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 using Newton's law of cooling, we will follow these steps:

### Step 1: Understand the formula
According to Newton's law of cooling, the time \( T \) required for cooling can be expressed as:
\[ T = \frac{1}{k} \ln\left(\frac{T_2 - T_s}{T_1 - T_s}\right) \]
where:
- \( T_1 \) is the initial temperature,
- \( T_2 \) is the final temperature,
- \( T_s \) is the surrounding temperature,
- \( k \) is the cooling constant.

### Step 2: Set up the first cooling scenario
In the first scenario, the body cools from \( 80^\circ C \) to \( 40^\circ C \) with an ambient temperature of \( 30^\circ C \). We can plug these values into the formula:
- \( T_1 = 80^\circ C \)
- \( T_2 = 40^\circ C \)
- \( T_s = 30^\circ C \)

The time taken for this cooling is given as \( 5 \) minutes. Thus, we can write:
\[ 5 = \frac{1}{k} \ln\left(\frac{40 - 30}{80 - 30}\right) \]
This simplifies to:
\[ 5 = \frac{1}{k} \ln\left(\frac{10}{50}\right) = \frac{1}{k} \ln\left(\frac{1}{5}\right) \]

### Step 3: Solve for \( k \)
Rearranging the equation to find \( k \):
\[ k = \frac{1}{5} \ln\left(5\right) \]

### Step 4: Set up the second cooling scenario
Now, we need to find the time taken to cool from \( 62^\circ C \) to \( 32^\circ C \) with the same ambient temperature of \( 30^\circ C \):
- \( T_1 = 62^\circ C \)
- \( T_2 = 32^\circ C \)
- \( T_s = 30^\circ C \)

Using the formula again:
\[ T = \frac{1}{k} \ln\left(\frac{32 - 30}{62 - 30}\right) = \frac{1}{k} \ln\left(\frac{2}{32}\right) = \frac{1}{k} \ln\left(\frac{1}{16}\right) \]

### Step 5: Substitute \( k \) into the equation
Substituting \( k \) from Step 3 into the equation for \( T \):
\[ T = 5 \cdot \frac{\ln(16)}{\ln(5)} \]

### Step 6: Calculate \( \ln(16) \)
We know that \( \ln(16) = \ln(2^4) = 4 \ln(2) \). Given that \( \ln(2) = 0.693 \):
\[ \ln(16) = 4 \times 0.693 = 2.772 \]

### Step 7: Substitute and calculate \( T \)
Now substituting \( \ln(16) \) back into the equation for \( T \):
\[ T = 5 \cdot \frac{2.772}{1.609} \]

Calculating this gives:
\[ T \approx 5 \cdot 1.724 = 8.62 \text{ minutes} \]

### Final Answer
Thus, the time taken to cool from \( 62^\circ C \) to \( 32^\circ C \) is approximately \( 8.62 \) minutes.

---