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A hot body, obeying Newton's law of cooling is cooling down from its peak value $80^{\circ} \mathrm{C}$ to an ambient temperature of $30^{\circ} \mathrm{C}$. It takes 5 minutes in cooling down from $80^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. How much time will it take to cool down from $62^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$ ? (Given $\operatorname{In} 2=0.693$, In $5=1.609$ )
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Verified Answer
The correct answer is:
$8.6$ minutes
$8.6$ minutes
From Newton's law of cooling,
$$
t=\frac{1}{k} \log _e\left(\frac{\theta_2-\theta_0}{\theta_1-\theta_0}\right)
$$
From question and above equation,
$$
5=\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}
$$
And, $t=\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}$
Dividing equation (2) by (1),
$$
\frac{t}{5}=\frac{\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}}{\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}}
$$
On solving we get, time taken to cool down from $62^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$, $t=8.6$ minutes.
$$
t=\frac{1}{k} \log _e\left(\frac{\theta_2-\theta_0}{\theta_1-\theta_0}\right)
$$
From question and above equation,
$$
5=\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}
$$
And, $t=\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}$
Dividing equation (2) by (1),
$$
\frac{t}{5}=\frac{\frac{1}{k} \log _e \frac{(32-30)}{(62-30)}}{\frac{1}{k} \log _e \frac{(40-30)}{(80-30)}}
$$
On solving we get, time taken to cool down from $62^{\circ} \mathrm{C}$ to $32^{\circ} \mathrm{C}$, $t=8.6$ minutes.
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