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(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.Since the spheres

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$2:1$

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(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.Since the spheres are of the same material,the emissivity $e$,specific heat $s$,and density $ho$ are the same.Mass $m = ho \cdot V = ho \cdot \frac{4}{3} \pi r^3$ and Surface Area $A = 4 \pi r^2$.Substituting these into the formula: $\frac{d	heta}{dt} \propto \frac{A}{m} \propto \frac{r^2}{r^3} \propto \frac{1}{r}$.Therefore,the ratio of the rates of cooling is $\frac{(d	heta/dt)_1}{(d	heta/dt)_2} = \frac{r_2}{r_1} = \frac{2r}{r} = 2:1$.

Two spheres of the same metal have radii $r$ and $2r$. They are heated to the same temperature and placed in the same environment. The ratio of their rates of cooling will be:

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