It takes `10 min` to cool a liquid from `61^(@)C` to `59^(@)C`. If room temperature is `30^(@)C` then find the time taken in cooling from `51^(@)C` to `49^(@)C`.

Options

A. `15 min`
B. `12 min`
C. `18 min`
D. `20 min`

Explanation

To solve the problem of cooling a liquid from 51°C to 49°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 ambient (room) temperature. The formula can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_0) \]
where:
- \( T \) is the temperature of the object,
- \( T_0 \) is the ambient temperature (room temperature),
- \( k \) is a constant.

### Step 2: Calculate the Cooling Constant \( k \)
From the given information, we know that it takes 10 minutes to cool from 61°C to 59°C when the room temperature is 30°C. We can use the average temperature during this interval to find \( k \).

- Initial temperature \( T_i = 61°C \)
- Final temperature \( T_f = 59°C \)
- Average temperature \( T_{avg} = \frac{61 + 59}{2} = 60°C \)
- Room temperature \( T_0 = 30°C \)

Using the formula:
\[ \frac{T_f - T_i}{t} = -k(T_{avg} - T_0) \]
Substituting the values:
\[ \frac{59 - 61}{10} = -k(60 - 30) \]
This simplifies to:
\[ \frac{-2}{10} = -k(30) \]
\[ -0.2 = -30k \]
\[ k = \frac{0.2}{30} = \frac{1}{150} \]

### Step 3: Calculate the Time for Cooling from 51°C to 49°C
Now, we need to find the time taken to cool from 51°C to 49°C.

- Initial temperature \( T_i = 51°C \)
- Final temperature \( T_f = 49°C \)
- Average temperature \( T_{avg} = \frac{51 + 49}{2} = 50°C \)

Using the same formula:
\[ \frac{T_f - T_i}{t} = -k(T_{avg} - T_0) \]
Substituting the values:
\[ \frac{49 - 51}{t} = -\frac{1}{150}(50 - 30) \]
This simplifies to:
\[ \frac{-2}{t} = -\frac{1}{150} \times 20 \]
\[ \frac{-2}{t} = -\frac{20}{150} \]
\[ \frac{-2}{t} = -\frac{2}{15} \]
Cross-multiplying gives:
\[ -2 \times 15 = -2t \]
\[ 30 = 2t \]
\[ t = 15 \text{ minutes} \]

### Conclusion
The time taken to cool the liquid from 51°C to 49°C is **15 minutes**.