Question

A body takes $T$ minutes to cool from $62^{\circ }C$ to $61^{\circ }C$ when the surrounding temperature is $30^{\circ }C$. The time taken by the body to cool from $46^{\circ }C$ to $45.5^{\circ }C$ is

• A. Greater than $T$ minutes
• B. Equal to $T$ minutes
• C. Less than $T$ minutes
• D. None of these

Answer 1

Answer: (B)

According to Newton's law of cooling, the rate of change of temperature will be,
$\frac{{\theta }_1-{\theta }_2}{t}\propto \left[\frac{{\theta }_1+{\theta }_2}{2}-\theta \right]$ ; where ${\theta }_1$ is initial temperature, ${\theta }_2$ is final temperature, $t$ is time and $\theta$ is the average temperature.
For the first condition,
$\frac{62-61}{T}=k\left[\frac{62+61}{2}-30\right]...\left(1\right)$
And for the second condition,
$\frac{46-45.5}{t}=k\left[\frac{46+45.5}{2}-30\right]...\left(2\right)$
By solving equation $\left(1\right)$ and $\left(2\right)$ , we get
$\frac{1\times t}{T\times 0.5}=\frac{61.5-30}{45.75-30}=\frac{31.5}{15.75}$
$t=T$ minutes .