Question

A body cools from ${62}^{o}C$ to ${50}^{o}C$ in $10$ minutes. If the temperature of the surroundings is ${26}^{o}C$, then the temperature of the body after next $10$ minutes will be:

• A. ${41}^{o}C$
• B. ${44}^{o}C$
• C. ${46}^{o}C$
• D. ${48}^{o}C$

Answer 1

The correct option is A ${41}^{o}C$

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 water, ${\theta }_1$ and ${\theta }_2$ are surrounding and average water temperatures and $dt$ is the time interval.
Given that for $d{\theta }_2={62}^{o}C-{50}^{o}C={12}^{o}C$, ${\theta }_1={26}^{o}C$, ${\theta }_2=\frac{62+50}{2}={56}^{o}C$, $dt=10min$
And for the first ten minutes
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{12}{56-26}=k\times 10k=\frac{12}{300}\frac{{}^{o}C}{min}$
For the next ten minutes
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{{\theta }_2-{\theta }_f}{\frac{{\theta }_2+{\theta }_f}{2}-26}=\frac{12}{300}\times 10\frac{50-{\theta }_f}{\frac{50+{\theta }_f}{2}-26}=\frac{12}{300}\times 1050-{\theta }_f=0.2({\theta }_f-2){\theta }_f={41.3}^{o}C$
option (A) is the correct answer