Question

An object kept in a large room having air temperature of $25^{\circ} C$ takes $12 \min$ to cool from $80^{\circ} C$ to $70^{\circ} C$. The time taken to cool for the same object from $70^{\circ} C$ to $60^{\circ} C$ would be nearly

• A. $10 \min$
• B. $12 \min$
• C. $20 \min$
• D. $15 \min$

Answer 1

Answer: (D)

From Newton's law of cooling, the time taken $t$ by a body to cool from $T_{1}$ to $T_{2}$ when placed in a medium of temperature $T_{0}$ can be calculated from relation
$-\frac{T_{1}-T_{2}}{t}=K\left(\frac{T_{1}+T_{2}}{2}-T_{0}\right)$
When the object cool from $80^{\circ} C$ to $70^{\circ} C$ in $12 \min$, then from Newton's law of cooling,
$-\frac{80-70}{12} =K\left(\frac{80+70}{2}-25\right) $
$\left[\because T_{0}=25^{\circ} C \right] $
$-\frac{5}{6} =K 50\,\,\,...(i)$
Similarly, when object cool from $70^{\circ} C$ to $60^{\circ} C$, we get
$-\frac{70-60}{t}=K\left(\frac{70+60}{2}-25\right) $
$-\frac{10}{t}=K 40\,\,\,...(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{5}{6} \times \frac{t}{10} =\frac{50}{40} $
$\Rightarrow t =\frac{5}{4} \times 12=15 \min$