Question

A body cools from $67°\mathrm{C}$ to $37°\mathrm{C}$. If this takes time $\mathrm{t}$ when the surrounding temperature is $27°\mathrm{C}$, what will be the time taken if the surrounding temperature is $7°\mathrm{C}$?

• A. $2\mathrm{t}$
• B. $\frac{\mathrm{t}}{3}$
• C. $\frac{\mathrm{t}}{2}$
• D. $\frac{\mathrm{t}}{4}$

Answer 1

The correct option is C

$\frac{\mathrm{t}}{2}$

Step 1. Given data

The initial temperature is, $67°\mathrm{C}$, final temperature is $37°\mathrm{C}$.

We have to find the time taken for surrounding temperature.

Step 2. Formula to be used.

Newton's law of cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the temperature of its surrounding.

$\log \left(\frac{{\mathrm{T}}_1-\mathrm{T}}{{\mathrm{T}}_2-\mathrm{T}}\right)=\mathrm{kt}$

Here, $\mathrm{T}$ is the surrounding temperature, ${\mathrm{T}}_1$ is the initial temperature, ${\mathrm{T}}_2$ is final temperature and $\mathrm{k}$ is constant.

Step 3. Find the time for cooling when surrounding temperature is $27°\mathrm{C}$.

For the surrounding temperature $27°\mathrm{C}$ is,

$\log \left(\frac{67-27}{37-27}\right)=\mathrm{kt}$

$\log 4=\mathrm{kt}$

$k=\frac{2\log 2}{t}$

Now, for the surrounding temperature $7°\mathrm{C}$ is,

$\log \left(\frac{67-7}{37-7}\right)=\mathrm{kt}\text{'}$

$\log 2=\mathrm{kt}\text{'}$

$\mathrm{t}\text{'}=\frac{\log 2}{\mathrm{k}}$

$\mathrm{t}\text{'}=\frac{\mathrm{t}}{2}$

Hence, option $\mathrm{C}$ is correct answer.