{\circ}C$,the rate of radiation of energy will become approximately ....... $Q \text{ kW m}^{-2}$.

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