The temperature of a piece of iron is ${27^o}C$ and it is radiating energy at the rate of $Q\;kW{m^{ - 2}}$. If its temperature is raised to ${151^o}C$, the rate of radiation of energy will become approximately ....... $Q\,kW\,{m^{ - 2}}$
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(b)${E_2} = {E_1}\frac{{T_2^4}}{{T_1^4}} = Q \times \frac{{{{(273 + 151)}^4}}}{{{{(273 + 27)}^4}}} = {\left( {\frac{{424}}{{300}}} \right)^4} = 3.99Q \approx 4Q$
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