Question

A body cools from ${80}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$ in $5\text{min}$. If the temperature of the surroundings is ${20}^{\circ }\text{C}$ then calculate the time it takes to cool from ${60}^{\circ }\text{C}$ to ${30}^{\circ }\text{C}$.

• A. $9\text{min}$
• B. $7\text{min}$
• C. $8\text{min}$
• D. $10\text{min}$

Answer 1

The correct option is A $9\text{min}$

Given,
Initial temperature $T_1={80}^{\circ }\text{C}$
Final temperature $T_2={50}^{\circ }\text{C}$
Temperature of the surroundings $T_0={20}^{\circ }\text{C}$
$t_1=5\text{min}$
According to Newton’s law of cooling.
Rate of cooling, $\frac{dT}{dt}=k[\frac{T_1+T_2}{2}-T_0]$
$\frac{(80-50)}{5}=k[\frac{80+50}{2}-20]$
$\frac{30}{5}=k(65-20)$
$6=k\times 45$
Or $k=\frac{6}{45}=\frac{2}{15}$ … (i)
In second condition,
intial temperature $T_1^{\prime }={60}^{\circ }\text{C}$
Final temperature $T_2^{\prime }={30}^{\circ }\text{C}$
$t^{\prime }=?$
Now, $\frac{(60-30)}{t^{\prime }}=\frac{2}{15}(\frac{60+30}{2}-20)$
$\frac{30}{t^{\prime }}=\frac{2}{15}(45-20)$
Or $t^{\prime }=\frac{30\times 15}{2\times 25}$
$=9\text{min}$