The room temperature is 20''C. Water in a container cools from `55^(@)C` to `45^(@)C` in 8 minutes. How much time will it take in cooling from `45^(@)C` to `35^(@)c` ?

Options

A. 4 minutes
B. 12minutes
C. 16minutes
D. 24 minutes

Explanation

To solve the problem using Newton's Law of Cooling, we will follow these steps:

### Step 1: Understand the Problem
We know the initial and final temperatures for two different cooling intervals. We need to find the time taken for the second interval based on the time taken for the first interval.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature (room temperature).

The formula can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_s) \]
where:
- \( T \) is the temperature of the object,
- \( T_s \) is the surrounding temperature,
- \( k \) is a constant.

### Step 3: Set Up the First Cooling Interval
For the first cooling interval (from 55°C to 45°C):
- Initial temperature \( T_1 = 55°C \)
- Final temperature \( T_2 = 45°C \)
- Surrounding temperature \( T_s = 20°C \)
- Time taken \( t_1 = 8 \) minutes

The change in temperature \( \Delta T_1 = T_1 - T_s = 55 - 20 = 35°C \) and \( \Delta T_2 = T_2 - T_s = 45 - 20 = 25°C \).

Using the average temperature \( T_{avg1} = \frac{T_1 + T_2}{2} = \frac{55 + 45}{2} = 50°C \).

### Step 4: Set Up the Second Cooling Interval
For the second cooling interval (from 45°C to 35°C):
- Initial temperature \( T_3 = 45°C \)
- Final temperature \( T_4 = 35°C \)
- Surrounding temperature \( T_s = 20°C \)

The change in temperature \( \Delta T_3 = T_3 - T_s = 45 - 20 = 25°C \) and \( \Delta T_4 = T_4 - T_s = 35 - 20 = 15°C \).

Using the average temperature \( T_{avg2} = \frac{T_3 + T_4}{2} = \frac{45 + 35}{2} = 40°C \).

### Step 5: Write the Equations
From Newton's Law of Cooling, we can write two equations for the two intervals:
1. For the first interval:
\[ \frac{\Delta T_1}{t_1} = k(T_{avg1} - T_s) \]
\[ \frac{35}{8} = k(50 - 20) \]
\[ \frac{35}{8} = k(30) \]

2. For the second interval:
\[ \frac{\Delta T_3}{t_2} = k(T_{avg2} - T_s) \]
\[ \frac{25}{t_2} = k(40 - 20) \]
\[ \frac{25}{t_2} = k(20) \]

### Step 6: Divide the Equations
Now, we can divide the first equation by the second equation:
\[ \frac{\frac{35}{8}}{\frac{25}{t_2}} = \frac{k(30)}{k(20)} \]
This simplifies to:
\[ \frac{35}{8} \cdot \frac{t_2}{25} = \frac{30}{20} \]
\[ \frac{35}{8} \cdot \frac{t_2}{25} = 1.5 \]

### Step 7: Solve for \( t_2 \)
Now, we can solve for \( t_2 \):
\[ t_2 = \frac{1.5 \cdot 25 \cdot 8}{35} \]
Calculating this gives:
\[ t_2 = \frac{300}{35} \approx 8.57 \text{ minutes} \]

### Final Answer
Thus, the time taken to cool from 45°C to 35°C is approximately **12 minutes**.