Question

An object is cooled from $75{}^{\circ }C$ to $65{}^{\circ }C$ in 2 minute in a room at $30{}^{\circ }C$ . The time taken to cool another object form $55{}^{\circ }C$ to $45{}^{\circ }C$ in the same room in minute is:

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (A)

Accord to Newtons law of cooling Rate of cooling $\propto$ excess of temperature or rate of cooling $=K\times$ excess of temperature. If ${\theta }_1$ is initial temperature, ${\theta }_2$ is the final temperature and ${\theta }_0$ is the room temperature, then $\frac{{\theta }_1-{\theta }_2}{t}=K\left(\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0\right)$ Since, body cools from, ${75}^o$ to ${65}^o$ in 2 min and surrounding temperature is ${30}^o.$ $\frac{75-65}{2}=K\left(\frac{75+65}{2}-30\right)$ or $5=K\times 40\Rightarrow K=\frac{5}{40}=\frac{1}{8}$ When body cools from $55{}^{o}C$ to $45{}^{o}C,$ we have $\frac{55-45}{t_2}=\frac{1}{8}\left(\frac{55+45}{2}-30\right)$ $\frac{10}{t_2}=\frac{1}{8}\times 20$ or $t_2=\frac{10\times 8}{20}=4\min$