A pan filled with hot food cools from `94^(@)C` to `86^(@)C` in 2 minutes when the room temperature is at `20^(@)C`. How long will it take to cool from `71^(@)C` to ` 69^(@)C`? Here cooling takes place according to Newton's law of cooling.

Options

A. 50 s
B. 40 s
C. 38 s
D. 42 s

Explanation

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 an object is proportional to the difference between its temperature and the ambient temperature. Mathematically, it can be expressed as:

\[ \frac{dT}{dt} \propto (T - T_a) \]

where \( T \) is the temperature of the object, \( T_a \) is the ambient temperature, and \( t \) is time.

### Step 2: Set Up the Proportionality for the First Cooling Interval
For the first cooling interval, the pan cools from \( 94^\circ C \) to \( 86^\circ C \) in \( 2 \) minutes when the ambient temperature \( T_a = 20^\circ C \).

The change in temperature \( \Delta T_1 = 94 - 86 = 8^\circ C \).

Using the formula, we can write:

\[ \frac{\Delta T_1}{t_1} \propto \left( \frac{T_1 + T_2}{2} - T_a \right) \]

Substituting the values:

\[ \frac{8}{2} \propto \left( \frac{94 + 86}{2} - 20 \right) \]

Calculating the average temperature:

\[ \frac{94 + 86}{2} = 90^\circ C \]

Thus,

\[ \frac{8}{2} \propto (90 - 20) = 70 \]

This gives us our first equation:

\[ \frac{8}{2} = k \cdot 70 \quad \text{(where \( k \) is the proportionality constant)} \]

### Step 3: Set Up the Proportionality for the Second Cooling Interval
Now, for the second cooling interval, the pan cools from \( 71^\circ C \) to \( 69^\circ C \).

The change in temperature \( \Delta T_2 = 71 - 69 = 2^\circ C \).

Using the same formula, we write:

\[ \frac{\Delta T_2}{t_2} \propto \left( \frac{71 + 69}{2} - 20 \right) \]

Calculating the average temperature:

\[ \frac{71 + 69}{2} = 70^\circ C \]

Thus,

\[ \frac{2}{t_2} \propto (70 - 20) = 50 \]

This gives us our second equation:

\[ \frac{2}{t_2} = k \cdot 50 \]

### Step 4: Relate the Two Equations
From the two equations, we can relate \( t_2 \) and \( t_1 \):

From the first equation:

\[ \frac{8}{2} = k \cdot 70 \implies k = \frac{4}{70} = \frac{2}{35} \]

From the second equation:

\[ \frac{2}{t_2} = k \cdot 50 \implies t_2 = \frac{2}{k \cdot 50} \]

Substituting \( k \):

\[ t_2 = \frac{2}{\left(\frac{2}{35}\right) \cdot 50} \]

### Step 5: Calculate \( t_2 \)
Calculating \( t_2 \):

\[ t_2 = \frac{2 \cdot 35}{2 \cdot 50} = \frac{35}{50} = 0.7 \text{ minutes} \]

### Step 6: Convert to Seconds
To convert \( t_2 \) into seconds:

\[ t_2 = 0.7 \times 60 = 42 \text{ seconds} \]

### Final Answer
It will take approximately **42 seconds** for the pan to cool from \( 71^\circ C \) to \( 69^\circ C \).

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