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A metal sphere cools at the rate of $1.5^{\circ} \mathrm{C} / \mathrm{min}$ when its temperature is $80^{\circ} \mathrm{C}$. At what rate will it cool when its temperature falls to $50^{\circ} \mathrm{C}$. [Temperature of surrounding is $30^{\circ} \mathrm{C}$ ]
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Verified Answer
The correct answer is:
$0.6^{\circ} \mathrm{C} / \mathrm{min}$
When temperature is $80^{\circ} \mathrm{C}$, by Newton's law of cooling, we have
$1.5=\mathrm{k}\left(\frac{80+30}{2}-30\right)=\mathrm{k}(55-30)=25 \mathrm{~K}$
When temperature is $50^{\circ} \mathrm{C}$, let $\mathrm{r}$ be the rate of cooling.
Then
$\begin{aligned}
& \mathrm{r}=\mathrm{k}\left(\frac{50+30}{2}-30\right)=\mathrm{k}(40-30)=10 \mathrm{~K} \\
& \therefore \frac{\mathrm{r}}{1.5}=\frac{10}{25} \\
& \mathrm{r}=0.6^{\circ} \mathrm{C} / \mathrm{min}
\end{aligned}$
$1.5=\mathrm{k}\left(\frac{80+30}{2}-30\right)=\mathrm{k}(55-30)=25 \mathrm{~K}$
When temperature is $50^{\circ} \mathrm{C}$, let $\mathrm{r}$ be the rate of cooling.
Then
$\begin{aligned}
& \mathrm{r}=\mathrm{k}\left(\frac{50+30}{2}-30\right)=\mathrm{k}(40-30)=10 \mathrm{~K} \\
& \therefore \frac{\mathrm{r}}{1.5}=\frac{10}{25} \\
& \mathrm{r}=0.6^{\circ} \mathrm{C} / \mathrm{min}
\end{aligned}$
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