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.
$\frac{60 - 50}{10}=K\left(\frac{60 + 50}{2} - \left(\theta \right)_{s}\right)$
$1=K\left(55 - \left(\theta \right)_{s}\right)...\left(\right.i\left.\right)$
Similarly, $\frac{50 - 42}{10}=K\left(\frac{50 + 42}{2} - \left(\theta \right)_{s}\right)$
$\frac{8}{10}=K\left(\right.46-\left(\theta \right)_{s}\left.\right)...\left(\right.ii\left.\right)$
Dividing Eq. $\left(i\right)$ by Eq. $\left(ii\right)$ , we get
$\frac{10}{8}=\frac{K \left(\right. 55 - \left(\theta \right)_{s} \left.\right)}{K \left(\right. 46 - \left(\theta \right)_{s} \left.\right)}$
$\Rightarrow $ $\theta _{s}=10℃$