Question

Hot water cools from $60℃$ to $50℃$ in the first $10min$ and to $42℃$ in the next $10min$ . Then the temperature of the surrounding is

• A. $20℃$
• B. $30℃$
• C. $15℃$
• D. $10℃$

Answer 1

Answer: (D)

According to Newton's law of cooling
$\frac{{\left(\theta \right)}_2-{\left(\theta \right)}_1}{t}=K\left(\frac{{\left(\theta \right)}_1+{\left(\theta \right)}_2}{2}-{\left(\theta \right)}_s\right)$
where, ${\theta }_s$ is the temperature of the surrounding.
$\frac{60-50}{10}=K\left(\frac{60+50}{2}-{\left(\theta \right)}_s\right)$
$1=K\left(55-{\left(\theta \right)}_s\right)...(i)$
Similarly, $\frac{50-42}{10}=K\left(\frac{50+42}{2}-{\left(\theta \right)}_s\right)$
$\frac{8}{10}=K(46-{\left(\theta \right)}_s)...(ii)$
Dividing Eq. $\left(i\right)$ by Eq. $\left(ii\right)$ , we get
$\frac{10}{8}=\frac{K(55-{\left(\theta \right)}_s)}{K(46-{\left(\theta \right)}_s)}$
$\Rightarrow$ ${\theta }_s=10℃$