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,
$-\frac{80-70}{12}=K\left(\frac{80+70}{2}-25\right)$
$\left[\because T_0=25^{\circ }C\right]$
$-\frac{5}{6}=K50...(i)$
Similarly, when object cool from $70^{\circ }C$ to $60^{\circ }C$, we get
$-\frac{70-60}{t}=K\left(\frac{70+60}{2}-25\right)$
$-\frac{10}{t}=K40...(ii)$
Dividing Eq. (i) by Eq. (ii), we get
$\frac{5}{6}\times \frac{t}{10}=\frac{50}{40}$
$\Rightarrow t=\frac{5}{4}\times 12=15\min$