A sphere is at temperature 600 text{ K}. In a | vedclass

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(D) According to Newton's law of cooling, the rate of cooling is proportional to the difference in temperature between the body and its surroundings,

Vedclass · Accepted Answer

$\frac{3}{16}R$

Vedclass · Answer

(D) According to Newton's law of cooling, the rate of cooling is proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small. However, for larger temperature differences, we use the Stefan-Boltzmann law for the net rate of heat loss: $P = \sigma A e (T^4 - T_0^4)$.Given:Initial temperature of the sphere $T_1 = 600 	ext{ K}$.Surrounding temperature $T_0 = 200 	ext{ K}$.Cooling rate $R \propto (T_1^4 - T_0^4)$.So, $R = k(600^4 - 200^4)$.When the temperature falls to $T_2 = 400 	ext{ K}$, the new cooling rate is $R' = k(400^4 - 200^4)$.Taking the ratio:$\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{(400^2 - 200^2)(400^2 + 200^2)}{(600^2 - 200^2)(600^2 + 200^2)}$$= \frac{(160000 - 40000)(160000 + 40000)}{(360000 - 40000)(360000 + 40000)} = \frac{120000 	imes 200000}{320000 	imes 400000}$$= \frac{12 	imes 20}{32 	imes 40} = \frac{240}{1280} = \frac{24}{128} = \frac{3}{16}$.Therefore, $R' = \frac{3}{16}R$.

$A$ sphere is at temperature $600 \text{ K}$. In an external environment of $200 \text{ K}$, its cooling rate is $R$. When the temperature of the sphere falls to $400 \text{ K}$, then the cooling rate $R'$ will become:

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