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. 12 s
B. 22 s
C. 32 s
D. 40 s

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 an object is proportional to the difference between its temperature and the surrounding temperature. The formula can be expressed as:
\[ \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right) \]
where:
- \(T_1\) is the initial temperature of the object,
- \(T_2\) is the final temperature of the object,
- \(t\) is the time taken,
- \(k\) is the cooling constant,
- \(T_s\) is the surrounding temperature.

### Step 2: Apply the Formula for the First Cooling Interval
For the first cooling interval where the temperature cools from \(94^\circ C\) to \(86^\circ C\) in \(2\) minutes:
- \(T_1 = 94^\circ C\)
- \(T_2 = 86^\circ C\)
- \(T_s = 20^\circ C\)
- \(t = 2\) minutes

Substituting these values into the formula:
\[ \frac{94 - 86}{2} = k \left( \frac{94 + 86}{2} - 20 \right) \]
Calculating the left side:
\[ \frac{8}{2} = 4 \]

Calculating the average temperature:
\[ \frac{94 + 86}{2} = 90 \]

Now substituting back into the equation:
\[ 4 = k (90 - 20) \]
\[ 4 = k \cdot 70 \]
Thus, we can solve for \(k\):
\[ k = \frac{4}{70} = \frac{2}{35} \]

### Step 3: Apply the Formula for the Second Cooling Interval
Now we need to find the time it takes to cool from \(71^\circ C\) to \(69^\circ C\):
- \(T_1 = 71^\circ C\)
- \(T_2 = 69^\circ C\)
- \(T_s = 20^\circ C\)
- Let \(t\) be the time we need to find.

Substituting these values into the formula:
\[ \frac{71 - 69}{t} = k \left( \frac{71 + 69}{2} - 20 \right) \]
Calculating the left side:
\[ \frac{2}{t} \]

Calculating the average temperature:
\[ \frac{71 + 69}{2} = 70 \]

Now substituting back into the equation:
\[ \frac{2}{t} = k (70 - 20) \]
\[ \frac{2}{t} = k \cdot 50 \]

### Step 4: Substitute \(k\) and Solve for \(t\)
Substituting \(k = \frac{2}{35}\):
\[ \frac{2}{t} = \frac{2}{35} \cdot 50 \]
Calculating the right side:
\[ \frac{2}{t} = \frac{100}{35} \]
Cross-multiplying gives:
\[ 2 \cdot 35 = 100t \]
\[ 70 = 100t \]
Thus, solving for \(t\):
\[ t = \frac{70}{100} = 0.7 \text{ minutes} \]

### Step 5: Convert Minutes to Seconds
To convert minutes to seconds:
\[ t = 0.7 \times 60 = 42 \text{ seconds} \]

### Final Answer
The time it will take to cool from \(71^\circ C\) to \(69^\circ C\) is **42 seconds**.

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