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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{e \sigma A}{ms} (	heta - 	heta_0)$.Since $m = \

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$2t$

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{e \sigma A}{ms} (	heta - 	heta_0)$.Since $m = ho V$,we have $\frac{d	heta}{dt} \propto \frac{A}{V}$.For a cube of side $a$,$A = 6a^2$ and $V = a^3$,so $\frac{A}{V} = \frac{6}{a}$.For a cube of side $2a$,$A' = 6(2a)^2 = 24a^2$ and $V' = (2a)^3 = 8a^3$,so $\frac{A'}{V'} = \frac{24}{8a} = \frac{3}{a}$.Comparing the rates: $\frac{(d	heta/dt)'}{(d	heta/dt)} = \frac{A'/V'}{A/V} = \frac{3/a}{6/a} = \frac{1}{2}$.Since the temperature range $(	heta_1 - 	heta_2)$ is the same,the rate of cooling is inversely proportional to the time taken: $\frac{1/t'}{1/t} = \frac{1}{2}$.Therefore,$t' = 2t$.

$A$ copper cube of side $a$ is heated and allowed to cool in a vacuum. It takes time $t$ to cool from temperature $\theta_1$ to $\theta_2$. Now,another copper cube of side $2a$ is allowed to cool in the same environment. How much time will it take to cool from $\theta_1$ to $\theta_2$?

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