A hot body, obeying Newton's law of cooling is cooling down from its peak value `80^(@)C` to an ambient temperature of `30^(@)C`. It takes 5 minutes in cooling down from `80^(@)C` to `40^(@)C`. How much time will it take to cool down from `62^(@)C` to `32^(@)C`? (Given In 2 = 0.693, In 5 = 1.609)

Options

A. 3.75 minutes
B. 8.6 minutes
C. 9.6 minutes
D. 6.5 minutes

Explanation

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula we will use is:

\[ t = \frac{1}{k} \ln\left(\frac{\theta_2 - \theta_0}{\theta_1 - \theta_0}\right) \]

Where:
- \( t \) is the time taken for the temperature change,
- \( k \) is the cooling constant,
- \( \theta_1 \) is the initial temperature,
- \( \theta_2 \) is the final temperature,
- \( \theta_0 \) is the ambient temperature.

### Step 1: Determine the cooling constant \( k \)

From the first part of the problem, we know that it takes 5 minutes to cool from \( 80^\circ C \) to \( 40^\circ C \). Here, the ambient temperature \( \theta_0 = 30^\circ C \), \( \theta_1 = 80^\circ C \), and \( \theta_2 = 40^\circ C \).

Substituting these values into the formula:

\[ 5 = \frac{1}{k} \ln\left(\frac{40 - 30}{80 - 30}\right) \]

This simplifies to:

\[ 5 = \frac{1}{k} \ln\left(\frac{10}{50}\right) = \frac{1}{k} \ln\left(\frac{1}{5}\right) \]

Thus, we can express it as:

\[ 5 = \frac{1}{k} \cdot (-\ln(5)) \]

Rearranging gives:

\[ k = \frac{-\ln(5)}{5} \]

### Step 2: Calculate the time to cool from \( 62^\circ C \) to \( 32^\circ C \)

Now, we need to find the time \( t \) to cool from \( 62^\circ C \) to \( 32^\circ C \). Here, \( \theta_1 = 62^\circ C \) and \( \theta_2 = 32^\circ C \).

Substituting into the formula:

\[ t = \frac{1}{k} \ln\left(\frac{32 - 30}{62 - 30}\right) \]

This simplifies to:

\[ t = \frac{1}{k} \ln\left(\frac{2}{32}\right) = \frac{1}{k} \ln\left(\frac{1}{16}\right) \]

Now substituting \( k \):

\[ t = \frac{5}{-\ln(5)} \ln\left(\frac{1}{16}\right) \]

We know that \( \ln\left(\frac{1}{16}\right) = -\ln(16) = -\ln(2^4) = -4\ln(2) \).

Thus,

\[ t = \frac{5}{-\ln(5)} \cdot (-4\ln(2)) = \frac{20 \ln(2)}{\ln(5)} \]

### Step 3: Substitute the values of \( \ln(2) \) and \( \ln(5) \)

Using the given values:

- \( \ln(2) = 0.693 \)
- \( \ln(5) = 1.609 \)

Substituting these values gives:

\[ t = \frac{20 \times 0.693}{1.609} \]

Calculating this:

\[ t \approx \frac{13.86}{1.609} \approx 8.61 \text{ minutes} \]

### Final Answer

Thus, the time taken to cool from \( 62^\circ C \) to \( 32^\circ C \) is approximately **8.61 minutes**.