A beaker full of hot water is kept in a room and it cools from `80^(@)C` to `75^(@)C` in `t_(1)` mminutes , from `75^(@)C` to `65^(@)C` in `t_(2)` minutes and from `70^(@)C` to `65^(@)C` in `t_(3)` min, then

Options

A. `t_(1) gt t_(2) gt t_(3)`
B. `t_(1) = t_(2) = t_(3)`
C. `t_(1) lt t_(2) = t_(3)`
D. `t_(1) lt t_(2) lt t_(3)`

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 directly proportional to the difference in temperature between the object and its surroundings.

### Step-by-Step Solution:

1. **Understanding the Cooling Process**:
- The beaker of hot water cools down from an initial temperature to a lower temperature in different time intervals. We have three intervals:
- From \(80^\circ C\) to \(75^\circ C\) in \(t_1\) minutes.
- From \(75^\circ C\) to \(65^\circ C\) in \(t_2\) minutes.
- From \(70^\circ C\) to \(65^\circ C\) in \(t_3\) minutes.

2. **Applying Newton's Law of Cooling**:
- According to Newton's Law of Cooling, we can express the rate of heat loss as:
\[ \frac{dQ}{dt} \propto (T - T_{room}) \]
- Here, \(T\) is the temperature of the water, and \(T_{room}\) is the room temperature (which we assume to be constant).

3. **Calculating Temperature Differences**:
- For the first interval \(t_1\):
- Initial temperature = \(80^\circ C\)
- Final temperature = \(75^\circ C\)
- Temperature difference = \(80 - T_{room}\) and \(75 - T_{room}\)
- Average temperature difference = \(\frac{(80 - T_{room}) + (75 - T_{room})}{2}\)
- For the second interval \(t_2\):
- Initial temperature = \(75^\circ C\)
- Final temperature = \(65^\circ C\)
- Temperature difference = \(75 - T_{room}\) and \(65 - T_{room}\)
- Average temperature difference = \(\frac{(75 - T_{room}) + (65 - T_{room})}{2}\)
- For the third interval \(t_3\):
- Initial temperature = \(70^\circ C\)
- Final temperature = \(65^\circ C\)
- Temperature difference = \(70 - T_{room}\) and \(65 - T_{room}\)
- Average temperature difference = \(\frac{(70 - T_{room}) + (65 - T_{room})}{2}\)

4. **Setting Up the Proportionality**:
- According to Newton's Law of Cooling, we can write:
\[ \frac{t_1}{\Delta T_1} = \frac{t_2}{\Delta T_2} = \frac{t_3}{\Delta T_3} \]
- Where \(\Delta T_1 = 5\), \(\Delta T_2 = 10\), and \(\Delta T_3 = 5\).

5. **Finding the Relationships**:
- From the above relationships:
\[ \frac{t_1}{5} = \frac{t_2}{10} = \frac{t_3}{5} \]
- This implies that:
\[ t_1 = k \cdot 5, \quad t_2 = k \cdot 10, \quad t_3 = k \cdot 5 \]
- Where \(k\) is a constant.

6. **Conclusion**:
- From the relationships, we can conclude that \(t_1\) and \(t_3\) are equal, and \(t_2\) is twice that of \(t_1\) and \(t_3\). Therefore, \(t_1 = t_3\) and \(t_2 = 2t_1\).

### Final Answer:
- The relationship among \(t_1\), \(t_2\), and \(t_3\) is that \(t_1 = t_3\) and \(t_2 = 2t_1\).