It takes `10 minutes` 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. 10 min
B. 11 min
C. 13 min
D. 15 min

Answer

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature (room temperature).

### Step-by-Step Solution:

1. **Identify the known values for the first cooling scenario:**
- Initial temperature (θ1) = 61°C
- Final temperature (θ2) = 59°C
- Room temperature (θs) = 30°C
- Time taken (t) = 10 minutes

2. **Apply Newton's Law of Cooling for the first scenario:**
\[ \frac{\theta_1 - \theta_2}{t} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right) \]
Substituting the known values:
\[ \frac{61 - 59}{10} = k \left( \frac{61 + 59}{2} - 30 \right) \]
Simplifying the left side:
\[ \frac{2}{10} = k \left( \frac{120}{2} - 30 \right) \]
\[ \frac{1}{5} = k (60 - 30) \]
\[ \frac{1}{5} = k \cdot 30 \]
Thus, we find:
\[ k = \frac{1}{5 \cdot 30} = \frac{1}{150} \]

3. **Identify the known values for the second cooling scenario:**
- Initial temperature (θ1) = 51°C
- Final temperature (θ2) = 49°C

4. **Apply Newton's Law of Cooling for the second scenario:**
\[ \frac{\theta_1 - \theta_2}{T} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right) \]
Substituting the known values:
\[ \frac{51 - 49}{T} = \frac{1}{150} \left( \frac{51 + 49}{2} - 30 \right) \]
Simplifying the left side:
\[ \frac{2}{T} = \frac{1}{150} \left( \frac{100}{2} - 30 \right) \]
\[ \frac{2}{T} = \frac{1}{150} (50 - 30) \]
\[ \frac{2}{T} = \frac{1}{150} \cdot 20 \]
\[ \frac{2}{T} = \frac{20}{150} = \frac{2}{15} \]

5. **Cross-multiply to find T:**
\[ 2 \cdot 15 = 2 \cdot T \]
\[ 30 = 2T \]
\[ T = 15 \text{ minutes} \]

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