Question

A body cools from ${80}^{o}C$ to ${50}^{o}C$ in $5$ minutes. Calculate the time it takes to cool from ${60}^{o}C$ to ${30}^{o}C$. The temperature of the surrounding is ${20}^{o}C$

• A. $9min$
• B. $9sec$
• C. $0.09sec$
• D. None

Answer 1

The correct option is A $9min$

$t=K\log \left(\frac{T_1-T_0}{T_2-T_0}\right)$

$T_1$ = Initial temperature

$T_2$ = Final temperature

$T_0$ = Surrounding temperature

So, $5\times 60=K\log \left(\frac{80-20}{50-20}\right)$

$300=K\log 2$

$\therefore K=\frac{300}{\log 2}$

Now, $t=K\log \left(\frac{60-20}{30-20}\right)$

$=K\log 4$

$=K\times 2\log 2$

$=\frac{300}{\log 2}\times 2\log 2$

$=600sec=10min$

If you use approximate formula which is

$\frac{T_1-T_2}{t}=\frac{1}{K}(\frac{T_1+T_2}{2}-T_0)$

Then you will get answer as $9minutes$