Two identical metal balls at temperatures 200 | vedclass

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(D) According to Stefan-Boltzmann law,the net rate of heat loss by radiation from a body at temperature $T$ to the surroundings at temperature $T_0$ i

Vedclass · Accepted Answer

$\frac{473^4 - 300^4}{673^4 - 300^4}$

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(D) According to Stefan-Boltzmann law,the net rate of heat loss by radiation from a body at temperature $T$ to the surroundings at temperature $T_0$ is given by $P = A\varepsilon \sigma (T^4 - T_0^4)$.Since the balls are identical,$A$ and $\varepsilon$ are the same for both.Therefore,the net heat loss $Q$ is proportional to $(T^4 - T_0^4)$.Given $T_1 = 200^{\circ}C = (200 + 273) K = 473 K$.Given $T_2 = 400^{\circ}C = (400 + 273) K = 673 K$.Given $T_0 = 27^{\circ}C = (27 + 273) K = 300 K$.The ratio of net heat loss is $\frac{Q_1}{Q_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.Substituting the values,we get $\frac{Q_1}{Q_2} = \frac{473^4 - 300^4}{673^4 - 300^4}$.

Two identical metal balls at temperatures $200^{\circ}C$ and $400^{\circ}C$ are kept in air at $27^{\circ}C$. The ratio of net heat loss by these bodies is:

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