A vessel contains water at `100^(@)` C. The temperature falls to `80^(@)` C in time `T_(1)` and to `60^(@)` C in further time `T_(2)` , then

Options

A. `T_(1)=T_(2)`
B. `T_(1)ltT_(2)`
C. `T_(1)ltT_(2)`
D. `T_(1)gtT_(2)`

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 in temperature between the object and its surroundings.

### Step-by-Step Solution:

1. **Identify Initial and Final Temperatures:**
- The initial temperature of the water, \( T_0 = 100^\circ C \).
- The first final temperature after time \( T_1 \) is \( T_1 = 80^\circ C \).
- The second final temperature after further time \( T_2 \) is \( T_2 = 60^\circ C \).

2. **Determine the Surrounding Temperature:**
- Let the surrounding temperature be \( T_s \).

3. **Apply Newton's Law of Cooling:**
- According to Newton's Law of Cooling, the time taken to cool from an initial temperature \( T_i \) to a final temperature \( T_f \) is given by:
\[ T_f - T_s = (T_i - T_s) e^{-kt} \]
- For the first cooling from \( 100^\circ C \) to \( 80^\circ C \):
\[ 80 - T_s = (100 - T_s) e^{-kT_1} \]
- For the second cooling from \( 80^\circ C \) to \( 60^\circ C \):
\[ 60 - T_s = (80 - T_s) e^{-kT_2} \]

4. **Rearranging the Equations:**
- From the first equation:
\[ e^{-kT_1} = \frac{80 - T_s}{100 - T_s} \]
- From the second equation:
\[ e^{-kT_2} = \frac{60 - T_s}{80 - T_s} \]

5. **Taking the Ratio of the Two Equations:**
- We can take the ratio of the two equations:
\[ \frac{e^{-kT_1}}{e^{-kT_2}} = \frac{80 - T_s}{100 - T_s} \cdot \frac{80 - T_s}{60 - T_s} \]
- This simplifies to:
\[ e^{-k(T_1 - T_2)} = \frac{(80 - T_s)^2}{(100 - T_s)(60 - T_s)} \]

6. **Analyzing the Temperature Differences:**
- The left side, \( e^{-k(T_1 - T_2)} \), indicates that if \( T_1 < T_2 \), then \( T_1 - T_2 < 0 \) and thus \( e^{-k(T_1 - T_2)} > 1 \).
- The right side is a positive quantity, and since both sides must equal, we can conclude that \( T_2 > T_1 \).

7. **Conclusion:**
- Therefore, the relation between the times is \( T_2 > T_1 \).

### Final Answer:
Thus, \( T_2 \) is greater than \( T_1 \).