Two black bodies at temperatures 327^{circ} C | vedclass

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(D) According to Stefan-Boltzmann Law,the rate of loss of heat $E$ from a black body at temperature $T$ in an environment at temperature $T_0$ is give

Vedclass · Accepted Answer

$\frac{243}{464}$

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(D) According to Stefan-Boltzmann Law,the rate of loss of heat $E$ from a black body at temperature $T$ in an environment at temperature $T_0$ is given by $E \propto (T^4 - T_0^4)$.Given temperatures are $T_1 = 327^{\circ} C = 600 \ K$,$T_2 = 427^{\circ} C = 700 \ K$,and $T_0 = 27^{\circ} C = 300 \ K$.The ratio of the rates of loss of heat is $\frac{E_1}{E_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.Substituting the values: $\frac{E_1}{E_2} = \frac{(600)^4 - (300)^4}{(700)^4 - (300)^4}$.Factoring out $(100)^4$: $\frac{E_1}{E_2} = \frac{6^4 - 3^4}{7^4 - 3^4} = \frac{1296 - 81}{2401 - 81}$.$\frac{E_1}{E_2} = \frac{1215}{2320}$.Dividing both numerator and denominator by $5$,we get $\frac{243}{464}$.

Two black bodies at temperatures $327^{\circ} C$ and $427^{\circ} C$ are kept in an evacuated chamber at $27^{\circ} C$. The ratio of their rates of loss of heat are :

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