Question

A body takes 4 minutes to cool from ${100}^{o}Cto{70}^{o}C$. To cool from ${70}^{o}Cto{40}^{o}C$ it will take (room temperature is ${15}^{o}C$ )

• A. 7 minutes
• B. 6 minutes
• C. 5 minutes
• D. 4 minutes

Answer 1

Answer: (B)

. 7minutes.

The Newtons law of cooling formula is given by:

$T_t=T_0+(T_0-T_S)e^{-kt}$

$T_t-T_0=(T_0-T_S)e^{-kt}$

$ln\frac{T_t-T_0}{(T_0-T_S)}=ln(e{)}^{-kt}$

$\Rightarrow -kt=ln\left(\frac{T_t-T_0}{T_0-T_S}\right)$

$k=-\frac{ln\left(\frac{70-15}{100-15}\right)}{4\times 60}=-0.001813$

Now for the second interval we get:

$t=\frac{ln\left(\frac{40-15}{70-15}\right)}{-0.001813}=434.9sec\approx 7mins$.