=-\mathrm{k}\left(\mathrm{T}-\mathrm{T}_0\right)$, where $\mathrm{k}>0$

$$ \begin{aligned} & \Rightarrow \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{~T}-290) \\ & \Rightarrow \frac{\mathrm{dT}}{\mathrm{~T}-290}=-\mathrm{kdt} \end{aligned} $$

$$ \begin{aligned} &\text { Integrating on both sides, we get }\\ &\begin{aligned} & \log (\mathrm{T}-290)=-\mathrm{kt}+\mathrm{c} \\ & \text { When } \mathrm{t}=0, \mathrm{~T}=370 \\ &\therefore \log (370-290)=0+\mathrm{c} \\ & \Rightarrow \mathrm{c}=\log 80 \\ &\therefore \log (\mathrm{~T}-290)=-\mathrm{kt}+\log 80 \,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $$

$$ \begin{aligned} & \text { When } \mathrm{t}=10, \mathrm{~T}=330 \\ & \therefore \quad \log (330-290)=-10 \mathrm{k}+\log 80 \\ & \Rightarrow \log 40=-10 \mathrm{k}+\log 80 \\ & \Rightarrow 10 \mathrm{k}=\log \frac{80}{40} \\ & \Rightarrow \mathrm{k}=\frac{1}{10} \log 2 \\ & \therefore \quad \log (\mathrm{~T}-290)=\frac{-\mathrm{t}}{10} \log 2+\log 80 \end{aligned} $$

…[From (i)]

When $T=295$, we have

$$ \begin{aligned} & \log (295-290)=\frac{-t}{10} \log 2+\log 80 \\ & \Rightarrow \frac{t}{10} \log 2=\log \frac{80}{5} \\ & \Rightarrow \frac{t}{10} \log 2=\log 16 \\ & \Rightarrow t=10 \frac{\log 16}{\log 2} \\ & \quad=10 \frac{\log 2^4}{\log 2}=40 \mathrm{~min} \end{aligned} $$

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