Question

A body in a laboratory takes 4 minutes to cool from ${61}^{o}C$ and ${59}^{o}C$. If the laboratory temperature is ${30}^{o}C$ then the time taken by it to cool from ${51}^{o}C$ to ${49}^{o}C$ will be:

• A. 8 min
• B. 6 min
• C. 5 min
• D. 4 min

Answer 1

Answer: (B)

6 min

By Newton's law of cooling
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt$
where $k$ is inversely proportional to product of mass and specific heat of the body, ${\theta }_1$ and ${\theta }_2$ are surrounding and average body temperatures and $dt$ is the time interval.
For the first 4 minutes
$d{\theta }_2={61}^{o}C-{59}^{o}C=2^{o}C$, ${\theta }_1={30}^{o}C$, ${\theta }_2=\frac{61+59}{2}={60}^{o}C$, $dt=4min$
So, $\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt$ $\Rightarrow$ $\frac{2}{60-30}=k\times 4k=\frac{1}{60}\frac{{}^{o}C}{min}$
Then time taken by the body to cool it from ${51}^{o}C$ to ${49}^{o}C:$
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{2}{50-30}=\frac{1}{60}\times dtdt=6min$