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 .