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A hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} \mathrm{C}$. The takes time $t_{1}$ to cool from $70^{\circ} \mathrm{C}$ to $68^{\circ} \mathrm{C}$ and time $t_{2}$ to cool from $60^{\circ} \mathrm{C}$ to $59.5^{\circ} \mathrm{C}$. Then
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Verified Answer
The correct answer is:
$\mathrm{t}_{2}=2 \mathrm{t}_{1}$
By Newton's law of cooling
$$
-\frac{\mathrm{dT}}{\mathrm{dt}}=\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)
$$
$\therefore \quad \frac{2}{t_{1}}=\alpha(90-30)=60 \alpha$
and $\frac{0.5}{t_{2}}=\alpha(60-30)=30 \alpha$
From Eqs. (i) and (ii),
$$
\mathrm{t}_{2}=2 \mathrm{t}_{1}
$$
$$
-\frac{\mathrm{dT}}{\mathrm{dt}}=\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)
$$
$\therefore \quad \frac{2}{t_{1}}=\alpha(90-30)=60 \alpha$
and $\frac{0.5}{t_{2}}=\alpha(60-30)=30 \alpha$
From Eqs. (i) and (ii),
$$
\mathrm{t}_{2}=2 \mathrm{t}_{1}
$$
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