A hot liquid kept in a beaker cools from `80^(@)C` to `70^(@)C` in two minutes. If the surrounding temperature is `30^(@)C`, find the time of coolilng of the same liquid from `60^(@)C` to `50^(@)C`.

Explanation

To solve the problem of cooling a hot liquid from 60°C to 50°C, we can 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 temperature and the surrounding temperature.

### Step-by-step solution:

1. **Identify the known values:**
- Initial temperature \( T_1 = 80°C \)
- Final temperature \( T_2 = 70°C \)
- Surrounding temperature \( T_0 = 30°C \)
- Time taken to cool from \( T_1 \) to \( T_2 \) is \( t_1 = 2 \) minutes.

2. **Calculate the temperature difference:**
- The change in temperature \( \Delta T_1 = T_1 - T_2 = 80°C - 70°C = 10°C \).

3. **Calculate the average temperature during the first cooling period:**
- Average temperature \( T_{avg1} = \frac{T_1 + T_2}{2} = \frac{80 + 70}{2} = 75°C \).

4. **Apply Newton's Law of Cooling:**
- According to Newton's Law, we can express the cooling process as:
\[ \frac{\Delta T_1}{t_1} = -k (T_{avg1} - T_0) \]
- Substituting the known values:
\[ \frac{10°C}{2 \text{ min}} = -k (75°C - 30°C) \]
\[ 5 = -k \times 45 \]
- Thus, we can express \( k \) as:
\[ k = -\frac{5}{45} = -\frac{1}{9} \text{ min}^{-1} \]

5. **Now consider the second cooling process:**
- Initial temperature \( T_3 = 60°C \)
- Final temperature \( T_4 = 50°C \)
- The change in temperature \( \Delta T_2 = T_3 - T_4 = 60°C - 50°C = 10°C \).

6. **Calculate the average temperature during the second cooling period:**
- Average temperature \( T_{avg2} = \frac{T_3 + T_4}{2} = \frac{60 + 50}{2} = 55°C \).

7. **Apply Newton's Law of Cooling for the second cooling period:**
- We can express the cooling process as:
\[ \frac{\Delta T_2}{t_2} = -k (T_{avg2} - T_0) \]
- Substituting the known values:
\[ \frac{10°C}{t_2} = -\left(-\frac{1}{9}\right) (55°C - 30°C) \]
\[ \frac{10}{t_2} = \frac{1}{9} \times 25 \]
\[ \frac{10}{t_2} = \frac{25}{9} \]

8. **Solve for \( t_2 \):**
- Rearranging gives:
\[ t_2 = \frac{10 \times 9}{25} = \frac{90}{25} = 3.6 \text{ minutes} \]

### Final Answer:
The time taken for the liquid to cool from 60°C to 50°C is **3.6 minutes**.