Question

A pan filled with hot food cools from $94 ^{\circ}C$ to $86^{\circ}C$ in $2$ minutes. When the room temperature is $20 ^{\circ}C$. How long will it cool from $74 ^{\circ}C$ to $66 ^{\circ}C$ ?

• A. 2 minutes
• B. 2.8 minutes
• C. 2.5 minutes
• D. 1.8 minutes

Answer 1

Answer: (B)

2.8 minutes

Answer 2

Key Idea When temperature difference is not large, rate of temperature change is given by Newton'8 law of cooling. In first case,
Average temperature
$=\frac{94+86}{2}=\frac{180}{2}=90^{\circ} C$
Excess temperature $=90^{\circ}-20^{\circ}=70^{\circ} C$
Rate of fall in temperature,
$\frac{d \theta_{1}}{d t} =-\left[\frac{94-86}{2}\right]=-\left[\frac{8}{2}\right]$
$=-4^{\circ} C \min ^{-1}$
$\frac{d \theta}{d t} \propto\left(\theta-\theta_{0}\right) =-k\left(\theta-\theta_{0}\right)$
$-4 =k \times 70 \dots$(i)
In second case,
Average temperature
$=\frac{74+66}{2}=\frac{140}{2}=70^{\circ} C$
Excess temperature
$=70^{\circ}-20^{\circ} C =50^{\circ} C$
Rate of fall in the temperaturre
$\frac{d \theta_{2}}{d t}=-\left[\frac{74-66}{t_{2}}\right]$
$=-\frac{8^{\circ} C }{t_{2}} min ^{-1}$
$\therefore \frac{-8}{t_{2}}=k \times 50 \dots$(ii)
On dividing E (i) by E (ii), we get
$\frac{t_{2}}{2}=\frac{7}{5}$
$t_{2}=\frac{14}{5}=2.8 \,min$