A black body at $227^\circ ext{C}$ radiates heat at the rate of $7 ext{ cal cm}^{-2} ext{s}^{-1}$. At a te

Question

A black body at $227^\circ	ext{C}$ radiates heat at the rate of $7 	ext{ cal cm}^{-2}	ext{s}^{-1}$. At a temperature of $727^\circ	ext{C}$, the rate of heat radiated in the same units will be:

Accepted Answer

According to the Stefan-Boltzmann law, the rate of heat radiated per unit area by a black body is directly proportional to the fourth power of its absolute temperature, i.e., $E \propto T^4$.
Given:
Initial temperature, $T_1 = 227^\circ	ext{C} = 227 + 273 = 500 	ext{ K}$
Initial rate of heat radiated, $E_1 = 7 	ext{ cal cm}^{-2}	ext{s}^{-1}$
Final temperature, $T_2 = 727^\circ	ext{C} = 727 + 273 = 1000 	ext{ K}$
Therefore, the ratio of the new rate of heat radiated ($E_2$) to the initial rate ($E_1$) is:
$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}ight)^4 = \left(\frac{1000}{500}ight)^4 = 2^4 = 16$
$E_2 = 16 	imes E_1 = 16 	imes 7 = 112 	ext{ cal cm}^{-2}	ext{s}^{-1}$

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