In a home experiment , Ram brings a new electric kettle with unknown power rating . He puts `1` litre water in the kettle and switches on . But to his dismay , the temperature becomes constants at `60^(@)C` after some time. The room temperature is `20^(@)C`. Ram gets bored and switches off the kittle . He sees that during first `20`s water cools down by `2^(@)C`.
What is the time taken for the water to cool to `40^(@)C`. (Approx)

Explanation

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 temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial temperature of water, \( \theta_1 = 60^\circ C \)
- Room temperature, \( \theta_0 = 20^\circ C \)
- Temperature after 20 seconds, \( \theta_2 = 58^\circ C \)
- Time interval, \( t = 20 \) seconds

2. **Calculate the Cooling Constant \( k \):**
Using Newton's Law of Cooling:
\[ \frac{\theta_1 - \theta_2}{t} = k \left( \frac{\theta_1 + \theta_2}{2} - \theta_0 \right) \]
Substituting the values:
\[ \frac{60 - 58}{20} = k \left( \frac{60 + 58}{2} - 20 \right) \]
Simplifying:
\[ \frac{2}{20} = k \left( \frac{118}{2} - 20 \right) \]
\[ \frac{1}{10} = k \left( 59 - 20 \right) \]
\[ \frac{1}{10} = k \cdot 39 \]
Therefore:
\[ k = \frac{1}{10 \cdot 39} = \frac{1}{390} \]

3. **Find Time to Cool from \( 60^\circ C \) to \( 40^\circ C \):**
We need to find the time \( T \) for the temperature to drop from \( 60^\circ C \) to \( 40^\circ C \).
Using the same formula:
\[ \frac{\theta_1 - \theta_3}{T} = k \left( \frac{\theta_1 + \theta_3}{2} - \theta_0 \right) \]
Here, \( \theta_3 = 40^\circ C \):
\[ \frac{60 - 40}{T} = \frac{1}{390} \left( \frac{60 + 40}{2} - 20 \right) \]
Simplifying:
\[ \frac{20}{T} = \frac{1}{390} \left( 50 - 20 \right) \]
\[ \frac{20}{T} = \frac{1}{390} \cdot 30 \]
\[ \frac{20}{T} = \frac{30}{390} \]
\[ \frac{20}{T} = \frac{1}{13} \]
Therefore:
\[ T = 20 \cdot 13 = 260 \text{ seconds} \]

### Final Answer:
The time taken for the water to cool to \( 40^\circ C \) is approximately **260 seconds**.