Question

A body takes 4 min. to cool from 61$^\circ$C to 59$^\circ$C. If the temperature of the surroundings is 30$^\circ$C, the time taken by the body to cool from 51$^\circ$C to 49$^\circ$C is :

• A. 3 min.
• B. 4 min.
• C. 6 min.
• D. 8 min.

Hint:

For small temperature differences, Newton's Law of Cooling can be approximated using the average temperature of the body during the cooling interval. This makes calculations much simpler than using the integral form.

Answer 1

The Correct Option is C

We will use the approximate form of Newton's Law of Cooling:
$\frac{\theta_1 - \theta_2}{t} = K \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right)$
where $\theta_1$ and $\theta_2$ are the initial and final temperatures, t is the time taken, $\theta_s$ is the surrounding temperature, and K is a constant.
Case 1: Cooling from 61$^\circ$C to 59$^\circ$C in 4 minutes.
$\theta_1 = 61^\circ$C, $\theta_2 = 59^\circ$C, $t = 4$ min, $\theta_s = 30^\circ$C.
The average temperature is $\frac{61+59}{2} = 60^\circ$C.
Substituting these values into the formula:
$\frac{61 - 59}{4} = K(60 - 30)$
$\frac{2}{4} = K(30) \implies \frac{1}{2} = 30K \implies K = \frac{1}{60}$ min$^{-1}$.
Case 2: Cooling from 51$^\circ$C to 49$^\circ$C. Let the time taken be t'.
$\theta_1 = 51^\circ$C, $\theta_2 = 49^\circ$C, $\theta_s = 30^\circ$C.
The average temperature is $\frac{51+49}{2} = 50^\circ$C.
Substituting into the formula with the value of K we found:
$\frac{51 - 49}{t'} = K(50 - 30)$
$\frac{2}{t'} = \frac{1}{60}(20)$
$\frac{2}{t'} = \frac{20}{60} = \frac{1}{3}$
Solving for t':
$t' = 2 \times 3 = 6$ min.