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(B) According to Stefan-Boltzmann law,the rate of heat loss $R$ is given by $R = e \sigma A (T^4 - T_0^4)$.Here,$T$ is the temperature of the sphere a

Vedclass · Accepted Answer

$5.3$

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(B) According to Stefan-Boltzmann law,the rate of heat loss $R$ is given by $R = e \sigma A (T^4 - T_0^4)$.Here,$T$ is the temperature of the sphere and $T_0$ is the temperature of the surroundings.Given: $T_1 = 627^{\circ} C = 627 + 273 = 900 \ K$,$T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$,and $T_0 = 27^{\circ} C = 27 + 273 = 300 \ K$.The ratio of the rates of heat loss is $\frac{R_1}{R_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.Substituting the values: $\frac{R_1}{R_2} = \frac{900^4 - 300^4}{600^4 - 300^4} = \frac{(300 	imes 3)^4 - 300^4}{(300 	imes 2)^4 - 300^4}$.$\frac{R_1}{R_2} = \frac{300^4 (3^4 - 1)}{300^4 (2^4 - 1)} = \frac{81 - 1}{16 - 1} = \frac{80}{15} = \frac{16}{3} \approx 5.33$.Thus,the ratio is approximately $5.3$.

Compare the rate of loss of heat from a metal sphere at $627^{\circ} C$ with the rate of loss of heat from the same sphere at $327^{\circ} C$,if the temperature of the surrounding is $27^{\circ} C$. (nearly)

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