An object kept in a large room having an air temperature of $25^\circ ext{C}$ takes $12 ext{ min}$ to cool

Question

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

Accepted Answer

According to the approximate form of Newton's law of cooling:
$\frac{\Delta T}{\Delta t} = K \left( \frac{T_1 + T_2}{2} - T_s ight)$
Where $T_s = 25^\circ 	ext{C}$ is the surrounding temperature.

Case 1: Cooling from $80^\circ 	ext{C}$ to $70^\circ 	ext{C}$ in $12 	ext{ min}$
$\frac{80 - 70}{12} = K \left( \frac{80 + 70}{2} - 25 ight)$
$\frac{10}{12} = K (75 - 25)$
$\frac{10}{12} = K (50)$
$\implies K = \frac{1}{60} 	ext{ min}^{-1}$

Case 2: Cooling from $70^\circ 	ext{C}$ to $60^\circ 	ext{C}$ in $t 	ext{ min}$
$\frac{70 - 60}{t} = K \left( \frac{70 + 60}{2} - 25 ight)$
$\frac{10}{t} = K (65 - 25)$
$\frac{10}{t} = K (40)$

Substituting the value of $K$ from Case 1:
$\frac{10}{t} = \frac{1}{60} 	imes 40$
$\frac{10}{t} = \frac{2}{3}$
$\implies t = \frac{30}{2} = 15 	ext{ min}$.

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