The temperature of a body falls from `40^(@)C` to `36(@)C` in 5 minutes when placed in a surrouning of constant temperature `16(@)C` Find the time taken for the temperature of the body to become `32(@)C`.

Explanation

To solve the problem of finding the time taken for the temperature of a body to fall from 36°C to 32°C using Newton's Law of Cooling, we can follow these steps:

### Step 1: Understand Newton's Law of Cooling
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the surrounding temperature. Mathematically, it is expressed as:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_n) \]
where:
- \(\theta\) is the temperature of the body,
- \(\theta_n\) is the surrounding temperature,
- \(k\) is a constant.

### Step 2: Set Up the Equation
In this problem:
- The surrounding temperature \(\theta_n = 16°C\).
- The initial temperature of the body is \(\theta_0 = 40°C\) and it cools to \(\theta_1 = 36°C\) in \(t_1 = 5\) minutes.

We need to find the time \(T\) taken for the temperature to drop from \(\theta_1 = 36°C\) to \(\theta_2 = 32°C\).

### Step 3: Integrate the Equation
We can rearrange and integrate the equation:
\[ \int \frac{d\theta}{\theta - \theta_n} = -k \int dt \]
Integrating from \(\theta_0 = 40°C\) to \(\theta_1 = 36°C\) gives:
\[ \ln(\theta - \theta_n) \Big|_{40}^{36} = -kt \Big|_{0}^{5} \]
This leads to:
\[ \ln(36 - 16) - \ln(40 - 16) = -5k \]
Calculating the logarithms:
\[ \ln(20) - \ln(24) = -5k \]
This simplifies to:
\[ \ln\left(\frac{20}{24}\right) = -5k \quad \Rightarrow \quad \ln\left(\frac{5}{6}\right) = -5k \]
Thus, we can express \(k\):
\[ k = -\frac{1}{5} \ln\left(\frac{5}{6}\right) \]

### Step 4: Find the Time \(T\) for the Next Interval
Now we need to find the time \(T\) for the temperature to drop from \(36°C\) to \(32°C\):
\[ \ln(\theta_2 - \theta_n) - \ln(\theta_1 - \theta_n) = -kT \]
Substituting the values:
\[ \ln(32 - 16) - \ln(36 - 16) = -kT \]
This leads to:
\[ \ln(16) - \ln(20) = -kT \]
This simplifies to:
\[ \ln\left(\frac{16}{20}\right) = -kT \quad \Rightarrow \quad \ln\left(\frac{4}{5}\right) = -kT \]

### Step 5: Substitute \(k\) and Solve for \(T\)
Substituting \(k\) into the equation:
\[ \ln\left(\frac{4}{5}\right) = \frac{1}{5} \ln\left(\frac{5}{6}\right) T \]
Rearranging gives:
\[ T = \frac{5 \ln\left(\frac{4}{5}\right)}{\ln\left(\frac{5}{6}\right)} \]

### Step 6: Calculate the Value of \(T\)
Using a calculator:
- \(\ln\left(\frac{4}{5}\right) \approx -0.2231\)
- \(\ln\left(\frac{5}{6}\right) \approx -0.1823\)

Now substituting these values:
\[ T = \frac{5 \times -0.2231}{-0.1823} \approx 6.1 \text{ minutes} \]

### Final Answer
The time taken for the temperature of the body to become \(32°C\) is approximately **6.1 minutes**.

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