Q100 / 125 · BITSAT 2011 · Physics
A copper sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 10 minutes and to $42^{\circ} \mathrm{C}$ in the next 10 minutes. Calculate the temperature of the surroundings.
- A$18.01^{\circ} \mathrm{C}$
- Correct answerB$26^{\circ} \mathrm{C}$
- C$10.6^{\circ} \mathrm{C}$
- D$20^{\circ} \mathrm{C}$
Solution
By Newton's law of cooling, $\frac{\theta_{1}-\theta_{2}}{t}=-k\left[\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right]$ ....(1)
A sphere cools from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$. $\frac{62-50}{10}=-\mathrm{k}\left[\frac{62+50}{2}-\theta_{0}\right]$ ....(2)
Now, sphere cools from $50^{\circ} \mathrm{C}$ to $42^{\circ} \mathrm{C}$ in next $10 \mathrm{~min} .$
$$
\frac{50-42}{10}=-\mathrm{k}\left[\frac{50+42}{2}-\theta_{0}\right] ....(3)
$$
Dividing eq $^{\text {n }}$. (2) by (3) we get, $\frac{56-\theta_{0}}{46-\theta_{0}}=\frac{1.2}{0.8}$ or $0.4 \theta_{0}=10.4$
Hence $\theta_{0}=26^{\circ} \mathrm{C}$
Hence, the correct answer is option: 'B'.