A body cool from `90^@C` to `70^@C` in 10 minutes if temperature of surrounding is `20^@C` find the time taken by body to cool from `60^@C` to `30^@C`. Assuming Newton’s law of cooling is valid.

Options

A. 10 min
B. 24 min
C. 36 min
D. 8 min

Answer

To solve the problem using Newton's law of cooling, we will 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 a body is directly proportional to the difference between its own temperature and the surrounding temperature. Mathematically, it can be expressed as:

\[ \frac{d\theta}{dt} \propto (\theta - \theta_s) \]

Where:
- \(\theta\) = temperature of the body
- \(\theta_s\) = temperature of the surroundings

### Step 2: Set Up the Equation for the First Cooling Interval
For the first interval, the body cools from \(90^\circ C\) to \(70^\circ C\) in \(10\) minutes. The surrounding temperature is \(20^\circ C\).

Using the average temperature for the body during this interval:

\[ \text{Average Temperature} = \frac{90 + 70}{2} = 80^\circ C \]

The change in temperature is:

\[ \Delta \theta = 90 - 70 = 20^\circ C \]

According to Newton's law, we can write:

\[ \frac{20}{10} = k \left(80 - 20\right) \]

This simplifies to:

\[ 2 = k \cdot 60 \]

### Step 3: Solve for \(k\)
Now, we can solve for \(k\):

\[ k = \frac{2}{60} = \frac{1}{30} \text{ per minute} \]

### Step 4: Set Up the Equation for the Second Cooling Interval
Now, we need to find the time taken for the body to cool from \(60^\circ C\) to \(30^\circ C\).

Using the average temperature for this interval:

\[ \text{Average Temperature} = \frac{60 + 30}{2} = 45^\circ C \]

The change in temperature is:

\[ \Delta \theta = 60 - 30 = 30^\circ C \]

Using Newton's law again:

\[ \frac{30}{\Delta T} = k (45 - 20) \]

Substituting the value of \(k\):

\[ \frac{30}{\Delta T} = \frac{1}{30} \cdot 25 \]

### Step 5: Solve for \(\Delta T\)
Now we can solve for \(\Delta T\):

\[ \frac{30}{\Delta T} = \frac{25}{30} \]

Cross-multiplying gives:

\[ 30 \cdot 30 = 25 \cdot \Delta T \]

\[ 900 = 25 \Delta T \]

\[ \Delta T = \frac{900}{25} = 36 \text{ minutes} \]

### Conclusion
The time taken by the body to cool from \(60^\circ C\) to \(30^\circ C\) is \(36\) minutes.

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